you see, this is another blog, that is here, on the internet, for people to see, some people, not really that many, very few people even read nowadays, damn irish........anyway, I've got little to no remembrance of this past week, except that I won the social studies fair, but I literally have no recollection of what happened last week
To find the area between two curves,
f(x) and g(x)
You take the integral of (the top graph minus the bottom graph)
So, if g(x) is on "top", it would be
⌠g(x) - f(x)
Sunday, January 30, 2011
Blog #23
Cross-Sections:
Ex. 1) Find the volume of a solid with triangular perpendicular cross-section for: y = 1 – x/2 & y = -1 + x/2.
To solve a cross-section problem, you simply do what we’ve been doing with area and volume this whole time, just with the area formula for whatever shape they give you. In this case, that would be a triangle.
The area of a triangle: ½ bh
So, you simply take the integral of base times height (the first equation times the second equation) and multiply it by ½. Oh, and don’t forget the dx.
½ S(1 – x/2)(-1 + x/2)dx = ½ S(-1 - x²/4 + x)dx
For this particular problem, they did not give you any bounds to work with, so you just leave the answer as an equation.
For other problems like this but with different shapes, use these area formulas:
Triangles >>> S ½ bh
Rectangles >>> S lw
Semi-cirlces >>> S ½ π r²
Cirles >>> S π r²
Squares >>> S s²
Ex. 2) Find the volume of a solid with square perpendicular cross-section for: y = x² & y = x + 1.
Given that our shape is a square, we are going to use S s². Because of the two equations, we are first going to subtract the two equations, then square it.
S ((x+1) – (x²))² dx
Be sure to put the graph that is one top first.
S (x + 1 – x²)² dx
Since we are not given any bounds, the answer stays as an equation.
Ex. 1) Find the volume of a solid with triangular perpendicular cross-section for: y = 1 – x/2 & y = -1 + x/2.
To solve a cross-section problem, you simply do what we’ve been doing with area and volume this whole time, just with the area formula for whatever shape they give you. In this case, that would be a triangle.
The area of a triangle: ½ bh
So, you simply take the integral of base times height (the first equation times the second equation) and multiply it by ½. Oh, and don’t forget the dx.
½ S(1 – x/2)(-1 + x/2)dx = ½ S(-1 - x²/4 + x)dx
For this particular problem, they did not give you any bounds to work with, so you just leave the answer as an equation.
For other problems like this but with different shapes, use these area formulas:
Triangles >>> S ½ bh
Rectangles >>> S lw
Semi-cirlces >>> S ½ π r²
Cirles >>> S π r²
Squares >>> S s²
Ex. 2) Find the volume of a solid with square perpendicular cross-section for: y = x² & y = x + 1.
Given that our shape is a square, we are going to use S s². Because of the two equations, we are first going to subtract the two equations, then square it.
S ((x+1) – (x²))² dx
Be sure to put the graph that is one top first.
S (x + 1 – x²)² dx
Since we are not given any bounds, the answer stays as an equation.
1/30/11
ok so this week we finished everything we needed to know about the study.
we learned how to find the area between 2 graphs
area: integral of top portion - bottom portion
what i have trouble with is determining which part is top or bottom but only when the 2 equations given are similar
for example, if f(x) is negative and g(x) is positive, more than likely f(x) will contain the top half, but if both are positive i get a little confused.
we also learned the PVA system which is AWESOME
Position is the integral of velocity
Velocity is the derivative of position
Acceleration is the derivative of velocity
with this system we can solve any problem involving P V or A (pretty cool right?) lol
Say an object is accelerating at 10t^2 m/s/s
what is the velocity?
To find velocity, take the integral of acceleration:
10/3t^3 m/s is your velocity
to sum this week up, it was very slow but we learned some pretty easy stuff so i guess all went well :P
we learned how to find the area between 2 graphs
area: integral of top portion - bottom portion
what i have trouble with is determining which part is top or bottom but only when the 2 equations given are similar
for example, if f(x) is negative and g(x) is positive, more than likely f(x) will contain the top half, but if both are positive i get a little confused.
we also learned the PVA system which is AWESOME
Position is the integral of velocity
Velocity is the derivative of position
Acceleration is the derivative of velocity
with this system we can solve any problem involving P V or A (pretty cool right?) lol
Say an object is accelerating at 10t^2 m/s/s
what is the velocity?
To find velocity, take the integral of acceleration:
10/3t^3 m/s is your velocity
to sum this week up, it was very slow but we learned some pretty easy stuff so i guess all went well :P
1/30/2011
Ok, this week went by extremelyyy too slow and I'm sure everyone else feels the same way, being that we had to come off of two 4-day weeks back to back. Anyway, this week in Calculus we did some UNO study packets, and to catch up with some of the other blogs I missed, we learned about finding the area and volume of two graphs.
To find the area between two curves,
f(x) and g(x)
You take the integral of (the top graph minus the bottom graph)
So, if g(x) is on "top", it would be
⌠g(x) - f(x)
Example:
f(x) = x^2 + 1
g(x) = -x^2 + 4
Find the area between these 2 graphs.
⌠(-x^2 + 4) - (x^2 + 1) dx
⌠ -2x^2 + 3 dx
x^2 + 1 = -x^2 + 4
2x^2 = 3
x^2 = 3/2
x = ± √3/2
-2/3 ((√3)/(2)) + 3((√3)/2) - [ -2/3 (-√3/2)^3 + 3(-√3/2))
= 4.899
1/30/11
Alrightttt, I think this week was the last week of that UNO program. The material really wasn't hard to understand at all. We covered LRAM,RRAM,MRAM,TRAM, finding areas between regions, disks&washers, and lastly we learned how to find the volume. I would explain volume since it's the most recent thing we've learned but since I don't have my notes I'll explain how to find the area between two curves.
AREA BETWEEN TWO CURVES:
This is when you're given two separate equations: you first graph each equation, then you have to decide which curve is on top, and which one is on the bottom(this will help when plugging into the formula, so you must graph them). After deciding which equation is the top, or bottom you use this formula: S top-bottom (where you subtract the two equations and then integrate it). After that (if you aren't given the bounds in the directions), to find them all you do is set the two equations equal to each other and solve for x. Then once you integrate(Don't forget there is a formula to integrate if you don't know how to do it :)), you plug in the numbers of the bounds into the area formula (Fundamental Theorem of Calculus)->f(b)-f(a). That's ittttt!
So here's an example:
1.) Find the area between the two curves for y = x^2 + 2x + 1 and y = 2x + 5.
*Okay so the first thing you have to do is draw this graph so that you can figure out which one of these curves is on top.
*So the graph of x^2+2x+1 is a parabola with vertex at x=-1 opening up...And the graph of 2x+5 is a line going up intersecting the parabola at points (-5/2,0) and (0,5). So that means that the line is on top and the parabola is on the bottom.
*Before we find the area between the two graphs, find the bounds since we were not given them...So to do that you just set your two equations equal to each other and solve for x like this:
x^2+2x+1 = 2x+5
x^2-4
(x+2)(x-2)
x=2, -2 <-that's the bounds: 2 going on the top, and -2 on the bottom
*Now we can use the formula to find the area. We already said that the line 2x+5 is on top so here's how to set up the equation:
S(2x+5)-(x^2+2x+1)
Now distribute the negative and simplify it to get this:
S -x^2+4
Now integrate that and you should end up with this:
-1/3x^3+4x
Now you can use the Fundamental Theorem of Calculus to plug in your bounds and find the area. So this is what you should get when you plug in:
-1/3(2)^3+4(2)-[-1/3(-2)^3+4(-2)]
= 32/3 <-AREA
That's about all there is too it! I think I understood everything we learned pretty well.. Now I guess were gonna start the AP stuff... ahhhh :\
AREA BETWEEN TWO CURVES:
This is when you're given two separate equations: you first graph each equation, then you have to decide which curve is on top, and which one is on the bottom(this will help when plugging into the formula, so you must graph them). After deciding which equation is the top, or bottom you use this formula: S top-bottom (where you subtract the two equations and then integrate it). After that (if you aren't given the bounds in the directions), to find them all you do is set the two equations equal to each other and solve for x. Then once you integrate(Don't forget there is a formula to integrate if you don't know how to do it :)), you plug in the numbers of the bounds into the area formula (Fundamental Theorem of Calculus)->f(b)-f(a). That's ittttt!
So here's an example:
1.) Find the area between the two curves for y = x^2 + 2x + 1 and y = 2x + 5.
*Okay so the first thing you have to do is draw this graph so that you can figure out which one of these curves is on top.
*So the graph of x^2+2x+1 is a parabola with vertex at x=-1 opening up...And the graph of 2x+5 is a line going up intersecting the parabola at points (-5/2,0) and (0,5). So that means that the line is on top and the parabola is on the bottom.
*Before we find the area between the two graphs, find the bounds since we were not given them...So to do that you just set your two equations equal to each other and solve for x like this:
x^2+2x+1 = 2x+5
x^2-4
(x+2)(x-2)
x=2, -2 <-that's the bounds: 2 going on the top, and -2 on the bottom
*Now we can use the formula to find the area. We already said that the line 2x+5 is on top so here's how to set up the equation:
S(2x+5)-(x^2+2x+1)
Now distribute the negative and simplify it to get this:
S -x^2+4
Now integrate that and you should end up with this:
-1/3x^3+4x
Now you can use the Fundamental Theorem of Calculus to plug in your bounds and find the area. So this is what you should get when you plug in:
-1/3(2)^3+4(2)-[-1/3(-2)^3+4(-2)]
= 32/3 <-AREA
That's about all there is too it! I think I understood everything we learned pretty well.. Now I guess were gonna start the AP stuff... ahhhh :\
Saturday, January 29, 2011
1/29/11
Alright well at the beginning of this week, we finished up with the UNO study learning how to revolve a graph about the x-axis using the washer and disk methods. This actually gives you the volume of a certain solid based off of the equation you're given in the problem. Then later in the week we went over position, velocity, and acceleration functions and how to apply integrals with them. For instance, displacement (which is the same thing as position) is the integral of velocity; and distance is the absolute value of the integral of velocity.
Here's an example:
1.) Meg accelerates her car, giving it a velocity of v = t^1/2 - 2, in feet per second, at time t, in seconds after she started accelerating.
A: Find the time(s) at which v = 0
*All they're asking for here is "t", so all you do is plug in 0 for v (i.e.-set the equation they gave you equal to zero and solve for t) So you should get this:
t^1/2 - 2 = 0
t^1/2 = 2 ...(*t^1/2 is the same thing as the squareroot of t, so to get rid of that all you do is square both sides of the equation)
(t^1/2)^2 = 2^2
t=4
B: Find her net displacement for the time interval [1,9].
*Soooo, they give you the time interval [1,9]..which is going to be your bounds for your integral
*Since they're asking for the displacement, you're going to simply take the integral of the velocity function with bounds 1,9
*So plugging into that formula you should get this:
S1,9 t^1/2-2 dt
Now taking the integral of that equation you should end up with:
2/3t^3/2 - 2t
Now use the Fundamental Theorem of Calculus to plug in 9 and 1 like this:
2/3(9)^3/2 - 2(9) - [2/3(1)^3/2 - 2(1)]
= 4/3 ft
is displacement
C: Find the total distance she travels for the time interval [1,9].
*So now since they're asking for distance, all we're going to do is use the distance formula..which says to take the absolute value of the integral of velocity
*So plugging that into your calculator with bounds [1,9] you should end up with 4
*So 4 ft is your total distance
*Again this week was a pretty good week because I understood everything :) haha....All except for #13 on the homework worksheet we had Thursday night..Oh and also, I have no clue how to integrate secant..or anything besides sine and cosine ha
Here's an example:
1.) Meg accelerates her car, giving it a velocity of v = t^1/2 - 2, in feet per second, at time t, in seconds after she started accelerating.
A: Find the time(s) at which v = 0
*All they're asking for here is "t", so all you do is plug in 0 for v (i.e.-set the equation they gave you equal to zero and solve for t) So you should get this:
t^1/2 - 2 = 0
t^1/2 = 2 ...(*t^1/2 is the same thing as the squareroot of t, so to get rid of that all you do is square both sides of the equation)
(t^1/2)^2 = 2^2
t=4
B: Find her net displacement for the time interval [1,9].
*Soooo, they give you the time interval [1,9]..which is going to be your bounds for your integral
*Since they're asking for the displacement, you're going to simply take the integral of the velocity function with bounds 1,9
*So plugging into that formula you should get this:
S1,9 t^1/2-2 dt
Now taking the integral of that equation you should end up with:
2/3t^3/2 - 2t
Now use the Fundamental Theorem of Calculus to plug in 9 and 1 like this:
2/3(9)^3/2 - 2(9) - [2/3(1)^3/2 - 2(1)]
= 4/3 ft
is displacement
C: Find the total distance she travels for the time interval [1,9].
*So now since they're asking for distance, all we're going to do is use the distance formula..which says to take the absolute value of the integral of velocity
*So plugging that into your calculator with bounds [1,9] you should end up with 4
*So 4 ft is your total distance
*Again this week was a pretty good week because I understood everything :) haha....All except for #13 on the homework worksheet we had Thursday night..Oh and also, I have no clue how to integrate secant..or anything besides sine and cosine ha
Sunday, January 23, 2011
Chads blogggggg
I think that the stuff we are doing now isnt that hard even though I missed a bunch of it being sick and all. We've learned how to find area dealing with definite integrals and how to find LRAM, RRAM, TRAP, AND MRAM. All of these are similar. To first start these problems it's best to draw the graph to give you a visual of how many shapes they have, and what sides are going to touch the x-axis. After figuring out the graph, for LRAM and RRAM your going to take the bounds and subtract them, then divide it by the amount of rectangles. This will be your delta x. Then, your going to take the the amount of rectangles and plug them into the the equation. The answers you get will be used to find LRAM and RRAM, and to find area you will take the delta x, and multiply them by the answers you got when plugging into the equation (Don't forget to add all the equation answers together and multiply each one by delta x). Now to find TRAP you plug into the formula 1/2(LRAM+RRAM). To find the area follow the formula: 1/2(b1+b2)h. Lastly for MRAM you divide the rectangles that you draw on the graph to find the midpoint(point in between the two whole number parts). Plug the numbers you find into the original equation, and add them togetherxdelta x.
TRAM:
Use the Trapezoidal Rule to approximate (-x^2+4) dx, with the bounds x=0, x=3. How can you get a better answer?
1.We're going to plug in -0,-1,-2, and -3 into the original equation to get 4,3,0, and -5.
2.Now plug into the formula 1/2(b1+b2)h, to get 5/2
3.The answer would be 5/2
TRAM:
Use the Trapezoidal Rule to approximate (-x^2+4) dx, with the bounds x=0, x=3. How can you get a better answer?
1.We're going to plug in -0,-1,-2, and -3 into the original equation to get 4,3,0, and -5.
2.Now plug into the formula 1/2(b1+b2)h, to get 5/2
3.The answer would be 5/2
Another blog
well, this week, we learned about stuff like LRAM and RRAM for that UNO study.......the least they could've done was deal out the cards right though, I got stuck with a damn Wild Card and a red 2...............they need to l2cut the cards -.-
Ex. Find the area of the region bounded by the curve f(x) = x² and the x-axis between x=0 and x=3. Use 3 rectangles.
First of all, x=0 and x=3 are your bounds. So to rewrite this equation, it would be: 3S0 x² dx. (*”S” is the integral symbol.)
Next, what is the width? To find it, subtract your upper and lower bounds, then divide by the number of rectangles: (3-0)/3 = 1
Now, to find the lengths, plug in to the original equation every number between 0 and 3 that is 1 width away. The numbers will be 0, 1, 2, and 3. To plug in: (0)² = 0; (1)² = 1; (2)² = 4; (3)² = 9
Finally, for LRAM, multiply the width by the first three lengths (all of the left points) and add them together: (1)(0) + (1)(1) + (1)(4) = 5
For RRAM, multiply the width by the last three lengths (all of the right points) and add them together: (1)(1) + (1)(4) + (1)(9) = 14
Your answer: 5 < area < 14
and don't forget, just bust a move!
Ex. Find the area of the region bounded by the curve f(x) = x² and the x-axis between x=0 and x=3. Use 3 rectangles.
First of all, x=0 and x=3 are your bounds. So to rewrite this equation, it would be: 3S0 x² dx. (*”S” is the integral symbol.)
Next, what is the width? To find it, subtract your upper and lower bounds, then divide by the number of rectangles: (3-0)/3 = 1
Now, to find the lengths, plug in to the original equation every number between 0 and 3 that is 1 width away. The numbers will be 0, 1, 2, and 3. To plug in: (0)² = 0; (1)² = 1; (2)² = 4; (3)² = 9
Finally, for LRAM, multiply the width by the first three lengths (all of the left points) and add them together: (1)(0) + (1)(1) + (1)(4) = 5
For RRAM, multiply the width by the last three lengths (all of the right points) and add them together: (1)(1) + (1)(4) + (1)(9) = 14
Your answer: 5 < area < 14
and don't forget, just bust a move!
taylor blog #22
Im going to review steps to the topics we have covered for this study
LRAM
Steps:
first you must create a number line with each of the bounds as the end points then you should put markers on the time line for each significant number between the bounds
the next step is to draw a specific number of rectangles as specified by the given n= across the numberline resembeling a bar graph
you must then plug in each significant number into the given equation
plug those results into the rectangles to serve as the height of the rectangle
then you must decipher what the width would be considering the number of rectangles needed in relation to the bounds.
then you will multiply each of the widths by the heights to get multiple solutions which you will add together to get the area of the bounded region
keep in mind that for lram you will multiply all widths by heights except for that of the last rectangle.
RRAM
Steps:
first you must create a number line with each of the bounds as the end points then you should put markers on the time line for each significant number between the bounds
the next step is to draw a specific number of rectangles as specified by the given n= across the numberline resembeling a bar graph
you must then plug in each significant number into the given equation
plug those results into the rectangles to serve as the height of the rectangle
then you must decipher what the width would be considering the number of rectangles needed in relation to the bounds.
then you will multiply each of the widths by the heights to get multiple solutions which you will add together to get the area of the bounded region
keep in mind that for Rram you will multiply all widths by heights except for that of the first rectangle.
TRAM
Steps:
first you must create a number line with each of the bounds as the end points then you should put markers on the time line for each significant number between the bounds
the next step is to draw a specific number of trapezoids as specified by the given n= across the numberline resembeling a bar graph
you must then plug in each significant number into the given equation
plug those results into the trapezoids to serve as the width of the trapezoid then you must decipher what the height would be considering the number of rectangles needed in relation to the bounds.
then you will plug in each of the numbers from the widths and heights into 1/2(b1+b2) to then get multiple solutions and then add the solutions all together to get the area when using the trapezodial rule.
LRAM
Steps:
first you must create a number line with each of the bounds as the end points then you should put markers on the time line for each significant number between the bounds
the next step is to draw a specific number of rectangles as specified by the given n= across the numberline resembeling a bar graph
you must then plug in each significant number into the given equation
plug those results into the rectangles to serve as the height of the rectangle
then you must decipher what the width would be considering the number of rectangles needed in relation to the bounds.
then you will multiply each of the widths by the heights to get multiple solutions which you will add together to get the area of the bounded region
keep in mind that for lram you will multiply all widths by heights except for that of the last rectangle.
RRAM
Steps:
first you must create a number line with each of the bounds as the end points then you should put markers on the time line for each significant number between the bounds
the next step is to draw a specific number of rectangles as specified by the given n= across the numberline resembeling a bar graph
you must then plug in each significant number into the given equation
plug those results into the rectangles to serve as the height of the rectangle
then you must decipher what the width would be considering the number of rectangles needed in relation to the bounds.
then you will multiply each of the widths by the heights to get multiple solutions which you will add together to get the area of the bounded region
keep in mind that for Rram you will multiply all widths by heights except for that of the first rectangle.
TRAM
Steps:
first you must create a number line with each of the bounds as the end points then you should put markers on the time line for each significant number between the bounds
the next step is to draw a specific number of trapezoids as specified by the given n= across the numberline resembeling a bar graph
you must then plug in each significant number into the given equation
plug those results into the trapezoids to serve as the width of the trapezoid then you must decipher what the height would be considering the number of rectangles needed in relation to the bounds.
then you will plug in each of the numbers from the widths and heights into 1/2(b1+b2) to then get multiple solutions and then add the solutions all together to get the area when using the trapezodial rule.
Alaina
This week, we talked about LRAM, RRAM, MRAM, and TRAP. We also talked about finding the area between two graphs..
LRAM, RRAM, MRAM and TRAP all circulate around Riemann Sums.
Anyway, LRAM is usually an underestimate; whereas, RRAM is usually an overestimate.
To find LRAM and RRAM, first you need to draw the graph or the number of rectangles within the bounds and figure out your delta X. Then, you need to plug in the numbers at the end of each rectangle within the bounds. This gives you your y-values. From there, for LRAM, you multiply each y-value by delta X except for the last and add them together. For RRAM, you multiply each y-value by delta X except for the first and add them together. These values are over and under estimates.
For the Trapezoidal rule, there are two ways to find it. You can add LRAM and RRAM together and multiply by one half, or you can 1/2 (base 1 + base 2)H where bases are y-values and h is like delta X. This method gets you closer to the true area.
Then, there's the midpoint rule (MRAM). Midpoint uses rectangles but touches at a midpoint. So if your directions say to find the area bounded by 0 and 1 using 1 rectangle, your midpoint would be 1/2. From there, you repeat the steps from LRAM and RRAM.
LRAM, RRAM, MRAM and TRAP all circulate around Riemann Sums.
Anyway, LRAM is usually an underestimate; whereas, RRAM is usually an overestimate.
To find LRAM and RRAM, first you need to draw the graph or the number of rectangles within the bounds and figure out your delta X. Then, you need to plug in the numbers at the end of each rectangle within the bounds. This gives you your y-values. From there, for LRAM, you multiply each y-value by delta X except for the last and add them together. For RRAM, you multiply each y-value by delta X except for the first and add them together. These values are over and under estimates.
For the Trapezoidal rule, there are two ways to find it. You can add LRAM and RRAM together and multiply by one half, or you can 1/2 (base 1 + base 2)H where bases are y-values and h is like delta X. This method gets you closer to the true area.
Then, there's the midpoint rule (MRAM). Midpoint uses rectangles but touches at a midpoint. So if your directions say to find the area bounded by 0 and 1 using 1 rectangle, your midpoint would be 1/2. From there, you repeat the steps from LRAM and RRAM.
Justin
Ok so this week we learned how to find the area between a graph and the x/y axis
This is part of the UNO study we have been working on.
The part of the study we are working on now are Rieman sums:
LRAM
RRAM
MRAM
TRAPAZOIDAL RULE
to begin, you plug in the intervals for x to the original function
ex: f(x)=x^2 [0,3]
f(0)=0
f(1)=1
f(2)=4
f(3)=9
To find LRAM, add all the values but leave out the last
To find RRAM, add all the values but leave out the first
Also, integrate the equation using the fundamental theory of calculus to find the actual value.
1/3x^3 3|0
This is part of the UNO study we have been working on.
The part of the study we are working on now are Rieman sums:
LRAM
RRAM
MRAM
TRAPAZOIDAL RULE
to begin, you plug in the intervals for x to the original function
ex: f(x)=x^2 [0,3]
f(0)=0
f(1)=1
f(2)=4
f(3)=9
To find LRAM, add all the values but leave out the last
To find RRAM, add all the values but leave out the first
Also, integrate the equation using the fundamental theory of calculus to find the actual value.
1/3x^3 3|0
Blog #22
Well, yet another blog. This week, we continued the UNO study and learned more about the LRAM, RRAM, MRAM, and TRAM. For this blog, I’m going to go over how to find LRAM and RRAM.
These methods are used to approximate the area of the region bounded by a curve and the x-axis.
LRAM is the left Reimann Sums. It uses the left points touching the graph.
RRAM is the right Reimann Sums. It uses the right points touching the graph.
In order to get an accurate estimation, you need to do both LRAM and RRAM because one will always be an underestimate and one will always be an overestimate.
Reimann Sums is the method of breaking up the region into smaller, easier shapes and adding the area of each one together. It is commonly used with rectangles.
First, the area of a rectangle is length times width. So, we must find the length and the width of each rectangle, then add all of them together.
Ex. Find the area of the region bounded by the curve f(x) = x² and the x-axis between x=0 and x=3. Use 3 rectangles.
First of all, x=0 and x=3 are your bounds. So to rewrite this equation, it would be: 3S0 x² dx. (*”S” is the integral symbol.)
Next, what is the width? To find it, subtract your upper and lower bounds, then divide by the number of rectangles: (3-0)/3 = 1
Now, to find the lengths, plug in to the original equation every number between 0 and 3 that is 1 width away. The numbers will be 0, 1, 2, and 3. To plug in: (0)² = 0; (1)² = 1; (2)² = 4; (3)² = 9
Finally, for LRAM, multiply the width by the first three lengths (all of the left points) and add them together: (1)(0) + (1)(1) + (1)(4) = 5
For RRAM, multiply the width by the last three lengths (all of the right points) and add them together: (1)(1) + (1)(4) + (1)(9) = 14
Your answer: 5 < area < 14
These methods are used to approximate the area of the region bounded by a curve and the x-axis.
LRAM is the left Reimann Sums. It uses the left points touching the graph.
RRAM is the right Reimann Sums. It uses the right points touching the graph.
In order to get an accurate estimation, you need to do both LRAM and RRAM because one will always be an underestimate and one will always be an overestimate.
Reimann Sums is the method of breaking up the region into smaller, easier shapes and adding the area of each one together. It is commonly used with rectangles.
First, the area of a rectangle is length times width. So, we must find the length and the width of each rectangle, then add all of them together.
Ex. Find the area of the region bounded by the curve f(x) = x² and the x-axis between x=0 and x=3. Use 3 rectangles.
First of all, x=0 and x=3 are your bounds. So to rewrite this equation, it would be: 3S0 x² dx. (*”S” is the integral symbol.)
Next, what is the width? To find it, subtract your upper and lower bounds, then divide by the number of rectangles: (3-0)/3 = 1
Now, to find the lengths, plug in to the original equation every number between 0 and 3 that is 1 width away. The numbers will be 0, 1, 2, and 3. To plug in: (0)² = 0; (1)² = 1; (2)² = 4; (3)² = 9
Finally, for LRAM, multiply the width by the first three lengths (all of the left points) and add them together: (1)(0) + (1)(1) + (1)(4) = 5
For RRAM, multiply the width by the last three lengths (all of the right points) and add them together: (1)(1) + (1)(4) + (1)(9) = 14
Your answer: 5 < area < 14
Devin's Blog
Definition of the Natural Logarithmic Function
-the natural logarithmic function is defined by: ln x = f1^x 1/t dt, x > 0.
the domain of the natural logarithmic function is the set of all positive real numbers.
Theorem 5.1 Properties of the Natural Logarithmic Function
The natural logarithmic function has the following properties.
1. The domain is (0, oo) and the range is (-oo,oo).
2. The function is continuous, increasing, and one to one.
3. The graph is concave downward.
Domain of f(x) = ln x^2
First derivative f’ = ln 2x
Second derivative f’’ = ln 2
Using the definition of the natural logarithmic function, you can prove several important properties involving operations with natural logarithms. If you are already familiar with logarithms, you will recognize that these properties are characteristic of all logarithms.
Theorem 5.2 Logarithmic Properties
If a and b are positive numbers and n is rational, then the following properties are true.
1. Ln(1) = 0
2. Ln(ab) = ln a + ln b
3. Ln(a^n) = n ln a
4. Ln(a/b) = ln a – ln b
-the natural logarithmic function is defined by: ln x = f1^x 1/t dt, x > 0.
the domain of the natural logarithmic function is the set of all positive real numbers.
Theorem 5.1 Properties of the Natural Logarithmic Function
The natural logarithmic function has the following properties.
1. The domain is (0, oo) and the range is (-oo,oo).
2. The function is continuous, increasing, and one to one.
3. The graph is concave downward.
Domain of f(x) = ln x^2
First derivative f’ = ln 2x
Second derivative f’’ = ln 2
Using the definition of the natural logarithmic function, you can prove several important properties involving operations with natural logarithms. If you are already familiar with logarithms, you will recognize that these properties are characteristic of all logarithms.
Theorem 5.2 Logarithmic Properties
If a and b are positive numbers and n is rational, then the following properties are true.
1. Ln(1) = 0
2. Ln(ab) = ln a + ln b
3. Ln(a^n) = n ln a
4. Ln(a/b) = ln a – ln b
Friday, January 21, 2011
1/21/11
Alright so this week we continued with the UNO study and I think the material is actually really simple now. Last week we learned how to find LRAM, RRAM, how to use the Trapezoidal Rule, and how to find the exact value of the area of a region using the Fundamental Theorem of Calculus. This week we learned MRAM, which is the Midpoint Rule. This rule uses rectangles that touch the graph at the midpoint of the rectangle instead of the left or right side. The only thing different that you have to do for this is plug the midpoint of each rectangle into the original equation to get your heights. And to find the midpoint of each, all you do is add the numbers on each side of the rectangle and divide by 2. Delta x would simply be the distance between each rectangle. And after you get the heights, all you do to find the area is multiply each height by delta x and add them all together....This week we also learned how to find the area between 2 curves. This is when you're given two separate equations: you first graph them on the same coordinate plane; then you have to decide which curve is on top; then you use this formula: Stop-bottom (where you subtract the two equations and then integrate it). After that (if you aren't given the bounds in the directions), to find them all you do is set the two equations equal to each other and solve for x. Then once you integrate, you plug in the numbers of the bounds into the area formula ("Fundamental Theorem of Calculus"). And that's it! So easy!
So here's an example:
1.) Find the area between the two curves for y = x^2 + 2x + 1 and y = 2x + 5.
*Okay so the first thing you have to do is draw this graph so that you can figure out which one of these curves is on top.
*So the graph of x^2+2x+1 is a parabola with vertex at x=-1 opening up...And the graph of 2x+5 is a line going up intersecting the parabola at points (-5/2,0) and (0,5). So that means that the line is on top and the parabola is on the bottom..And the area in between those two graphs is what you're going to be finding.
*But first, let's find our bounds since we were not given them...So to do that you just set your two equations equal to each other and solve for x like this:
x^2+2x+1 = 2x+5
x^2-4 ...(subtracted 2x and 5 over to the left)
(x+2)(x-2) ...(difference of two squares)
x=2, -2 And these are your bounds! 2 going on the top, and -2 on the bottom
*Now we can use the formula to find the area. We already said that the line 2x+5 is on top so here's how to set up the equation:
S(2x+5)-(x^2+2x+1)
Now distribute the negative and simplify it to get this:
S -x^2+4
Now integrate that and you should end up with this: (before plugging in your bounds)
-1/3x^3+4x
Now you can use the Fundamental Theorem of Calculus to plug in your bounds and find the area. So this is what you should get when you plug in:
-1/3(2)^3+4(2)-[-1/3(-2)^3+4(-2)]
= 32/3 ..And that's your area!
**Soooooooooo, this week was really good; I understood everything we went over and honestly this material is extremely easy!
So here's an example:
1.) Find the area between the two curves for y = x^2 + 2x + 1 and y = 2x + 5.
*Okay so the first thing you have to do is draw this graph so that you can figure out which one of these curves is on top.
*So the graph of x^2+2x+1 is a parabola with vertex at x=-1 opening up...And the graph of 2x+5 is a line going up intersecting the parabola at points (-5/2,0) and (0,5). So that means that the line is on top and the parabola is on the bottom..And the area in between those two graphs is what you're going to be finding.
*But first, let's find our bounds since we were not given them...So to do that you just set your two equations equal to each other and solve for x like this:
x^2+2x+1 = 2x+5
x^2-4 ...(subtracted 2x and 5 over to the left)
(x+2)(x-2) ...(difference of two squares)
x=2, -2 And these are your bounds! 2 going on the top, and -2 on the bottom
*Now we can use the formula to find the area. We already said that the line 2x+5 is on top so here's how to set up the equation:
S(2x+5)-(x^2+2x+1)
Now distribute the negative and simplify it to get this:
S -x^2+4
Now integrate that and you should end up with this: (before plugging in your bounds)
-1/3x^3+4x
Now you can use the Fundamental Theorem of Calculus to plug in your bounds and find the area. So this is what you should get when you plug in:
-1/3(2)^3+4(2)-[-1/3(-2)^3+4(-2)]
= 32/3 ..And that's your area!
**Soooooooooo, this week was really good; I understood everything we went over and honestly this material is extremely easy!
1/21/11
We're still doing the UNO study so i'll talk about what we've learned so far this week dealing with that. We've learned how to find area dealing with definite integrals. The definite integral of a function represents the area between the curve and x-axis. We also learned how to find LRAM, RRAM, TRAP, AND MRAM. All of these are similar to the same process, it's just following steps. To first start these problems it's best to draw the graph to give you a visual of how many shapes they have, and what sides are going to touch the x-axis. **With LRAM and RRAM you want to make sure you know which side(left or right) is touching the x-axis. Most of the time for LRAM when plugging in you're going to leave out 0, but with RRAM you're going to put 0 and leave out the last number. So after figuring out the graph, for LRAM and RRAM your going to take the bounds and subtract them, then divide it by the amount of rectangles. This will be your delta x. Then, your going to take the the amount of rectangles and plug them into the the equation. The answers you get will be used to find LRAM and RRAM, and to find area you will take the delta x, and multiply them by the answers you got when plugging into the equation (Don't forget to add all the equation answers together and multiply each one by delta x). Now to find TRAP you plug into the formula 1/2(LRAM+RRAM)<-**This is the short cut. To do it the long way, you have to plug into the original formula just like LRAM and RRAM, however to find the area follow the formula: 1/2(b1+b2)h. Lastly for MRAM you divide the rectangles that you draw on the graph to find the midpoint(point in between the two whole number parts). Plug the numbers you find into the original equation, and add them togetherxdelta x.
Here is an example of TRAM:
Use the Trapezoidal Rule to approximate (-x^2+4) dx, with the bounds x=0, x=3. How can you get a better answer?
*First things first, draw the picture out, which is going to be 3 trapezoids. We're going to plug in -0,-1,-2, and -3 into the original equation to get 4,3,0, and -5.
*Now plug into the formula 1/2(b1+b2)h, to get 5/2
**The answer would be 5/2
Here is an example of TRAM:
Use the Trapezoidal Rule to approximate (-x^2+4) dx, with the bounds x=0, x=3. How can you get a better answer?
*First things first, draw the picture out, which is going to be 3 trapezoids. We're going to plug in -0,-1,-2, and -3 into the original equation to get 4,3,0, and -5.
*Now plug into the formula 1/2(b1+b2)h, to get 5/2
**The answer would be 5/2
Monday, January 17, 2011
taylor blog #21
since this week we mainly focused on the uno study i am going to review again before we start doing the ap review tests.
I want to give a review on all things derivative because to me majority of the calculus we have learned has some form of derivative involved.
SLOPE OF A TANGENT LINE
There are two formulas which need to be memorized
(((& means delta)
• The first formula is
f(x+&x)- f(x)/&x
This is the formula for a derivative. This formula is known as the secant line formula.
• The second formula is only a tiny bit different from the first
Lim f(x+&x)- f(x)/&x
&x -> 0
This formula is known as the slope of a tangent line
Solving for the problems we’ve had thus far in chapter 2 have consisted of plugging into these formulas and solving.
When given an equation you must plug x+&x into all x’s of the given equation and fill in the rest of the formula by placing – f(given equation exactly how its given)/ &x
Ex:
Find the slope of the graph of f(x)= 2x-3 at (2,1)
First you would plug in:
f(x) 2(x+&x)-3-(2x-3)/ &x
&x-> 0
Expand:
2x+2&x – 3 – 2x + 3 / &x
Take out what cancels:
2x + 2&x -3 -2x +3/&x
And you’re left with:
2&x/&x
Simplify:
2 &x/ &x
Therefore the answer is 2
((because there are no x’s left in the equation you can ignore the point (2,1) however if there had been any x’s you would have also plugged in 2 for the x’s before you simplified then solved as normal.))
There are a few helpful hints that need to be remembered
**remember that for each of these formulas any variable can be used for delta x
** remember that slope of the tangent line means to use the derivative formula
** there are many was to ask for a derivative these ways are:
Dy/dx
Y^1
F’(x)
d/dx {f(x)}
Dx[y]
D/dx
There are two rules to follow when using the derivative shortcut
These rules are : * The Constant Rule
• The Power Rule
The constant rule states that the derivative of a constant is 0
Therefore when asked to find a derivative of an equation with a constant the constant will automatically become 0.
The Power rule states that every time you take a derivative you lose a power.
Therefore if you are not taking the derivative of a constant you bring the exponent to the front of the variable and subtract one from the exponent.
The formula that displays this is
d/dx [x^n ] = nx^n-1
Example: d/dx[x^3] = 3x^2
There are many variations of equations that will call for simplification before you can take the derivative
When taking the derivative of a fraction you must simplify the equation by taking the numerator and placing it in front of the x and the exponent behind the denominator will become a negative
Example: d/dx [1/x^2] = x^-2
Then you would proceed to use the shortcut method to find the derivative.
When taking the derivative of a root you will turn the value of the root into a fraction with the exponent of x over the value of the root.
Example: d/dx [cuberoot x] = X^1/3
Then you would proceed to use the shortcut method to find the derivative.
**do not forget that after finding the derivative of a simplified equation you must convert the derivative back to an unsimplified form
Example: d/dx [ x^1/3] = 1/3x^-2/3 = 1/3cuberoot x^2
For both the product rule and the quotient there is a recognizable format that will allow you tho know which rule you will need to use and for each rule there is a formula to memorize and put into effect to find the derivative.
The product rule:
The product rule is recognized as F(x)G(x)
The formula for solving with the product rule is
D/Dx [F(x)G(x)] = F(x) Gprime(x) + G(x) Fprime(x)
Example: (3x-2x^2) (5+4x)
First I take the derivative e of each so that when I plug in I already know what the derivatives are to be plugged in
D/Dx [3x-2x^2]= 3-4x
Therefore F prime= 3-4x
D/Dx [5+4x]= 4
Therefore G prime= 4
Now you plug into the formula
Therefore
3x-2x^2(4)+ 5+4x (3-4x)
Distribute
12x^2x-8x^2+15-20x+12x-16
Simplify
-24x^2 + 4x +15
Because this equation cannot be simplified any further
Dx= -24x^2+4x+15
The quotient rule:
The quotient rule is recognized as F(x)/G(x)
The formula for solving with the quotient rule is
D/Dx [F(x)/G(x)] = G(x) Fprime(x)- F(x) Gprime(x)/ [G(x)]^2
Example: 5x-2/x^2-1
First I take the derivative of each so that when I plug in I already know what the derivatives are to be plugged in
D/dx [ 5x-2] = 5
Therefore F prime= 5
D/dx [x^2 – 1] = 2x
Therefore G prime = 2x
Now you plug into the formula
Therefore
(x^2+1)(5)-[(5x-2)(2x)]/(x^2+1)^2
Distribute
5x^2+5-[10x^2-4x]/(x^2+1)^2
Distribute the negative
5x^2+5-10x^2+4x/(x^2+1)^2
Simplify
-5x^2 + 4x+5/ (x^2+1)^2
Because this equation cannot be simplified any further
DX= -5x^2 + 4x+5/ (x^2+1)^2
Some things to remember:
**don’t forget that when solving with the quotient rule the negative must be distributed to the entire product of f(x)G prime(X)
** don’t forget which order F(X)G prime(X) and G(X)F prime(X) go in for each equation because it DOES matter.
The chain rule could loosely be defined as an order of operations that is used when solving composite functions.
A very important key concept to remember when using the chain rule is to work from the outside to the inside.
The “Formula” for the chain rule is
d/dx [fg(x)] = f ’(g(x)) (g’(x))
The most common procedure to using the chain rule is
• First, take the derivative of the outside
• Second, recopy just the inside
• Finally, multiply by the derivative of the inside
Example:
Square root of 3x^2-X+1
= (3x^2-X+1)^1/2
First, take the derivative of the outside:
½(________)^-1/2
Second, recopy just the inside
½(3x^2-X+1)^-1/2
Finally, multiply by the derivative of the inside
½(3x^2-X+1)^-1/2 (6x-1)
Simplify
6x-1/2(3x^2-x+1)^1/2
The key thing to remember when taking implicit derivatives is that the steps are the same as taking a regular derivative.
“with respect to” is a phrase you will see often when dealing with implicit derivatives.
The phrase “with respect to” gives you a clue as to what you will be solving for
When you see “d/dx” you also must be able to recognize that this is a symbol for “with respect to x”
When you take a derivative of anything besides x you will write it in the asterisk spot of the following example: D*/Dx
In other words, the top letter next to the d is the variable you are taking the derivative of
For example:
With respect to y: dy/dx
With respect to r: dr/dx
With respect to s: ds/dx
To solve a problem you must follow three general steps:
First take the derivative
Second gather all d*/dx
Third solve for d*/dx
Example:::
Sqrtxy= x^2y+1
(xy)1/2=x^2y+1
X^ 1/2 y^ ½ = x^2 y +1
First take the derivative
X^1/2 (1/2y^-1/2 dy/dx) + y^1/2 (1/2x^-1/2)= x^2 (1dy/dx+y(2x))
1/2x^1/2y^-1/2 dy/dx + 1/2x ^-1/2 y^1/2 = x^2 dy/dx + 2xy
Second gather all d*/dx
1/2x^1/2y^-1/2 dy/dx- X^2 dy/dx= 2xy - 1/2x^-1/2y^1y^1/2
Third solve for d*/dx
Dy/dx = 2xy -1/2x^-1/2y^1/2 / ½ x^1/2 y^-1/2 – x^2
Example with trig:
4cosx siny=1
4[cosx(cosy dy/dx)+ siny (-sinx)] = 0
4cosxx cosy dy/dx – 4 siny sin x =0
4 cosx cosy dy/dx= 4siny sinx
Dy/dx= 4 siny sinx / 4 cosx cos y
Dy/dx= tan y tan x
We also learned how to take the second derivative when dealing with implicit derivatives
The steps to solving the second derivative of an implicit derivative are:
Take the first derivative
Solve for dy/dx
Take the derivative of dy/dx (this will be d^2y/dx^2)
Plug in the result of when you solved for dy/dx into the d^2y/dx^2 equation for dy/dx
Solve again
**** sometimes when you solve the numerator will turn out to be your original equation, if so, plug in what the original equation equaled and simplify.
If not then d^2y/dx^2 will be what you found when you “solved again”
Example:
x^2 + y^2 = 25
Take the first derivative
2x + 2y dy/dx = 0
Solve for dy/dx
Dy/dx= -2x/2y therefore -x/y
Take the derivative of dy/dx
- Y(1)- [ x(1dy/dx)]/y^2
- Y-x dy/dx / y^2
Plug in the result of when you solved for dy/dx into the d^2y/dx^2 equation for dy/dx
- Y-x ( -x/y)/ y^2
Solve again
- Y^2 + x^2/y/y^2/1
-y^2 +x^2/ y^3
**** remember that sometimes when you solve the numerator will turn out to be your original equation, if so, plug in what the original equation equaled and simplify.
Therefore:
-25/y^3
I want to give a review on all things derivative because to me majority of the calculus we have learned has some form of derivative involved.
SLOPE OF A TANGENT LINE
There are two formulas which need to be memorized
(((& means delta)
• The first formula is
f(x+&x)- f(x)/&x
This is the formula for a derivative. This formula is known as the secant line formula.
• The second formula is only a tiny bit different from the first
Lim f(x+&x)- f(x)/&x
&x -> 0
This formula is known as the slope of a tangent line
Solving for the problems we’ve had thus far in chapter 2 have consisted of plugging into these formulas and solving.
When given an equation you must plug x+&x into all x’s of the given equation and fill in the rest of the formula by placing – f(given equation exactly how its given)/ &x
Ex:
Find the slope of the graph of f(x)= 2x-3 at (2,1)
First you would plug in:
f(x) 2(x+&x)-3-(2x-3)/ &x
&x-> 0
Expand:
2x+2&x – 3 – 2x + 3 / &x
Take out what cancels:
2x + 2&x -3 -2x +3/&x
And you’re left with:
2&x/&x
Simplify:
2 &x/ &x
Therefore the answer is 2
((because there are no x’s left in the equation you can ignore the point (2,1) however if there had been any x’s you would have also plugged in 2 for the x’s before you simplified then solved as normal.))
There are a few helpful hints that need to be remembered
**remember that for each of these formulas any variable can be used for delta x
** remember that slope of the tangent line means to use the derivative formula
** there are many was to ask for a derivative these ways are:
Dy/dx
Y^1
F’(x)
d/dx {f(x)}
Dx[y]
D/dx
There are two rules to follow when using the derivative shortcut
These rules are : * The Constant Rule
• The Power Rule
The constant rule states that the derivative of a constant is 0
Therefore when asked to find a derivative of an equation with a constant the constant will automatically become 0.
The Power rule states that every time you take a derivative you lose a power.
Therefore if you are not taking the derivative of a constant you bring the exponent to the front of the variable and subtract one from the exponent.
The formula that displays this is
d/dx [x^n ] = nx^n-1
Example: d/dx[x^3] = 3x^2
There are many variations of equations that will call for simplification before you can take the derivative
When taking the derivative of a fraction you must simplify the equation by taking the numerator and placing it in front of the x and the exponent behind the denominator will become a negative
Example: d/dx [1/x^2] = x^-2
Then you would proceed to use the shortcut method to find the derivative.
When taking the derivative of a root you will turn the value of the root into a fraction with the exponent of x over the value of the root.
Example: d/dx [cuberoot x] = X^1/3
Then you would proceed to use the shortcut method to find the derivative.
**do not forget that after finding the derivative of a simplified equation you must convert the derivative back to an unsimplified form
Example: d/dx [ x^1/3] = 1/3x^-2/3 = 1/3cuberoot x^2
For both the product rule and the quotient there is a recognizable format that will allow you tho know which rule you will need to use and for each rule there is a formula to memorize and put into effect to find the derivative.
The product rule:
The product rule is recognized as F(x)G(x)
The formula for solving with the product rule is
D/Dx [F(x)G(x)] = F(x) Gprime(x) + G(x) Fprime(x)
Example: (3x-2x^2) (5+4x)
First I take the derivative e of each so that when I plug in I already know what the derivatives are to be plugged in
D/Dx [3x-2x^2]= 3-4x
Therefore F prime= 3-4x
D/Dx [5+4x]= 4
Therefore G prime= 4
Now you plug into the formula
Therefore
3x-2x^2(4)+ 5+4x (3-4x)
Distribute
12x^2x-8x^2+15-20x+12x-16
Simplify
-24x^2 + 4x +15
Because this equation cannot be simplified any further
Dx= -24x^2+4x+15
The quotient rule:
The quotient rule is recognized as F(x)/G(x)
The formula for solving with the quotient rule is
D/Dx [F(x)/G(x)] = G(x) Fprime(x)- F(x) Gprime(x)/ [G(x)]^2
Example: 5x-2/x^2-1
First I take the derivative of each so that when I plug in I already know what the derivatives are to be plugged in
D/dx [ 5x-2] = 5
Therefore F prime= 5
D/dx [x^2 – 1] = 2x
Therefore G prime = 2x
Now you plug into the formula
Therefore
(x^2+1)(5)-[(5x-2)(2x)]/(x^2+1)^2
Distribute
5x^2+5-[10x^2-4x]/(x^2+1)^2
Distribute the negative
5x^2+5-10x^2+4x/(x^2+1)^2
Simplify
-5x^2 + 4x+5/ (x^2+1)^2
Because this equation cannot be simplified any further
DX= -5x^2 + 4x+5/ (x^2+1)^2
Some things to remember:
**don’t forget that when solving with the quotient rule the negative must be distributed to the entire product of f(x)G prime(X)
** don’t forget which order F(X)G prime(X) and G(X)F prime(X) go in for each equation because it DOES matter.
The chain rule could loosely be defined as an order of operations that is used when solving composite functions.
A very important key concept to remember when using the chain rule is to work from the outside to the inside.
The “Formula” for the chain rule is
d/dx [fg(x)] = f ’(g(x)) (g’(x))
The most common procedure to using the chain rule is
• First, take the derivative of the outside
• Second, recopy just the inside
• Finally, multiply by the derivative of the inside
Example:
Square root of 3x^2-X+1
= (3x^2-X+1)^1/2
First, take the derivative of the outside:
½(________)^-1/2
Second, recopy just the inside
½(3x^2-X+1)^-1/2
Finally, multiply by the derivative of the inside
½(3x^2-X+1)^-1/2 (6x-1)
Simplify
6x-1/2(3x^2-x+1)^1/2
The key thing to remember when taking implicit derivatives is that the steps are the same as taking a regular derivative.
“with respect to” is a phrase you will see often when dealing with implicit derivatives.
The phrase “with respect to” gives you a clue as to what you will be solving for
When you see “d/dx” you also must be able to recognize that this is a symbol for “with respect to x”
When you take a derivative of anything besides x you will write it in the asterisk spot of the following example: D*/Dx
In other words, the top letter next to the d is the variable you are taking the derivative of
For example:
With respect to y: dy/dx
With respect to r: dr/dx
With respect to s: ds/dx
To solve a problem you must follow three general steps:
First take the derivative
Second gather all d*/dx
Third solve for d*/dx
Example:::
Sqrtxy= x^2y+1
(xy)1/2=x^2y+1
X^ 1/2 y^ ½ = x^2 y +1
First take the derivative
X^1/2 (1/2y^-1/2 dy/dx) + y^1/2 (1/2x^-1/2)= x^2 (1dy/dx+y(2x))
1/2x^1/2y^-1/2 dy/dx + 1/2x ^-1/2 y^1/2 = x^2 dy/dx + 2xy
Second gather all d*/dx
1/2x^1/2y^-1/2 dy/dx- X^2 dy/dx= 2xy - 1/2x^-1/2y^1y^1/2
Third solve for d*/dx
Dy/dx = 2xy -1/2x^-1/2y^1/2 / ½ x^1/2 y^-1/2 – x^2
Example with trig:
4cosx siny=1
4[cosx(cosy dy/dx)+ siny (-sinx)] = 0
4cosxx cosy dy/dx – 4 siny sin x =0
4 cosx cosy dy/dx= 4siny sinx
Dy/dx= 4 siny sinx / 4 cosx cos y
Dy/dx= tan y tan x
We also learned how to take the second derivative when dealing with implicit derivatives
The steps to solving the second derivative of an implicit derivative are:
Take the first derivative
Solve for dy/dx
Take the derivative of dy/dx (this will be d^2y/dx^2)
Plug in the result of when you solved for dy/dx into the d^2y/dx^2 equation for dy/dx
Solve again
**** sometimes when you solve the numerator will turn out to be your original equation, if so, plug in what the original equation equaled and simplify.
If not then d^2y/dx^2 will be what you found when you “solved again”
Example:
x^2 + y^2 = 25
Take the first derivative
2x + 2y dy/dx = 0
Solve for dy/dx
Dy/dx= -2x/2y therefore -x/y
Take the derivative of dy/dx
- Y(1)- [ x(1dy/dx)]/y^2
- Y-x dy/dx / y^2
Plug in the result of when you solved for dy/dx into the d^2y/dx^2 equation for dy/dx
- Y-x ( -x/y)/ y^2
Solve again
- Y^2 + x^2/y/y^2/1
-y^2 +x^2/ y^3
**** remember that sometimes when you solve the numerator will turn out to be your original equation, if so, plug in what the original equation equaled and simplify.
Therefore:
-25/y^3
alaina's blog, 17 jan 2011
This weekend, i kind of got lost in the days. So, we started the uno thing this week and didn't really learn much, so i'm reviewing the basics.
Product and quotient rules:
Product Rule:[f(x)g(x)]=f(x)g1(x)+g(x)f1(x)
Quotient Rule:[f(x)/g(x)]= g(x)f1(x)-[f(x)g1(x)]/g(x)2
The product and quotient rules are pretty simple.
All you do is plug into them.
Ex 1. [(x2+4)(x-2)]=(x2+4)(1)+(x-2)(2x)=(x2+4)+(2x2-4x)=3x2-4x+4
Ex 2. [(x2+4)/(x-2)]=(x-2)(2x)-[(x2+4)(1)]/(x-2)2=2x2-4x-x2-4/(x-2)2=x2-4x-4/(x-2)2=(x-4)(x+1)/(x-2)2
The last thing we went over was taking the 2nd and 3rd derivative; all you do is take a derivative two or three times. Second and third derivative are represented like say g-prime or g', but instead of ', it is '' or '''-> g'' or g'''.
Ex 1. y''[x2]; y'=2x; y''=2
Ex 2. y'''[x2]; y'=2x; y''=2; y'''=0
Product and quotient rules:
Product Rule:[f(x)g(x)]=f(x)g1(x)+g(x)f1(x)
Quotient Rule:[f(x)/g(x)]= g(x)f1(x)-[f(x)g1(x)]/g(x)2
The product and quotient rules are pretty simple.
All you do is plug into them.
Ex 1. [(x2+4)(x-2)]=(x2+4)(1)+(x-2)(2x)=(x2+4)+(2x2-4x)=3x2-4x+4
Ex 2. [(x2+4)/(x-2)]=(x-2)(2x)-[(x2+4)(1)]/(x-2)2=2x2-4x-x2-4/(x-2)2=x2-4x-4/(x-2)2=(x-4)(x+1)/(x-2)2
The last thing we went over was taking the 2nd and 3rd derivative; all you do is take a derivative two or three times. Second and third derivative are represented like say g-prime or g', but instead of ', it is '' or '''-> g'' or g'''.
Ex 1. y''[x2]; y'=2x; y''=2
Ex 2. y'''[x2]; y'=2x; y''=2; y'''=0
Sunday, January 16, 2011
Blog #21
This week we learned a few new things, but we also went over the Fundamental Theorem of Calculus and such.
So, I’m just going to go over Integrals and the Fundamental Theorem of Calculus.
*The integral symbol will be represented by “S” and the bounds will be represent as “3S1”.
Integrals:
First of all, an antiderivative is a general solution, while an integral is more specific where you must solve for c.
Second, an antiderivative is just the derivative backwards. You will be given f’(x) and asked to find f(x).
Third, how to find an antiderivative: If your given ax^b (a being a constant and b being an exponent), your formula is (a/b+1)(x^b+1).
*Note: When solving for the antiderivative, always add “+ c” in place of a constant. When solving for an integral, you will be given something like “f(1) = 2”, in that case, plug in 1 for x into your antiderivative and set it equal to 2 and that’s your constant.
**Note: When given the second derivative, just solve twice.
***Note: When given an integral, it is always used with the integral symbol in front (“S”) and dx afterwards.
****Note: Shortcut: If you have a fractional exponent, add one and multiply by the reciprocal.
*****Note: Trig functions: You just do the opposite of the derivative.
Next, the Fundamental Theorem of Calculus:
This theorem is very simple; it just states that “bSa f(x) dx = F(b) – F(a) = area where F(x) is the integrated function”.
Examples: Find the area of the shape between the boundaries given.
Ex. 1) 4S1 6 dx
F(x) = 6x
Area = 6(4) – 6(1) = 24 – 6
= 18
Ex. 2) 2S0 (x + 1) dx
F(x) = ½ x² + x
Area = (½ (2)² + (2)) – (½ (0)² + (0)) = 4 – 0
= 4
Ex. 3) 3S3 2x dx
= 0
**Whenever the upper and lower bounds are the same, the area is always 0.
So, I’m just going to go over Integrals and the Fundamental Theorem of Calculus.
*The integral symbol will be represented by “S” and the bounds will be represent as “3S1”.
Integrals:
First of all, an antiderivative is a general solution, while an integral is more specific where you must solve for c.
Second, an antiderivative is just the derivative backwards. You will be given f’(x) and asked to find f(x).
Third, how to find an antiderivative: If your given ax^b (a being a constant and b being an exponent), your formula is (a/b+1)(x^b+1).
*Note: When solving for the antiderivative, always add “+ c” in place of a constant. When solving for an integral, you will be given something like “f(1) = 2”, in that case, plug in 1 for x into your antiderivative and set it equal to 2 and that’s your constant.
**Note: When given the second derivative, just solve twice.
***Note: When given an integral, it is always used with the integral symbol in front (“S”) and dx afterwards.
****Note: Shortcut: If you have a fractional exponent, add one and multiply by the reciprocal.
*****Note: Trig functions: You just do the opposite of the derivative.
Next, the Fundamental Theorem of Calculus:
This theorem is very simple; it just states that “bSa f(x) dx = F(b) – F(a) = area where F(x) is the integrated function”.
Examples: Find the area of the shape between the boundaries given.
Ex. 1) 4S1 6 dx
F(x) = 6x
Area = 6(4) – 6(1) = 24 – 6
= 18
Ex. 2) 2S0 (x + 1) dx
F(x) = ½ x² + x
Area = (½ (2)² + (2)) – (½ (0)² + (0)) = 4 – 0
= 4
Ex. 3) 3S3 2x dx
= 0
**Whenever the upper and lower bounds are the same, the area is always 0.
blog for 1/16
wow, uhhh, I really don't remember anything from this week, even what the homework was
I have no clue what to put on here tbh.......
I have no clue what to put on here tbh.......
1/16/11
Okay so this week we only really learned stuff on Wednesday and Thursday..And it was for that UNO study. So first we reviewed Sigma notation and how to evaluate the sums and whatnot. Then we went over the different methods one can use to find the area of a region bounded by a curve and the x-axis--You can use Riemann sums (LRAM & RRAM)-(break up the region into shapes, find the area of each shape and then add them all together); and you can use the Fundamental Theorem of Calculus (which is when you take the integral of the function and then use the formula).
Here's an example:
1.) Use left and right endpoints and the given number of rectangles to find two approximations of the area of the region between the graph of the function and the x-axis over the given interval.
f(x) = 2x+5 [0,2]; 4 rectangles
*Well you're going to need to find LRAM and RRAM using only 4 rectangles so the first thing you have to figure out is the width of each rectangle. To do this you can use a little formula where you subtract the numbers of the interval over the number of rectangles, which would be this:
2-0/4 and that gives you a width (delta x) of 1/2
*Now you have to plug in the numbers between 0 and 2 into the original equation. (The numbers you're going to plug in are 0, 1/2, 1, 3/2, and 2...because you add 1/2 each time and there's 4 rectangles)
*So here's what you get when you plug in those numbers:
2(0)+5 = 5
2(1/2)+5 = 6
2(1)+5 = 7
2(3/2)+5 = 8
2(2)+5 = 9
*Now you need to find LRAM & RRAM. So first, let's start with LRAM. To find this, you're going to start with the numbers to the left (starting with 0) and you're NOT going to plug in the last number on the right. And you're going to multiply each bolded number by 1/2 because that's the width of each shape, and then you'll add them all together to get the area
*So for LRAM you should have this:
1/2(5)+1/2(6)+1/2(7)+1/2(8)
=13 (which is an underestimate)
*And for RRAM you should have this:
1/2(6)+1/2(7)+1/2(8)+1/2(9)
= 15 (which is an overestimate)..and you don't plug in what you got for 0 because you're looking at values from the right
*And if you go on to use the Fundamental Theorem of Calculus you get that the area is 14, which is right in the middle of your estimates
*I'm not sure if I explained this all that well, but I'm just getting the hang of it haha..
Here's an example:
1.) Use left and right endpoints and the given number of rectangles to find two approximations of the area of the region between the graph of the function and the x-axis over the given interval.
f(x) = 2x+5 [0,2]; 4 rectangles
*Well you're going to need to find LRAM and RRAM using only 4 rectangles so the first thing you have to figure out is the width of each rectangle. To do this you can use a little formula where you subtract the numbers of the interval over the number of rectangles, which would be this:
2-0/4 and that gives you a width (delta x) of 1/2
*Now you have to plug in the numbers between 0 and 2 into the original equation. (The numbers you're going to plug in are 0, 1/2, 1, 3/2, and 2...because you add 1/2 each time and there's 4 rectangles)
*So here's what you get when you plug in those numbers:
2(0)+5 = 5
2(1/2)+5 = 6
2(1)+5 = 7
2(3/2)+5 = 8
2(2)+5 = 9
*Now you need to find LRAM & RRAM. So first, let's start with LRAM. To find this, you're going to start with the numbers to the left (starting with 0) and you're NOT going to plug in the last number on the right. And you're going to multiply each bolded number by 1/2 because that's the width of each shape, and then you'll add them all together to get the area
*So for LRAM you should have this:
1/2(5)+1/2(6)+1/2(7)+1/2(8)
=13 (which is an underestimate)
*And for RRAM you should have this:
1/2(6)+1/2(7)+1/2(8)+1/2(9)
= 15 (which is an overestimate)..and you don't plug in what you got for 0 because you're looking at values from the right
*And if you go on to use the Fundamental Theorem of Calculus you get that the area is 14, which is right in the middle of your estimates
*I'm not sure if I explained this all that well, but I'm just getting the hang of it haha..
Saturday, January 15, 2011
1/15/11
Alrighttttt, this week we started to learn definite integrals. The definite integral of a function represents the area between the curve and the x-axis. *Definite integrals can also be calculated by counting squares and Riemann Sums. Sounds complicated? Not at all. Basically what you're doing is drawing the graph of the function, putting your points, drawing however many rectangles there asking for, looking to see which rectangle (right or left) touches the graph, or goes over, and plugging in..Or finding the area haha :)
Here's an example:
What are 2 types of Riemann sums that we could use to find the area of the region bounded by the graph of f(x)=1/2^2 and the x-axis between x=0 and x=3. Find the area ofthis region using all methods.
*So switching this to an integral we get the bounds of 3 as the upper, and 0 as the lower bound 1/2x^2=1/6x^3 bar 3 at top and 0 at the bottom.
*Now that we know 1/6x^3 is our formula were gonna plug in the bounds 3 and 0. 1/6(3)^3-1/6(0)^3=9/2 (this is usuing the fundamental theorm of calculus)
**Now draw the LRRAM
We know the function is a parabola, and the x axis goes up to 3, with 3 triangles, and the left point of the rectangles are touching the graph(which means it does NOT go over)
*Now plug in the 3 rectangle points into the original function (1/2x^2)
We're only going to plug in 0,1, and 2 because 3 is not touching the graph.
*1/2(0)^2=0
1/2(1)^2=1/2
1/2(2)^2=2
*Add them all together to get 5/2
**Now draw the RRAM
It's drawn the exact same except we know the rectangles go over, and when we plug in points were going to plug in 1,2, and 3.
That's pretty much all there is to it!
And now i'm off to state, goodluck sweetheartssssss! :D
Here's an example:
What are 2 types of Riemann sums that we could use to find the area of the region bounded by the graph of f(x)=1/2^2 and the x-axis between x=0 and x=3. Find the area ofthis region using all methods.
*So switching this to an integral we get the bounds of 3 as the upper, and 0 as the lower bound 1/2x^2=1/6x^3 bar 3 at top and 0 at the bottom.
*Now that we know 1/6x^3 is our formula were gonna plug in the bounds 3 and 0. 1/6(3)^3-1/6(0)^3=9/2 (this is usuing the fundamental theorm of calculus)
**Now draw the LRRAM
We know the function is a parabola, and the x axis goes up to 3, with 3 triangles, and the left point of the rectangles are touching the graph(which means it does NOT go over)
*Now plug in the 3 rectangle points into the original function (1/2x^2)
We're only going to plug in 0,1, and 2 because 3 is not touching the graph.
*1/2(0)^2=0
1/2(1)^2=1/2
1/2(2)^2=2
*Add them all together to get 5/2
**Now draw the RRAM
It's drawn the exact same except we know the rectangles go over, and when we plug in points were going to plug in 1,2, and 3.
That's pretty much all there is to it!
And now i'm off to state, goodluck sweetheartssssss! :D
Sunday, January 9, 2011
That Real Blogg, Yerd?
This week we discussed Definite Integrals and then the Fundamental Theorem of Calculus and also the Properties of Integrals. Starting with the definite integrals-that means the integral has upper and lower bounds and it represents an areabetween the graph and the x-axis. Well first, you write the integral symbol and your bounds are going to be what's shaded on the graph. Then you put the equation of the graph after the S followed by dx to complete your integral. OR for definite integrals they can do it the other way around and give you the integral function and they'll ask you to sketch the graph based off of that.
Example: S2-4 (x-9) dx:
1/2(4)^2-9(4)-[1/2(2)^2-9/2)]
=8-36-(2-18)
=-28+16
=-12
1/2(4)^2-9(4)-[1/2(2)^2-9/2)]
=8-36-(2-18)
=-28+16
=-12
another blog, another buck
this blog is gonna be short and sweet
I'm just gonna talk about something I don't know too much about, integrals
I think it's something like ∫(x)^2 dx = 1/2(x)^3
I must've missed the days that we learned this, because I really have no clue, but I'm getting there
I'm just gonna talk about something I don't know too much about, integrals
I think it's something like ∫(x)^2 dx = 1/2(x)^3
I must've missed the days that we learned this, because I really have no clue, but I'm getting there
Taylor blog #20
because we will soon be moving on to ap review and i am scared to death im going to expand upon the basics
In that i mean I am going to expand upon that work sheet we were given to help remember what processes we use when solving various types of problems
1. Slope of a normal line: Take a derivative and plug in the x-‐value. Take the negative reciprocal of the number. (not covered)
2. Average rate of change/ average velocity: slope where a,b are x-values
Find the slope of the graph of f(x) = x^2 + 3x + 1 at (1, 3)
lim as Dx approaches 0: ( f(x + Dx) - f(x) ) / (Dx)
What you have to do is plug in (x+ Dx) for every x in the formula.
(x + Dx)^2 + 3(x + Dx) + 1 - (x^2 + 3x + 1) / (Dx)
That simplifies to...
x^2 + 2xDx + Dx^2 + 3x + 3Dx + 1 - x^2 - 3x -1 / Dx
2xDx + Dx^2 + 3Dx / Dx
2x + Dx + 3
plug in 0 for Dx
2x + 3 is your equation
Now just plug in the x coordinate to find the slope.
2(1) + 3 = 5
The slope is 5.
3. rate of change: derivative. if given an x value plug in to get a number
Example:
A Pebble is dropped into a calm pond causing ripples in the form of concentric circles. The radius of the outer circle is increasing at a rate of 1 foot PER second when the radius is 4 ft at what rate is the total area (A) of the disturbed water changing?
#1 dr/dt= 1ft/sec
R=4 ft
dA/dt= ?
#2 A= pi r ^2
#3 dA/dt= pi{2r dr/dt}
#4 dA/dt= 2pi (4)(1)
Therefore 8 pi ft/sec
4. Slope of a horizontal tangent line
take derivative set equal to zero and solve for x
Ex:
Find the slope of the graph of f(x)= 2x-3 at (2,1)
First you would plug in:
f(x) 2(x+&x)-3-(2x-3)/ &x
&x-> 0
Expand:
2x+2&x – 3 – 2x + 3 / &x
Take out what cancels:
2x + 2&x -3 -2x +3/&x
And you’re left with:
2&x/&x
Simplify:
2 &x/ &x
Therefore the answer is 2
5. equation of a tangent line
take the dericative and plug in the x value
if you are not given a y value plug the original equation to get the y=value
then plug these values into point slope formula
EXAMPLE:
Find the equation of the tangent line to the graph of the function f(x)= sqrt x-3 at the point (39,6)
Sqrt x-3
Simplifies to (x-3)^1/2
Take the derivative
d/dx= 1/2( x-3)^-1/2 (1)
convert back to a fraction 1/ 2(x-3)^1/2
plug in for x 1/ 2 (39-3)^1/2
Solve = 1/ 2 (36)^1/2
Simplify= 1/ 2 sqrt 36
Solve= 1/12
Therefore the derivative of f(x)= sqrt x-3 at the point (39,6) is 1/12
Because this is the equation of a tangent line you must plug into the equation of a line formula to answer the question of the problem.
Therefore you will plug into y-y1 = m(x-x1)
plug in= y-6 = 1/12 (x-39)
solve: y-6 = 1/12(x-39)
+6 +6
Y= 1/12x + 11/4
Therefore the equation of the tangent line to the graph of the function f(x)= sqrt x-3 at the point (39,6) is y=1/12x + 11/4
6. average value ((not yet covered..)
7. maximum minimum critical values increasing decreasing
take derivative set equal to zero and solve for x
set up intervals using these numbers
plug in numbers on each interval
if it changes from positive to negative it is a maximum. If it changes from negative to positive it is a minimum
all potential maximums and minimums are called critical values. If the interval is positive the interval is increasing
id the interval is negative then the interval is decreasing
Find any critical numbers of the function f(x) = 4x / x^2+1
*Since they're asking you to find critical numbers, the first thing you have to do is take the derivative of the function
*So, using the quotient rule, you get:
*(x^2+1)(4) - [(4x)(2x)] / (x^2+1)^2
Simplifying that you get:
-4x^2+4 / (x^2+1)^2
*To find the critical numbers, you take the numerator of the fraction and set it equal to zero.
So you get:
-4x^2 + 4 = 0
x^2 = 1
*So your critical numbers are x = 1, -1
8. point of inflection concave up or down
take second derivative set equal to zero and solve for x
set up intervals using these numbers plug in numbers on each interval if the intervalsa change signs then there is a point of inflection at that value if it does not change sign then there is not a point of inflection
if the interval is positive then the interval is concave up is the interval is negative then the interval is concave down
Ex. 1) Determine the open intervals on which the graph f(x) = -x^3 + 6x^2 – 9x – 1 is concave up or down.
f’(x) = -3x^2 + 12x – 9
f’’(x) = -6x + 12
-6x + 12 = 0
-6x = -12
x = 2
It is differentiable.
(-infinity, 2) u (2, infinity)
f’’(1) = positive #
f’’(3) = negative #
concave up: (-infinity, 2)
concave down: (2, infinity)
*Note that the question does not ask for the point of inflection, but if it did, you would say: x = 2 is a point of inflection.
9. linearization ((have not covered))
10. infinite limits
id the degree of the top is greater than the degree of the bottom the limit is infinity. If the degree of the top is less than the degree of the bottom the limit is zero is the degree of the top is equal to the degree of the bottom the limit is the coefficients of the terms with the highest exponents
ex:
2x/3x therefore 2/3
11.vertical asymptotes
factor the top and the bottom of the fraction make cancellations if possible set what is left in the bottom equal to zero and solve for x
vertical asymptotes are given in the form of x= a number
x^2+2x-3/x^2 -5x -6
x2 – 5x – 6 = 0
(x – 6)(x + 1) = 0
x = 6 or –1
12. removables
factor the top and the bottom of the fraction make cancellations set each thing that cancelled equal to zero and solve for x removables are given in the form of x= a number
13. horizontal asymptotes
take the limit as x goes to infinty of the equation
if you get a number then your answer is y= a number
if you get infinity there are no horizontal asymptotes
14. area if only one equation is given (not covered)
15. area between two equations (not covered)
16. volume by disks (not covered)
17. volume by washers (not covered)
In that i mean I am going to expand upon that work sheet we were given to help remember what processes we use when solving various types of problems
1. Slope of a normal line: Take a derivative and plug in the x-‐value. Take the negative reciprocal of the number. (not covered)
2. Average rate of change/ average velocity: slope where a,b are x-values
Find the slope of the graph of f(x) = x^2 + 3x + 1 at (1, 3)
lim as Dx approaches 0: ( f(x + Dx) - f(x) ) / (Dx)
What you have to do is plug in (x+ Dx) for every x in the formula.
(x + Dx)^2 + 3(x + Dx) + 1 - (x^2 + 3x + 1) / (Dx)
That simplifies to...
x^2 + 2xDx + Dx^2 + 3x + 3Dx + 1 - x^2 - 3x -1 / Dx
2xDx + Dx^2 + 3Dx / Dx
2x + Dx + 3
plug in 0 for Dx
2x + 3 is your equation
Now just plug in the x coordinate to find the slope.
2(1) + 3 = 5
The slope is 5.
3. rate of change: derivative. if given an x value plug in to get a number
Example:
A Pebble is dropped into a calm pond causing ripples in the form of concentric circles. The radius of the outer circle is increasing at a rate of 1 foot PER second when the radius is 4 ft at what rate is the total area (A) of the disturbed water changing?
#1 dr/dt= 1ft/sec
R=4 ft
dA/dt= ?
#2 A= pi r ^2
#3 dA/dt= pi{2r dr/dt}
#4 dA/dt= 2pi (4)(1)
Therefore 8 pi ft/sec
4. Slope of a horizontal tangent line
take derivative set equal to zero and solve for x
Ex:
Find the slope of the graph of f(x)= 2x-3 at (2,1)
First you would plug in:
f(x) 2(x+&x)-3-(2x-3)/ &x
&x-> 0
Expand:
2x+2&x – 3 – 2x + 3 / &x
Take out what cancels:
2x + 2&x -3 -2x +3/&x
And you’re left with:
2&x/&x
Simplify:
2 &x/ &x
Therefore the answer is 2
5. equation of a tangent line
take the dericative and plug in the x value
if you are not given a y value plug the original equation to get the y=value
then plug these values into point slope formula
EXAMPLE:
Find the equation of the tangent line to the graph of the function f(x)= sqrt x-3 at the point (39,6)
Sqrt x-3
Simplifies to (x-3)^1/2
Take the derivative
d/dx= 1/2( x-3)^-1/2 (1)
convert back to a fraction 1/ 2(x-3)^1/2
plug in for x 1/ 2 (39-3)^1/2
Solve = 1/ 2 (36)^1/2
Simplify= 1/ 2 sqrt 36
Solve= 1/12
Therefore the derivative of f(x)= sqrt x-3 at the point (39,6) is 1/12
Because this is the equation of a tangent line you must plug into the equation of a line formula to answer the question of the problem.
Therefore you will plug into y-y1 = m(x-x1)
plug in= y-6 = 1/12 (x-39)
solve: y-6 = 1/12(x-39)
+6 +6
Y= 1/12x + 11/4
Therefore the equation of the tangent line to the graph of the function f(x)= sqrt x-3 at the point (39,6) is y=1/12x + 11/4
6. average value ((not yet covered..)
7. maximum minimum critical values increasing decreasing
take derivative set equal to zero and solve for x
set up intervals using these numbers
plug in numbers on each interval
if it changes from positive to negative it is a maximum. If it changes from negative to positive it is a minimum
all potential maximums and minimums are called critical values. If the interval is positive the interval is increasing
id the interval is negative then the interval is decreasing
Find any critical numbers of the function f(x) = 4x / x^2+1
*Since they're asking you to find critical numbers, the first thing you have to do is take the derivative of the function
*So, using the quotient rule, you get:
*(x^2+1)(4) - [(4x)(2x)] / (x^2+1)^2
Simplifying that you get:
-4x^2+4 / (x^2+1)^2
*To find the critical numbers, you take the numerator of the fraction and set it equal to zero.
So you get:
-4x^2 + 4 = 0
x^2 = 1
*So your critical numbers are x = 1, -1
8. point of inflection concave up or down
take second derivative set equal to zero and solve for x
set up intervals using these numbers plug in numbers on each interval if the intervalsa change signs then there is a point of inflection at that value if it does not change sign then there is not a point of inflection
if the interval is positive then the interval is concave up is the interval is negative then the interval is concave down
Ex. 1) Determine the open intervals on which the graph f(x) = -x^3 + 6x^2 – 9x – 1 is concave up or down.
f’(x) = -3x^2 + 12x – 9
f’’(x) = -6x + 12
-6x + 12 = 0
-6x = -12
x = 2
It is differentiable.
(-infinity, 2) u (2, infinity)
f’’(1) = positive #
f’’(3) = negative #
concave up: (-infinity, 2)
concave down: (2, infinity)
*Note that the question does not ask for the point of inflection, but if it did, you would say: x = 2 is a point of inflection.
9. linearization ((have not covered))
10. infinite limits
id the degree of the top is greater than the degree of the bottom the limit is infinity. If the degree of the top is less than the degree of the bottom the limit is zero is the degree of the top is equal to the degree of the bottom the limit is the coefficients of the terms with the highest exponents
ex:
2x/3x therefore 2/3
11.vertical asymptotes
factor the top and the bottom of the fraction make cancellations if possible set what is left in the bottom equal to zero and solve for x
vertical asymptotes are given in the form of x= a number
x^2+2x-3/x^2 -5x -6
x2 – 5x – 6 = 0
(x – 6)(x + 1) = 0
x = 6 or –1
12. removables
factor the top and the bottom of the fraction make cancellations set each thing that cancelled equal to zero and solve for x removables are given in the form of x= a number
13. horizontal asymptotes
take the limit as x goes to infinty of the equation
if you get a number then your answer is y= a number
if you get infinity there are no horizontal asymptotes
14. area if only one equation is given (not covered)
15. area between two equations (not covered)
16. volume by disks (not covered)
17. volume by washers (not covered)
Blog #20
This week, besides the AP review, we learned about Definite Integrals, the Fundamental Theorem of Calculus, and the Properties of Integrals.
But, I’m just going to go over the Fundamental Theorem of Calculus and the Integral Properties.
*The integral symbol will be represented by “S” and the bounds will be represent as “1S3”.
Firstly, the Fundamental Theorem of Calculus:
This theorem is very simple; it just states that “bSa f(x) dx = F(b) – F(a) = area where F(x) is the integrated function”.
Examples: Find the area of the shape between the boundaries given.
Ex. 1) 4S1 6 dx
F(x) = 6x
Area = 6(4) – 6(1) = 24 – 6
= 18
Ex. 2) 2S0 (x + 1) dx
F(x) = ½ x² + x
Area = (½ (2)² + (2)) – (½ (0)² + (0)) = 4 – 0
= 4
Ex. 3) 3S3 2x dx
= 0
**Whenever the upper and lower bounds are the same, the area is always 0.
Secondly, Integral Properties:
1. bSa f(x) dx = - (aSb) f(x) dx
2. Integrals can be broken up into many smaller integrals.
Ex. 7S0 f(x) dx = 3S0 f(x) dx + 5S3 f(x) dx + 7S5 f(x) dx
3. S k f(x) dx = k S f(x) dx where k stands for a constant.
Examples: Rewrite.
Ex. 4) 2S4 2x dx <-- Property 1
= - 4S2 2x dx
Ex. 5) 2S-1 x dx <-- Property 2
= 0S-1 x dx + 1S0 x dx + 2S1 x dx
Ex. 6) 7 6S3 2x dx <-- Property 3
= 6S3 14x dx
But, I’m just going to go over the Fundamental Theorem of Calculus and the Integral Properties.
*The integral symbol will be represented by “S” and the bounds will be represent as “1S3”.
Firstly, the Fundamental Theorem of Calculus:
This theorem is very simple; it just states that “bSa f(x) dx = F(b) – F(a) = area where F(x) is the integrated function”.
Examples: Find the area of the shape between the boundaries given.
Ex. 1) 4S1 6 dx
F(x) = 6x
Area = 6(4) – 6(1) = 24 – 6
= 18
Ex. 2) 2S0 (x + 1) dx
F(x) = ½ x² + x
Area = (½ (2)² + (2)) – (½ (0)² + (0)) = 4 – 0
= 4
Ex. 3) 3S3 2x dx
= 0
**Whenever the upper and lower bounds are the same, the area is always 0.
Secondly, Integral Properties:
1. bSa f(x) dx = - (aSb) f(x) dx
2. Integrals can be broken up into many smaller integrals.
Ex. 7S0 f(x) dx = 3S0 f(x) dx + 5S3 f(x) dx + 7S5 f(x) dx
3. S k f(x) dx = k S f(x) dx where k stands for a constant.
Examples: Rewrite.
Ex. 4) 2S4 2x dx <-- Property 1
= - 4S2 2x dx
Ex. 5) 2S-1 x dx <-- Property 2
= 0S-1 x dx + 1S0 x dx + 2S1 x dx
Ex. 6) 7 6S3 2x dx <-- Property 3
= 6S3 14x dx
1/9/11
So this week we learned about definite integrals, I thinkkk this is somewheres in chapter four. Surprisingly it is very simple! Sooo to start off a definite integral is an integral with upper and lower bounds that represents an area between the graph and x-axis, bounded by numbers. So say there's an integral and the number at the top is b and the bottom is a. A would be the lower bound, or the smaller number. B would be the upper bound, or the larger number. **Note that it has to be this way to be correct, if it is the opposite and the larger number is at the bottom then you have to switch them by making the function negative, then switching the numbers.**Also, if both of the numbers are the same then the function will automatically be zero.
Example:
Sketch the region whose area is given by the intergral
Integral with 5 on top and 1 on the bottom x^2 dx
Now to start this off you would draw a graph, the x-axis would go to five, and the graph itself would be a parabola because in the function there is an x^2. After drawing the parabola you would shade from x=5 to till the x-axis reaches the parabola.
We also learned the fundamental theorem of calculus. Which is if theres an intergral with b on the top and a on the bottom, f(x)dx=f(b)=f(a)=area where f(x) is the integrated function.
It sounds complicating but here's an example.
Example:
Integral with 3 on the top and 1 on the bottom 4dx
Evaluate: First figure out what the derivative was to get 4, it would be 4x |^3 <-top number of integral. Then plug into the formula f(b)-f(a), so 4(3)-4(1)=8. Then graph, so make the x-axis up to 3, and the y-axis up to 4, then shade.
The only thing that I really find pretty confusing is all the AP stuff, I tried doing that binder thing and I don't really know what i'm doing. I guess it will be easier to understand once we go over it in class.
Example:
Sketch the region whose area is given by the intergral
Integral with 5 on top and 1 on the bottom x^2 dx
Now to start this off you would draw a graph, the x-axis would go to five, and the graph itself would be a parabola because in the function there is an x^2. After drawing the parabola you would shade from x=5 to till the x-axis reaches the parabola.
We also learned the fundamental theorem of calculus. Which is if theres an intergral with b on the top and a on the bottom, f(x)dx=f(b)=f(a)=area where f(x) is the integrated function.
It sounds complicating but here's an example.
Example:
Integral with 3 on the top and 1 on the bottom 4dx
Evaluate: First figure out what the derivative was to get 4, it would be 4x |^3 <-top number of integral. Then plug into the formula f(b)-f(a), so 4(3)-4(1)=8. Then graph, so make the x-axis up to 3, and the y-axis up to 4, then shade.
The only thing that I really find pretty confusing is all the AP stuff, I tried doing that binder thing and I don't really know what i'm doing. I guess it will be easier to understand once we go over it in class.
Saturday, January 8, 2011
1/8/11
Alright well this week we discussed Definite Integrals and then the Fundamental Theorem of Calculus and also the Properties of Integrals. Starting with the definite integrals-that means the integral has upper and lower bounds (where the upper bound is the larger # and the lower bound is the smaller #) and it represents an AREA between the graph and the x-axis. For problems like this they'll typically give you a picture to look at where there is a specific shaded area (on a coordinate plane of course) and they're going to ask you to write an integral based off of that graph. Well first, you write the integral symbol (the big S thing) and your bounds are going to be what's shaded on the graph. Then you put the equation of the graph after the "S" followed by "dx" to complete your integral. OR for definite integrals they can do it the other way around and give you the integral function and they'll ask you to sketch the graph based off of that. *Now the Fundamental Theorem of Calculus finds the actual AREA of the shaded region of whatever graph you have of an integral. You simply integrate the function you are given and then plug the bound #'s into this formula: F(b)-F(a)..and that gives you the area.
Here are a few simple examples: (obviously not involving any graphs because there's no way of doing that on here haha)
(*whenever I put numbers like 2-4 or something like that, that means it's the bounds for the integral going from lower bound to upper bound..and I'll just put whatever the function is in parenthesis so it's easier to distinguish..)
Ex. 1) S2-4 (x) dx
*Okay ha let me explain what this function actually is first because it looks very confusing. It's the integral from 2 to 4 (lower to upper bound) of x
*So first you need to take the integral of x which would be 1/2x^2
*Then after that you can but a symbol that looks like this: l 2-4 (like a vertical bar) and that lets you know that you've integrated the function, you just need to plug into the formula to get the area
*So when you plug in you get:
1/2(2)^2 - 1/2(4)^2
=2-8
=-6
Ex. 2) S2-4 (x-9) dx
*Again, this means it's the integral with bounds from 2 to 4 of the function (x-9)
*So first, take the integral of x-9 and you should end up with 1/2x^2-9x l2-4
*Now all you do is plug into the formula like this:
1/2(4)^2-9(4)-[1/2(2)^2-9/2)]
=8-36-(2-18)
=-28+16
=-12
**I think I pretty much understood all of what we went over this week. I could use more practice on it though..Oh, and the AP stuff that we're about to do sounds VERY intimidating so I'm kinda scared for that haha..just throwing that out there.
Here are a few simple examples: (obviously not involving any graphs because there's no way of doing that on here haha)
(*whenever I put numbers like 2-4 or something like that, that means it's the bounds for the integral going from lower bound to upper bound..and I'll just put whatever the function is in parenthesis so it's easier to distinguish..)
Ex. 1) S2-4 (x) dx
*Okay ha let me explain what this function actually is first because it looks very confusing. It's the integral from 2 to 4 (lower to upper bound) of x
*So first you need to take the integral of x which would be 1/2x^2
*Then after that you can but a symbol that looks like this: l 2-4 (like a vertical bar) and that lets you know that you've integrated the function, you just need to plug into the formula to get the area
*So when you plug in you get:
1/2(2)^2 - 1/2(4)^2
=2-8
=-6
Ex. 2) S2-4 (x-9) dx
*Again, this means it's the integral with bounds from 2 to 4 of the function (x-9)
*So first, take the integral of x-9 and you should end up with 1/2x^2-9x l2-4
*Now all you do is plug into the formula like this:
1/2(4)^2-9(4)-[1/2(2)^2-9/2)]
=8-36-(2-18)
=-28+16
=-12
**I think I pretty much understood all of what we went over this week. I could use more practice on it though..Oh, and the AP stuff that we're about to do sounds VERY intimidating so I'm kinda scared for that haha..just throwing that out there.
1/8/11
So this week we got back into a little new stuff, which was kind of easy. We learned how to integrate properly and find the area of a section on a graph. For this blog, I'm going to review about natural logs and some integration.
The derivative of a natural log is 1/f(x) x the derivative of the bottom
ex: lnx^5 = 1/x^5 x 5x4 = 5x^4/x5 = 5/x
ex: ln2x = 1/2x x 2 = 2/2x = 1/x
ex: lnx^3-2x+3 = 1/x^3-2x+3 x 3x^2-2 = 3x^2-2/x^3-2x+3
Integration, in simpler terms, is taking the "backwards derivative" wherein you're given f'(x) and you're looking for f(x).
ex: f'(x) = x^2
add one to the exponent then multiply x by its reciprocal and add the variable C, because we are not sure if there is a constant in the original function.
solution: f(x)=1/3x^3 + C
ex: f'(x)=x^2 + 4x + 3
1/3/x^3 + 2x^2 + 3x + C
The derivative of a natural log is 1/f(x) x the derivative of the bottom
ex: lnx^5 = 1/x^5 x 5x4 = 5x^4/x5 = 5/x
ex: ln2x = 1/2x x 2 = 2/2x = 1/x
ex: lnx^3-2x+3 = 1/x^3-2x+3 x 3x^2-2 = 3x^2-2/x^3-2x+3
Integration, in simpler terms, is taking the "backwards derivative" wherein you're given f'(x) and you're looking for f(x).
ex: f'(x) = x^2
add one to the exponent then multiply x by its reciprocal and add the variable C, because we are not sure if there is a constant in the original function.
solution: f(x)=1/3x^3 + C
ex: f'(x)=x^2 + 4x + 3
1/3/x^3 + 2x^2 + 3x + C
chase's blog
lets talk about differentiability and stuffs
how to tell if something isn't differentiable:
1. theres a curve - when absolute values or piecewises are in the function you are deriving; |4-x|: set the inside equal to 0 and solve for x. |4-x| is not differentiable at x = 4
2. it is discontinuous - if it ain't continuous, it ain't differentiable. period.
3. theres a cusp - a cusp occurs when you have x^4/5 for example; or x^even #/odd #
4. a vertical asymptote - a vertical asymptote occurs when you have x^3/5 for example; or x^odd #/odd #
how to tell if something isn't differentiable:
1. theres a curve - when absolute values or piecewises are in the function you are deriving; |4-x|: set the inside equal to 0 and solve for x. |4-x| is not differentiable at x = 4
2. it is discontinuous - if it ain't continuous, it ain't differentiable. period.
3. theres a cusp - a cusp occurs when you have x^4/5 for example; or x^even #/odd #
4. a vertical asymptote - a vertical asymptote occurs when you have x^3/5 for example; or x^odd #/odd #
Monday, January 3, 2011
taylor holiday blog #4
Chapter 5 section one-
covers taking the derivative of a natural log.
For completing this task there is a simple formula to follow.
this formula is
D/dx ln(u)= 1/u X (du/dx)
This formula is read as the derivative of natural log u is found by putting one over what is in parenthesis multiplied by the derivative of what had been in parenthesis.
Therefore you would plug into the formula
Simplify
and
solve to get your result.
DO NOT FORGET THAT YOUR DERIVATIVE RULES SUCH AS THE CHAIN RULE THE PRODUCT RULE AND THE QUOTIENT RULE STILL APPLY.
Some examples of chapter five section one include:
'd/dx ln(2x) = 1/2x (2)
d/dx ln(2x)= 2/2x
Therefore ln(2x)= 1/x
d/dc ln (x^2+1)= 1/x^2+1 (2x)
Therefore ln(x^2+1)= 2x/x^2+1
d/dx x ln (x) = x(1/x (1)) + lnx (1)
d/dx x ln (x)= x(1/x)+ln x
d/dx x ln (x)= x/x+lnx
Therefore d/dx x ln (x)= 1 + lnx
d/dx (ln(x))^3= 3(ln x)^2 (1/x) (1)
d/dx (ln(x))^3= 3(ln x)^2 (1/x)
Therefore d/dx (ln(x))^3= 3(ln x)^2/ x
Chapter five section two
covers taking the derivative of an ecponential function.
For completing this task there is also a simple formula to follow.
the formula for solving these problems is
D/dx e^u = e^u X (du/dx)
This formula is read as the derivative of the exponential function e is found by recopying what the problem is asking to take the derivative of and multiplying it by the derivative of the exponent.
Therefore when trying to solve these problem too you would plug into the formula
Simplify
and
solve to get your result.
AGAIN!
DO NOT FORGET THAT YOUR DERIVATIVE RULES SUCH AS THE CHAIN RULE THE PRODUCT RULE AND THE QUOTIENT RULE STILL APPLY.
Some examples of these type problems include
d/dx e^2x-1 = e^2x-1 (2)
Therefore d/dx e^2x-1 = 2e^2x-1
d/dx e^-3/x= e^-3/x (-3x^-1)
d/dx e^-3/x= e^-3/x (3x^-2)
d/dx e^-3/x= e^-3/x (3/x^2)
Therefore d/dx e^-3/x= 3e^-3/x / x^2
Chapter five section four
covered taking the derivative for exponents other than e
For completing this task there are a few simple formulas to follow.
one of the formula for solving these problems is
D/dx A^u = A^u(ln (a))(du/dx)
This formula is read as the derivative of a number raised to a variable as a power is found by multiplying what the problem is asking to take the derivative of by ln of the number being raised to the variable and then multiplying all of that by the derivative of the exponent.
Therefore when trying to solve these problem too you would plug into the formula
Simplify
and
solve to get your result.
Some examples of these type problems include
d/dx 2^x= 2^xln2 (1)
Therefore the derivative of 2^x is 2^x ln 2
d/dx 2^3x= 2^3x ln 2 (3)
Threfore the derivative of d/dx 2^3x is 3 (2^3xln2)
d/dx 7 ^x^2= 7^x^2ln7 (2x)
Therefore the derivative of d/dx 7 ^x^2 is 2x(7^x^2ln7)
The other formula for solving these problems is
d/dx logaU= 1/U(ln(a)) X (du/dx)
This formula is read as the derivative of a log is found by placing what the problem is asking for in asking for the derivative of the log under one then multiplying it by the derivative of what the problem is asking for in asking for the derivative of the log.
Therefore when trying to solve these problem too you would plug into the formula
Simplify
and
solve to get your result.
Some examples of these type problems include
d/dx log cosx = 1/cosx(ln10) (-sin x)
d/dx log cosx = -sinx/ cosx(ln10)
Therefore the derivative of log cos x= -tanx/ ln10
d/dx Log5 x^3+4x= 1/x^3+4x(ln5) (3x^2+4)
Therefore d/dx Log5 x^3+4x= 3x^2+4/ (x^3+4x)(ln5)
STEPS TO USING THE SECOND DERIVATIVE TEST:
Step #1:
take the first derivative set equal to zero and solve. ((keep in mind that this step will not always be necessary but can always be performed anyway. This step is not necessary when one looking for concavity or points of inflection.))
Step #2:
take the second derivative set = 0 and solve
Step #3:
Check differentiability
Step #4:
set up intervals
Step #5:
pick numbers on the interval and plug them into the SECOND derivative
Step #6:
if POSITIVE: the function is concave up on that interval
if NEGATIVE: the function is concave down on that interval
Step #7:
If there is a change it is a point of inflection ((This step only needs to be done when the problem is asking for the points of inflection specifically))
Example #1:
Determine the open intervals on which the graph is concave up or concave down:
F(x)= 6/ x^2+3
Step #1:
take the first derivative set equal to zero and solve. (( But this step is not necessary because the problem is looking for concavity.))
Step #2:
take the second derivative set = 0 and solve
(x^2+3)(0)-[(6)(2x)]/ (x^2+3)^2
F'(x)= -12x/ (x^2+3)^2
(x^2+3)^2(-12)-[(-12x)(2(x^2+3)(2x))]/ (x^2+3)^2
(x^2+3)(-12)+48x^2
-12x^2-36+48x^2
F"(x)= -36x^2-36=0
-36x^2=36
x^2= 1
x= +/- 1
Step #3:
Check differentiability
DIFFERENTIABLE?
yes.
Step #4:
set up intervals
(-infinity,-1)U(-1,1)U(1, infinity)
Step #5:
pick numbers on the interval and plug them into the SECOND derivative
F"(-2)= 36(-2)^2-36/ (-2^2+3)^2 = POSITIVE
F"(0)= 36(0)^2-36/ (0^2+3)^2 = NEGATIVE
F"(2)= 36(2)^2-36/ (2^2+3)^2= POSITIVE
Step #6:
if POSITIVE: the function is concave up on that interval
Therefore the function is concave up on the intervals (-infinity, -1)U(1,Infinity)
if NEGATIVE: the function is concave down on that interval
Therefore the function is concave down on the interval (-1,1)
Step #7:
If there is a change it is a point of inflection.
This step only needs to be done when the problem is asking for the points of inflection specifically therefore this step is not necessary for this problem because it is not asking for the points of inflection.
THE SHORTCUT TO THE FIRST DERIVATIVE TEST
Steps to the shortcut for the first derivative test:
Step #1:
take the derivative, set equal to zero, and solve
Step #2:
Check to see if differentiable
Step #3:
Plug in critical points into the second derivative
Step #4
If POSITIVE there is a min
If NEGATIVE there is a max
If 0 the test is false
(((NOTICE THAT FOR THE SHORTCUT WHETHER THE PLUG IN IS POSITIVE OR NEGATIVE HAS A REVERSED MEANING)))
Example:
Step #1:
take the derivative, set equal to zero, and solve
-15x^4+15x^2=0
15x^2(-x^2+1)
x= 0, 1,-1
Step #2:
Check to see if differentiable
differentiable?
yes.
Step #3:
Plug in critical points into the second derivative
F"(X)=60x^3 + 30x
-60(0)^3+30(0)=0
-60(-1)^3+30(-1)=POSITIVE
-60(1)^3+30(1)=NEGATIVE
Step #4
If POSITIVE there is a min
Therefore, at x= -1 there is a min
If NEGATIVE there is a max
Therefore, at x= 1 there is a max
If 0 the test is false
Therefore, at x=0 the test fails
Horizontal asymptotes and curve sketching
To find a horizontal asymptote there are three rules to follow
these rules apply to the degree of the leading coefficients of the polynomials in both the numerator and denominator
These rules are:
* If the top degree of the leading coefficient is equal to the bottom degree of the leading coefficient then you take the coefficient of the highest degree in the numerator and put it over the coefficient of the highest degree in the denominator
Example:
What is the horizontal asymptote of
2x^2+5 / 3x^2+1
because the degrees of the leading coefficients are equivalent there is a horizontal asymptote at 2/3
*If the top degree of the leading coefficient is greater than bottom degree of the leading coefficient then there is a horizontal asymptote at - infinity , infinity. In other words there is no horizontal asymptote.
Example:
What is the horizontal asymptote of
2x^2+5 / 3x+1
because the degree of the leading coefficient in the top is greater than the leading coefficient in the bottom the result is
-infinity, infinity and therefore there are no horizontal asymptotes.
*If the top degree of the leading coefficient is less than bottom degree of the leading coefficient then there is a horizontal asymptote at 0.
Example:
What is the horizontal asymptote of
2x+5 / 3x^2+1
because the degree of the leading coefficient in the top is less than the leading coefficient in the bottom the horizontal asymptote is 0.
For curve sketching problems there are five steps to follow
these steps are:
#1
find the domain and find the range
The domain is found by setting the bottom equal to zero, solving for x, and setting up intervals
The range is found by finding the horizontal asymptotes and setting up intervals
#2
Find the x intercepts, y intercepts, vertical asymptotes, and horizontal asymptotes.
X intercepts are found by setting the original equal to zero and solving for x to get the point (result,0)
Y intercepts are found by plugging in 0 for x and solving to get the point (0,result)
Vertical asymptotes and horizontal asymptotes have already been found at this point
your vertical asymptote is the x= from when you found the domain before you set up your intervals
your horizontal asymptote is the y= from when you found the range before you set up your intervals
#3
Do the first derivative test by
Taking the derivative and setting it equal to zero
Solve for x
check differentiability
Set up intervals (-infinity, pt)U(pt, infinity)
Plug in value on the interval into the derivative and solve
((if you get a positive number the function is increasing, If you get a negative number the function is decreasing))
(( A max is increasing then decreasing, A min is decreasing then increasing))
#4
Do second derivative test
take the derivative of the first derivative
set it equal to zero
solve for x
check differentiability
set up intervals (-infinity, pt)U(pt,infinity)
plug into the second derivative and solve
((if you get a positive number the function is concave up, If you get a negative number the function is concave down))
((any point on the interval where a number switches from concave up to concave down or vise versa there is a point of inflection))
#5
Plot all important information on a sketched graph.
IMPORTANT INFORMATION INCLUDES:
domain intervals
range intervals
x intercepts
y intercepts
vertical asymptotes
horizontal asymptotes
whether the intervals of the first derivative test are increasing or decreasing
where there is a max or a min
whether the intervals of the second derivative test are concave up or concave down
Example:
y= 2(x^2-9)/x^2-4
#1
find the domain and find the range
The domain:
x^2-4=0
x^2=4
x = +/- 2
(-infinity,-2)(-2,2)(2,infinity)
The range is found by finding the horizontal asymptotes and setting up intervals
y= 2
(-infinity,2)(2,infinity)
#2
X intercepts are found by setting the original equal to zero and solving for x to get the point (result,0)
2(x^2-9)=0 therefore: x=+/-3
Y intercepts are found by plugging in 0 for x and solving to get the point (0,result)
2(0^2-9)/0^2-4
2(-9)/-4
-18/-4
9/2
(0,9/2)
your vertical asymptote is the x= from when you found the domain before you set up your intervals
your horizontal asymptote is the y= from when you found the range before you set up your intervals
#3
Do the first derivative test by
Taking the derivative and setting it equal to zero
x^2-4(4x)-[2x^2-18(2x)]/(2^2-4)^2
20x/(x^2-4)^2
Solve for x
20x=0
x=0
check differentiability
vertical asymptotes at +/- 2
Set up intervals (-infinity, pt)U(pt, infinity)
(-infinity,-2)(-2,0)(0,2)(2,infinity)
Plug in value on the interval into the derivative and solve
f(-3) negative therefore decreasing
f(-1) negative therefore decreasing
f(1) positive therefore increasing
f(3) positive therefore increasing
MIN @ x=0
((if you get a positive number the function is increasing, If you get a negative number the function is decreasing))
(( A max is increasing then decreasing, A min is decreasing then increasing))
#4
Do second derivative test
take the derivative of the first derivative
-20(3x^2+4)/(x^2-4)^3
set it equal to zero
-20(3x^2+14)=0
solve for x
x=sqrt -4/3 therefore does not apply
check differentiability
vertical asymptotes at +/-2
set up intervals (-infinity, pt)U(pt,infinity)
(-infinity, -2)(-2,2)(2, infinity)
plug into the second derivative and solve
f(-3) negative concave down
f(0) positive concave up
f(3) negative concave down
THE STEPS TO OPTIMIZATION: (are also listed on page 219 of our text book)
#1
list what is given to you in the problem
#2
list the primary equation
#3
List the secondary equation
#4
Solve for y in the secondary equation
#5
plug y= into the primary equation for y
#6
set = 0 and solve
#7
perform the first derivative test
meaning:
take the derivative
set = 0
solve for x
set up intervals
plug in for F(x) between intervals into the derivative
determine if result is positive or negative and thereby increasing or decreasing respectivley on the intervals
therefore enabling you to determine whther x= is a max or a min
EXAMPLE:
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. what dimensions will produce a maximum volume?
#1
given: SA=108in^2 SA= x^2+4xy
Primary Equation: V= x^2y
Secondary equation: x^2+4xy=108
4xy=108-x^2
y= 108-x^2/ 4x
Plug in: V= X^2(108-x^2/4x)
V= 108x-x^3/4
27x-1/4x^3
first derivative test:
27-3/4x^2=0
-3/4x^2= -27
x^2=36
x= + - 6
Set up intervals
(0,6) U (6,sqrt108)
F(1) is positive and thereby increasing
F(7) is negative and thereby decreasing
Therefore X=6 is a maximum
Therefore V= LXWXH
V= 6X6XH
6^2 x y = 108
y= 108/ 36
y=3
V= 6X6X3
covers taking the derivative of a natural log.
For completing this task there is a simple formula to follow.
this formula is
D/dx ln(u)= 1/u X (du/dx)
This formula is read as the derivative of natural log u is found by putting one over what is in parenthesis multiplied by the derivative of what had been in parenthesis.
Therefore you would plug into the formula
Simplify
and
solve to get your result.
DO NOT FORGET THAT YOUR DERIVATIVE RULES SUCH AS THE CHAIN RULE THE PRODUCT RULE AND THE QUOTIENT RULE STILL APPLY.
Some examples of chapter five section one include:
'd/dx ln(2x) = 1/2x (2)
d/dx ln(2x)= 2/2x
Therefore ln(2x)= 1/x
d/dc ln (x^2+1)= 1/x^2+1 (2x)
Therefore ln(x^2+1)= 2x/x^2+1
d/dx x ln (x) = x(1/x (1)) + lnx (1)
d/dx x ln (x)= x(1/x)+ln x
d/dx x ln (x)= x/x+lnx
Therefore d/dx x ln (x)= 1 + lnx
d/dx (ln(x))^3= 3(ln x)^2 (1/x) (1)
d/dx (ln(x))^3= 3(ln x)^2 (1/x)
Therefore d/dx (ln(x))^3= 3(ln x)^2/ x
Chapter five section two
covers taking the derivative of an ecponential function.
For completing this task there is also a simple formula to follow.
the formula for solving these problems is
D/dx e^u = e^u X (du/dx)
This formula is read as the derivative of the exponential function e is found by recopying what the problem is asking to take the derivative of and multiplying it by the derivative of the exponent.
Therefore when trying to solve these problem too you would plug into the formula
Simplify
and
solve to get your result.
AGAIN!
DO NOT FORGET THAT YOUR DERIVATIVE RULES SUCH AS THE CHAIN RULE THE PRODUCT RULE AND THE QUOTIENT RULE STILL APPLY.
Some examples of these type problems include
d/dx e^2x-1 = e^2x-1 (2)
Therefore d/dx e^2x-1 = 2e^2x-1
d/dx e^-3/x= e^-3/x (-3x^-1)
d/dx e^-3/x= e^-3/x (3x^-2)
d/dx e^-3/x= e^-3/x (3/x^2)
Therefore d/dx e^-3/x= 3e^-3/x / x^2
Chapter five section four
covered taking the derivative for exponents other than e
For completing this task there are a few simple formulas to follow.
one of the formula for solving these problems is
D/dx A^u = A^u(ln (a))(du/dx)
This formula is read as the derivative of a number raised to a variable as a power is found by multiplying what the problem is asking to take the derivative of by ln of the number being raised to the variable and then multiplying all of that by the derivative of the exponent.
Therefore when trying to solve these problem too you would plug into the formula
Simplify
and
solve to get your result.
Some examples of these type problems include
d/dx 2^x= 2^xln2 (1)
Therefore the derivative of 2^x is 2^x ln 2
d/dx 2^3x= 2^3x ln 2 (3)
Threfore the derivative of d/dx 2^3x is 3 (2^3xln2)
d/dx 7 ^x^2= 7^x^2ln7 (2x)
Therefore the derivative of d/dx 7 ^x^2 is 2x(7^x^2ln7)
The other formula for solving these problems is
d/dx logaU= 1/U(ln(a)) X (du/dx)
This formula is read as the derivative of a log is found by placing what the problem is asking for in asking for the derivative of the log under one then multiplying it by the derivative of what the problem is asking for in asking for the derivative of the log.
Therefore when trying to solve these problem too you would plug into the formula
Simplify
and
solve to get your result.
Some examples of these type problems include
d/dx log cosx = 1/cosx(ln10) (-sin x)
d/dx log cosx = -sinx/ cosx(ln10)
Therefore the derivative of log cos x= -tanx/ ln10
d/dx Log5 x^3+4x= 1/x^3+4x(ln5) (3x^2+4)
Therefore d/dx Log5 x^3+4x= 3x^2+4/ (x^3+4x)(ln5)
STEPS TO USING THE SECOND DERIVATIVE TEST:
Step #1:
take the first derivative set equal to zero and solve. ((keep in mind that this step will not always be necessary but can always be performed anyway. This step is not necessary when one looking for concavity or points of inflection.))
Step #2:
take the second derivative set = 0 and solve
Step #3:
Check differentiability
Step #4:
set up intervals
Step #5:
pick numbers on the interval and plug them into the SECOND derivative
Step #6:
if POSITIVE: the function is concave up on that interval
if NEGATIVE: the function is concave down on that interval
Step #7:
If there is a change it is a point of inflection ((This step only needs to be done when the problem is asking for the points of inflection specifically))
Example #1:
Determine the open intervals on which the graph is concave up or concave down:
F(x)= 6/ x^2+3
Step #1:
take the first derivative set equal to zero and solve. (( But this step is not necessary because the problem is looking for concavity.))
Step #2:
take the second derivative set = 0 and solve
(x^2+3)(0)-[(6)(2x)]/ (x^2+3)^2
F'(x)= -12x/ (x^2+3)^2
(x^2+3)^2(-12)-[(-12x)(2(x^2+3)(2x))]/ (x^2+3)^2
(x^2+3)(-12)+48x^2
-12x^2-36+48x^2
F"(x)= -36x^2-36=0
-36x^2=36
x^2= 1
x= +/- 1
Step #3:
Check differentiability
DIFFERENTIABLE?
yes.
Step #4:
set up intervals
(-infinity,-1)U(-1,1)U(1, infinity)
Step #5:
pick numbers on the interval and plug them into the SECOND derivative
F"(-2)= 36(-2)^2-36/ (-2^2+3)^2 = POSITIVE
F"(0)= 36(0)^2-36/ (0^2+3)^2 = NEGATIVE
F"(2)= 36(2)^2-36/ (2^2+3)^2= POSITIVE
Step #6:
if POSITIVE: the function is concave up on that interval
Therefore the function is concave up on the intervals (-infinity, -1)U(1,Infinity)
if NEGATIVE: the function is concave down on that interval
Therefore the function is concave down on the interval (-1,1)
Step #7:
If there is a change it is a point of inflection.
This step only needs to be done when the problem is asking for the points of inflection specifically therefore this step is not necessary for this problem because it is not asking for the points of inflection.
THE SHORTCUT TO THE FIRST DERIVATIVE TEST
Steps to the shortcut for the first derivative test:
Step #1:
take the derivative, set equal to zero, and solve
Step #2:
Check to see if differentiable
Step #3:
Plug in critical points into the second derivative
Step #4
If POSITIVE there is a min
If NEGATIVE there is a max
If 0 the test is false
(((NOTICE THAT FOR THE SHORTCUT WHETHER THE PLUG IN IS POSITIVE OR NEGATIVE HAS A REVERSED MEANING)))
Example:
Step #1:
take the derivative, set equal to zero, and solve
-15x^4+15x^2=0
15x^2(-x^2+1)
x= 0, 1,-1
Step #2:
Check to see if differentiable
differentiable?
yes.
Step #3:
Plug in critical points into the second derivative
F"(X)=60x^3 + 30x
-60(0)^3+30(0)=0
-60(-1)^3+30(-1)=POSITIVE
-60(1)^3+30(1)=NEGATIVE
Step #4
If POSITIVE there is a min
Therefore, at x= -1 there is a min
If NEGATIVE there is a max
Therefore, at x= 1 there is a max
If 0 the test is false
Therefore, at x=0 the test fails
Horizontal asymptotes and curve sketching
To find a horizontal asymptote there are three rules to follow
these rules apply to the degree of the leading coefficients of the polynomials in both the numerator and denominator
These rules are:
* If the top degree of the leading coefficient is equal to the bottom degree of the leading coefficient then you take the coefficient of the highest degree in the numerator and put it over the coefficient of the highest degree in the denominator
Example:
What is the horizontal asymptote of
2x^2+5 / 3x^2+1
because the degrees of the leading coefficients are equivalent there is a horizontal asymptote at 2/3
*If the top degree of the leading coefficient is greater than bottom degree of the leading coefficient then there is a horizontal asymptote at - infinity , infinity. In other words there is no horizontal asymptote.
Example:
What is the horizontal asymptote of
2x^2+5 / 3x+1
because the degree of the leading coefficient in the top is greater than the leading coefficient in the bottom the result is
-infinity, infinity and therefore there are no horizontal asymptotes.
*If the top degree of the leading coefficient is less than bottom degree of the leading coefficient then there is a horizontal asymptote at 0.
Example:
What is the horizontal asymptote of
2x+5 / 3x^2+1
because the degree of the leading coefficient in the top is less than the leading coefficient in the bottom the horizontal asymptote is 0.
For curve sketching problems there are five steps to follow
these steps are:
#1
find the domain and find the range
The domain is found by setting the bottom equal to zero, solving for x, and setting up intervals
The range is found by finding the horizontal asymptotes and setting up intervals
#2
Find the x intercepts, y intercepts, vertical asymptotes, and horizontal asymptotes.
X intercepts are found by setting the original equal to zero and solving for x to get the point (result,0)
Y intercepts are found by plugging in 0 for x and solving to get the point (0,result)
Vertical asymptotes and horizontal asymptotes have already been found at this point
your vertical asymptote is the x= from when you found the domain before you set up your intervals
your horizontal asymptote is the y= from when you found the range before you set up your intervals
#3
Do the first derivative test by
Taking the derivative and setting it equal to zero
Solve for x
check differentiability
Set up intervals (-infinity, pt)U(pt, infinity)
Plug in value on the interval into the derivative and solve
((if you get a positive number the function is increasing, If you get a negative number the function is decreasing))
(( A max is increasing then decreasing, A min is decreasing then increasing))
#4
Do second derivative test
take the derivative of the first derivative
set it equal to zero
solve for x
check differentiability
set up intervals (-infinity, pt)U(pt,infinity)
plug into the second derivative and solve
((if you get a positive number the function is concave up, If you get a negative number the function is concave down))
((any point on the interval where a number switches from concave up to concave down or vise versa there is a point of inflection))
#5
Plot all important information on a sketched graph.
IMPORTANT INFORMATION INCLUDES:
domain intervals
range intervals
x intercepts
y intercepts
vertical asymptotes
horizontal asymptotes
whether the intervals of the first derivative test are increasing or decreasing
where there is a max or a min
whether the intervals of the second derivative test are concave up or concave down
Example:
y= 2(x^2-9)/x^2-4
#1
find the domain and find the range
The domain:
x^2-4=0
x^2=4
x = +/- 2
(-infinity,-2)(-2,2)(2,infinity)
The range is found by finding the horizontal asymptotes and setting up intervals
y= 2
(-infinity,2)(2,infinity)
#2
X intercepts are found by setting the original equal to zero and solving for x to get the point (result,0)
2(x^2-9)=0 therefore: x=+/-3
Y intercepts are found by plugging in 0 for x and solving to get the point (0,result)
2(0^2-9)/0^2-4
2(-9)/-4
-18/-4
9/2
(0,9/2)
your vertical asymptote is the x= from when you found the domain before you set up your intervals
your horizontal asymptote is the y= from when you found the range before you set up your intervals
#3
Do the first derivative test by
Taking the derivative and setting it equal to zero
x^2-4(4x)-[2x^2-18(2x)]/(2^2-4)^2
20x/(x^2-4)^2
Solve for x
20x=0
x=0
check differentiability
vertical asymptotes at +/- 2
Set up intervals (-infinity, pt)U(pt, infinity)
(-infinity,-2)(-2,0)(0,2)(2,infinity)
Plug in value on the interval into the derivative and solve
f(-3) negative therefore decreasing
f(-1) negative therefore decreasing
f(1) positive therefore increasing
f(3) positive therefore increasing
MIN @ x=0
((if you get a positive number the function is increasing, If you get a negative number the function is decreasing))
(( A max is increasing then decreasing, A min is decreasing then increasing))
#4
Do second derivative test
take the derivative of the first derivative
-20(3x^2+4)/(x^2-4)^3
set it equal to zero
-20(3x^2+14)=0
solve for x
x=sqrt -4/3 therefore does not apply
check differentiability
vertical asymptotes at +/-2
set up intervals (-infinity, pt)U(pt,infinity)
(-infinity, -2)(-2,2)(2, infinity)
plug into the second derivative and solve
f(-3) negative concave down
f(0) positive concave up
f(3) negative concave down
THE STEPS TO OPTIMIZATION: (are also listed on page 219 of our text book)
#1
list what is given to you in the problem
#2
list the primary equation
#3
List the secondary equation
#4
Solve for y in the secondary equation
#5
plug y= into the primary equation for y
#6
set = 0 and solve
#7
perform the first derivative test
meaning:
take the derivative
set = 0
solve for x
set up intervals
plug in for F(x) between intervals into the derivative
determine if result is positive or negative and thereby increasing or decreasing respectivley on the intervals
therefore enabling you to determine whther x= is a max or a min
EXAMPLE:
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. what dimensions will produce a maximum volume?
#1
given: SA=108in^2 SA= x^2+4xy
Primary Equation: V= x^2y
Secondary equation: x^2+4xy=108
4xy=108-x^2
y= 108-x^2/ 4x
Plug in: V= X^2(108-x^2/4x)
V= 108x-x^3/4
27x-1/4x^3
first derivative test:
27-3/4x^2=0
-3/4x^2= -27
x^2=36
x= + - 6
Set up intervals
(0,6) U (6,sqrt108)
F(1) is positive and thereby increasing
F(7) is negative and thereby decreasing
Therefore X=6 is a maximum
Therefore V= LXWXH
V= 6X6XH
6^2 x y = 108
y= 108/ 36
y=3
V= 6X6X3
Devin's Blog
A function is not differentiable if:
1. There is a corner
a. Absolute value or a piecewise
2. Vertical tangent line
a. X^odd/odd
3. If it is not continuous
a. Jumps
b. Removables
c. Vertical asymptotes
4. There is a cusp
a. X^even/odd
Examples:
a. x+2/(X^2)
=there is a corner
b. X^5/3/(x)
=there is a vertical tangent line
c. X+1/x+1
=the function is not continuous
d. X^4/3/(x)
=there is a cusp
-If a function is not differentiable at a point you can take a derivative, but you cannot plug in “x”.
Using the steps and skills above you can educationally answer the following questions.
If a function is differentiable, is it continuous?
Yes
If a function is not differentiable, is it continuous?
Sometimes
If something is not differentiable, does it have a tangent line at that point?
No
1. There is a corner
a. Absolute value or a piecewise
2. Vertical tangent line
a. X^odd/odd
3. If it is not continuous
a. Jumps
b. Removables
c. Vertical asymptotes
4. There is a cusp
a. X^even/odd
Examples:
a. x+2/(X^2)
=there is a corner
b. X^5/3/(x)
=there is a vertical tangent line
c. X+1/x+1
=the function is not continuous
d. X^4/3/(x)
=there is a cusp
-If a function is not differentiable at a point you can take a derivative, but you cannot plug in “x”.
Using the steps and skills above you can educationally answer the following questions.
If a function is differentiable, is it continuous?
Yes
If a function is not differentiable, is it continuous?
Sometimes
If something is not differentiable, does it have a tangent line at that point?
No
Sunday, January 2, 2011
holiday blog 3
in this blog, I'll talk about how to adding and subtracting logs
When you are adding logs your numbers are multiplied if both have the same base.
Ex. log5+log7
You can condense the equation by multiplying and getting log35 at which point you simplify.
When subtracting logs you do just like addition except you divide instead of multiplying.
Ex.log35-log7
You condense by dividing and get log5 and then you simplify.
D.A.N.C.E. by Justice
believe me, most people here should like this especially if they watch the video, lol
http://www.youtube.com/watch?v=daMuNH1lcQE
When you are adding logs your numbers are multiplied if both have the same base.
Ex. log5+log7
You can condense the equation by multiplying and getting log35 at which point you simplify.
When subtracting logs you do just like addition except you divide instead of multiplying.
Ex.log35-log7
You condense by dividing and get log5 and then you simplify.
D.A.N.C.E. by Justice
believe me, most people here should like this especially if they watch the video, lol
http://www.youtube.com/watch?v=daMuNH1lcQE
holiday blog 2
This one is gonna be about derivatives.
When taking a derivative, you are given a primary function referred to as f. The derivative, or F', is basically taking the exponent of each term, moving it to the front of each term, multiplying the coefficient of each term by that exponent, and reducing the exponent by 1.
Ex: 4x^2
F'=(2)(4)x^(2-1)
F'=8x
Holy Wars by Megadeath
even if you don't like rock-style music, at least listen to the meaning behind the song and the lyrics
http://www.youtube.com/watch?v=9d4ui9q7eDM
When taking a derivative, you are given a primary function referred to as f. The derivative, or F', is basically taking the exponent of each term, moving it to the front of each term, multiplying the coefficient of each term by that exponent, and reducing the exponent by 1.
Ex: 4x^2
F'=(2)(4)x^(2-1)
F'=8x
Holy Wars by Megadeath
even if you don't like rock-style music, at least listen to the meaning behind the song and the lyrics
http://www.youtube.com/watch?v=9d4ui9q7eDM
holiday blog 1 & my gift to all yall in 3 parts
there's a little twist to my next few blogs
I'm going to include an awesome song that people should listen to at the bottom of the blog, because people need to learn new music, and I actually think people on here would like them
anyway, this blog is gonna be about differentiation
these are ways to figure out if something is not differentiable:
1) if there is a curve
curves occur when absolute values and/or piecewises are found in the function that you are trying to take the derivative of
an example of a function with a curve is:
1) |3-x| you set what's inside of the absolute value equal to 0 and solve for x, so |3-x| is not differentiable at x=3
2) if there is a vertical asymptote
vertical asymptotes occur when x is raised to the (odd/odd)
x^5/9 would be a vertical asymptote, so x would not be differentiable at x=1
3) if it's not continuous
if a function is not continuous, then it's not differentiable, it's as simple as that
4) if there's a cusp
a cusp is when you have x^(even/odd)
x^2/7 would not be differentiable at x=0
first song:
Cthulu Sleeps by Deadmau5
(pronounced dead mouse)
http://www.youtube.com/watch?v=1vtV1ArrHuQ
I'm going to include an awesome song that people should listen to at the bottom of the blog, because people need to learn new music, and I actually think people on here would like them
anyway, this blog is gonna be about differentiation
these are ways to figure out if something is not differentiable:
1) if there is a curve
curves occur when absolute values and/or piecewises are found in the function that you are trying to take the derivative of
an example of a function with a curve is:
1) |3-x| you set what's inside of the absolute value equal to 0 and solve for x, so |3-x| is not differentiable at x=3
2) if there is a vertical asymptote
vertical asymptotes occur when x is raised to the (odd/odd)
x^5/9 would be a vertical asymptote, so x would not be differentiable at x=1
3) if it's not continuous
if a function is not continuous, then it's not differentiable, it's as simple as that
4) if there's a cusp
a cusp is when you have x^(even/odd)
x^2/7 would not be differentiable at x=0
first song:
Cthulu Sleeps by Deadmau5
(pronounced dead mouse)
http://www.youtube.com/watch?v=1vtV1ArrHuQ
tres and done
okay may stupid computer completely erased my last version of this blog and im a bit ticked so this one's gunna be shorter. It was about the rate of change of the area of a square. if the rate of change of the sides of a square is six meters per second while the area of the square is four the rate of change is 24 meters per second squared.
toll free numba tue
derivatives are used to find the slope of a line tangent to the graph of an equation. but sometimes a problem will ask you to take it a step further than that. you may occasionally be asked to find the equation of a line tangent to a graph at a certain point. for example, find the equation of the line tangent to f (x) = x^2 at th point (2,3).
first you must find the derivative of the function in order to determine the slope of the line tangent to your graph. the derivative is 2x. and you also must know that the equation of a line is y-y1 = m(x-x1). then replace y1 and x1 with the points that the problem has given you. you then get the equation y = 2x^2 - 4x +3. tada you now have your answer.
numero one-o
chain rule is always kind of nice, right? well its a magical process that requires nothing but a screwdriver, some type of tablet, a hammer, and many years of math tutelage.
anywhen the way you do it is you take the exponent of a set of numbers that are normally within parentheses and you place it in front of the thing the subtract one like this:
(x+y)^3
f '(x) = 3(x+y)^2
then you multiply the equation by the derivative of whats inside the parentheses which in this particular case is one so this would be the end of the problem.
Chad Blog 3
differentiability is the ability to take a derivative from a function
there are many ways that a function cannot be differentiable
those ways are:
1) if there is a curve
curves occur when absolute values and/or piecewises are found in the function that you are trying to take the derivative of
an example of a function with a curve is:
1) |3-x| you set what's inside of the absolute value equal to 0 and solve for x, so |3-x| is not differentiable at x=3
2) if there is a vertical asymptote
vertical asymptotes occur when x is raised to the (odd/odd)
x^5/9 would be a vertical asymptote, so x would not be differentiable at x=1
3) if it's not continuous
if a function is not continuous, then it's not differentiable, it's as simple as that
4) if there's a cusp
a cusp is when you have x^(even/odd)
x^2/7 would not be differentiable at x=0
there are many ways that a function cannot be differentiable
those ways are:
1) if there is a curve
curves occur when absolute values and/or piecewises are found in the function that you are trying to take the derivative of
an example of a function with a curve is:
1) |3-x| you set what's inside of the absolute value equal to 0 and solve for x, so |3-x| is not differentiable at x=3
2) if there is a vertical asymptote
vertical asymptotes occur when x is raised to the (odd/odd)
x^5/9 would be a vertical asymptote, so x would not be differentiable at x=1
3) if it's not continuous
if a function is not continuous, then it's not differentiable, it's as simple as that
4) if there's a cusp
a cusp is when you have x^(even/odd)
x^2/7 would not be differentiable at x=0
Chad Blog 2
Extreme Values are defined as maximums and minimums
Absolute Max is defined as the highest point on an interval
** notice that absolute max is the highest point on a given interval not the highest point on the graph
Absolute min is defined as the lowest point on an interval
**notice that absolute min is the lowest point of a given interval not the lowest point on the graph
Relative Max (also referred to as local max) is defined as any max NOT on an interval
Relative Min (also referred to a local min) is defined as any min NOT on an interval
Critical Numbers are defined as any max or min typically referred to as x=c
When trying to solve problems concerning a critical number, extreme value, max, min, or horizontal tangents take the derivative and set equal to zero.
Example #1:
Find the critical values of f(x)= 9(x^2-3)/x^3
9x^2-27/x^3
Take the derivative
X^3(18x)-[(9x^2-27)(3x^2)]/ x^6
18x^4-[27x^4-81x^2]/x^6
18x^4-27x^4+81x^2/x^6
-9x^4+81x^2/ x^6
9x^2(-x^2+9)/X^6
9(-x^2+9)/x^4
Set equal to zero and solve
((when setting a fraction equal to zero you are really only setting what is in the numerator equal to zero))
9(-x^2+9)=0
-9x^2+81=0
-9x^2=-81
X^2=9
X=+/- 3
Example #2:
Find the extrema of f(x)= 3x^4-4x^3 on the interval [-1,2]
4(3)x^3-3(4)x^2
12x^3-12x^2=0
12x^2(x-1)=0
X=0,1
To determine max or min on an interval plug into original and the biggest result will be the max
and the smallest result will be the min
f(-1)= 3(-1)^4-4(-1)^3=7
f(0)=3(0)^4-4(0)^3=0
f(1)= 3(1)^4 – 4(1)^3= -1 ((smallest so absolute min))
f(2)= 3(2)^4-4(2)^3= 16 ((largest so absolute max))
Absolute Max is defined as the highest point on an interval
** notice that absolute max is the highest point on a given interval not the highest point on the graph
Absolute min is defined as the lowest point on an interval
**notice that absolute min is the lowest point of a given interval not the lowest point on the graph
Relative Max (also referred to as local max) is defined as any max NOT on an interval
Relative Min (also referred to a local min) is defined as any min NOT on an interval
Critical Numbers are defined as any max or min typically referred to as x=c
When trying to solve problems concerning a critical number, extreme value, max, min, or horizontal tangents take the derivative and set equal to zero.
Example #1:
Find the critical values of f(x)= 9(x^2-3)/x^3
9x^2-27/x^3
Take the derivative
X^3(18x)-[(9x^2-27)(3x^2)]/ x^6
18x^4-[27x^4-81x^2]/x^6
18x^4-27x^4+81x^2/x^6
-9x^4+81x^2/ x^6
9x^2(-x^2+9)/X^6
9(-x^2+9)/x^4
Set equal to zero and solve
((when setting a fraction equal to zero you are really only setting what is in the numerator equal to zero))
9(-x^2+9)=0
-9x^2+81=0
-9x^2=-81
X^2=9
X=+/- 3
Example #2:
Find the extrema of f(x)= 3x^4-4x^3 on the interval [-1,2]
4(3)x^3-3(4)x^2
12x^3-12x^2=0
12x^2(x-1)=0
X=0,1
To determine max or min on an interval plug into original and the biggest result will be the max
and the smallest result will be the min
f(-1)= 3(-1)^4-4(-1)^3=7
f(0)=3(0)^4-4(0)^3=0
f(1)= 3(1)^4 – 4(1)^3= -1 ((smallest so absolute min))
f(2)= 3(2)^4-4(2)^3= 16 ((largest so absolute max))
Chads first blog
chain ruleeee is coolllll unlike a tool that goes to school and swims in a pool. Its basically the same as the power rule just used at different times with one difference. We also learned the general power rule, which is a form of chain rule.
The chain rule is used when you have two things going on. For example:
(x^2+x+3)^2
First step is like normal, you would move the exponent to the front.
2(x^2+x+3)^2
Next, you subtract one from the original exponent to form the new exponent.
2(x^2+x+3)
The next step is to take the derivative of the inside and multiply by it.
2(x^2+x+3)(2x+1)
Then you factor/simplify if you can, but this one is as simplified as it needs to be.
So, that's about it for what we learned this week.
The chain rule is used when you have two things going on. For example:
(x^2+x+3)^2
First step is like normal, you would move the exponent to the front.
2(x^2+x+3)^2
Next, you subtract one from the original exponent to form the new exponent.
2(x^2+x+3)
The next step is to take the derivative of the inside and multiply by it.
2(x^2+x+3)(2x+1)
Then you factor/simplify if you can, but this one is as simplified as it needs to be.
So, that's about it for what we learned this week.
alaina's fourth holiday blog
The second derivative test. It has most of the same steps as the first derivative test, but it also has some of its own.
First, you take the first derivative, set it equal to zero and solve for x.
Then, you take the second derivative, set it equal to zero, and solve for x.
Next, you check for differentiability. If it is not differentiable, the second derivative test fails.
Then, you set up intervals with your x value from the second derivative test, if it works.
Next, you plug values from your intervals into the second derivative.
If the output value is positive, it is concave up. If the output value is negative, it is concave down.
Ex: determine the intervals on which the graphis concave up or concave down.
*Because it is only asking for the intervals on which the graph is concave up or concave down, you do not need to complete the first step (solve the first derivative for x).
1. f'(x)=(-3x^2)+(12x)-9
2. f''(x)=(-6x)+12=0
-6x=-12
x=2
3. yes, it is differentiable
4. (infinity, 2)u (2, infinity)
5. f''(0)=positive, concave up
f''(3)=negative, concave down.
The function is concave up on the interval (-infinity,2) and concave down on the interval (2,infinity).
First, you take the first derivative, set it equal to zero and solve for x.
Then, you take the second derivative, set it equal to zero, and solve for x.
Next, you check for differentiability. If it is not differentiable, the second derivative test fails.
Then, you set up intervals with your x value from the second derivative test, if it works.
Next, you plug values from your intervals into the second derivative.
If the output value is positive, it is concave up. If the output value is negative, it is concave down.
Ex: determine the intervals on which the graphis concave up or concave down.
*Because it is only asking for the intervals on which the graph is concave up or concave down, you do not need to complete the first step (solve the first derivative for x).
1. f'(x)=(-3x^2)+(12x)-9
2. f''(x)=(-6x)+12=0
-6x=-12
x=2
3. yes, it is differentiable
4. (infinity, 2)u (2, infinity)
5. f''(0)=positive, concave up
f''(3)=negative, concave down.
The function is concave up on the interval (-infinity,2) and concave down on the interval (2,infinity).
alaina's third holiday blog
First, i'm going to talk about chapter 3 section 5.
limits at infinity.
-to find a horizontal asymptote, you take the limit as x=>infinity
y=ans is an asymptote
degree of top=degree of bottom -> coefficient of highest exponent
degree of top>degree of bottom ->-infinity or infinity
degree of top0
Ex 1: 2x+5/(3x^2+1)
lim x->infinity 2x+5/(3x^2+1)
lim x->infinity is 0
0 is a horizontal asymptote.
Chapter 3 section 6
Curve sketching
Steps
1. identify the domain and range
2. find the x and y intercepts, identify the vertical and horizontal asymptotes
3. do the first derivative test
4. do the second derivative test
5. plot the information and sketch
Ex: y=2(x^2-9)(x^2-4)
1. domain: x^2-4=0; x=+/- 2
domain: (-infinity, -2)u(-2,2)u(2,infinity)
range: find the horizontal asymptotes (limits approaching infinity) and set up intervals y=2; (-infinity,2)u(2,infinity)
2. vertal asymptotes: x=+/-2
horizontal asymptotes: y=2
x intercepts:(3,0), (-3,0)
y intercepts:(0,9/2)
3. a)use quotient rule
f'=20x/(x^2=4)^2=0
20x=0
x=0
b)x=+/-2
c)(-infinity,-2)u(-2,0)u(0,2)u(2,infinity)
d)f'(-3)=-ve; f'(-1)=-ve;f'(1)=+ve;f'(3)=+ve
e)decreasing; increasin; decreasing
f)concave down; concave up; concave down.
x=-2,2 are points of inflection
limits at infinity.
-to find a horizontal asymptote, you take the limit as x=>infinity
y=ans is an asymptote
degree of top=degree of bottom -> coefficient of highest exponent
degree of top>degree of bottom ->-infinity or infinity
degree of top
Ex 1: 2x+5/(3x^2+1)
lim x->infinity 2x+5/(3x^2+1)
lim x->infinity is 0
0 is a horizontal asymptote.
Chapter 3 section 6
Curve sketching
Steps
1. identify the domain and range
2. find the x and y intercepts, identify the vertical and horizontal asymptotes
3. do the first derivative test
4. do the second derivative test
5. plot the information and sketch
Ex: y=2(x^2-9)(x^2-4)
1. domain: x^2-4=0; x=+/- 2
domain: (-infinity, -2)u(-2,2)u(2,infinity)
range: find the horizontal asymptotes (limits approaching infinity) and set up intervals y=2; (-infinity,2)u(2,infinity)
2. vertal asymptotes: x=+/-2
horizontal asymptotes: y=2
x intercepts:(3,0), (-3,0)
y intercepts:(0,9/2)
3. a)use quotient rule
f'=20x/(x^2=4)^2=0
20x=0
x=0
b)x=+/-2
c)(-infinity,-2)u(-2,0)u(0,2)u(2,infinity)
d)f'(-3)=-ve; f'(-1)=-ve;f'(1)=+ve;f'(3)=+ve
e)decreasing; increasin; decreasing
f)concave down; concave up; concave down.
x=-2,2 are points of inflection
alaina's second holiday blog
I'm going to talk about antiderivatives. An antiderivative is a general solution, while an integral is more specific and you must solved for c (the constant). Second, an antiderivative is just a backwards derivative. Your given is F' or F'' and you must find F.
Third, how to find an antiderivative: if you're given ax^b, where a is your constant, x is your variable and b is your exponent, your formula is (a/(b+1))(x^(b+1))
*when solving for the antiderivative, always add "+c" in place of a constant
**when solving for an integral, you will be given something like f(1)=2, in that case, you plug 1 in for x in your antiderivative and set it equal to 2 and solve to find the constant.
***When given the second derivative, just take the antiderivative twice.
****When given an integral, it is always used with the integral symbol (a large S) before and dx after.
*****Shortcut: if you have a fractional exponent, add one and multiply the coefficient by the reciprocal.
******Trig Functions: you just do the opposite of the derivative.
Ex 1. S dx =x+c
Ex 2. S (x+2)dx = (1/1+1)(x^1+1)=x^2/2+2x+c
Ex 3. S (3x^4-5x^2+x)dx =3x^5/5-5x^3/3+x^2/2+c
Ex 4. S sinxdx = -cosx
Third, how to find an antiderivative: if you're given ax^b, where a is your constant, x is your variable and b is your exponent, your formula is (a/(b+1))(x^(b+1))
*when solving for the antiderivative, always add "+c" in place of a constant
**when solving for an integral, you will be given something like f(1)=2, in that case, you plug 1 in for x in your antiderivative and set it equal to 2 and solve to find the constant.
***When given the second derivative, just take the antiderivative twice.
****When given an integral, it is always used with the integral symbol (a large S) before and dx after.
*****Shortcut: if you have a fractional exponent, add one and multiply the coefficient by the reciprocal.
******Trig Functions: you just do the opposite of the derivative.
Ex 1. S dx =x+c
Ex 2. S (x+2)dx = (1/1+1)(x^1+1)=x^2/2+2x+c
Ex 3. S (3x^4-5x^2+x)dx =3x^5/5-5x^3/3+x^2/2+c
Ex 4. S sinxdx = -cosx
alaina's first blog post from 12/16/10
well, the week before the holidays, we didn't learn anything new. that week, we actually did study guides. the study guides helped a lot with studying for the exam.
So, i'm going to talk about something EVERYONE in calculus should know how to do....
derivatives.
When taking a derivative, you are given a primary function referred to as f. The derivative, or F', is basically taking the exponent of each term, moving it to the front of each term, multiplying the coefficient of each term by that exponent, and reducing the exponent by 1.
Ex: 4x^2
F'=(2)(4)x^(2-1)
F'=8x
Remember, the derivative of a constant, some number, is 0.
So, i'm going to talk about something EVERYONE in calculus should know how to do....
derivatives.
When taking a derivative, you are given a primary function referred to as f. The derivative, or F', is basically taking the exponent of each term, moving it to the front of each term, multiplying the coefficient of each term by that exponent, and reducing the exponent by 1.
Ex: 4x^2
F'=(2)(4)x^(2-1)
F'=8x
Remember, the derivative of a constant, some number, is 0.
Dustins Third and final holiday blog
For this post I will go over the basics of derivatives and integrals. These are probably the two most important things in calculus and if you don't know these simple things about them you will die...i mean fail, then die.
First things first:
The derivative of a constant is always 0.
Ex. Derive 1. Answer=0
To derive a variable you take the degree and move it to the front, then subtract one from the original degree.
Ex. Derive x. Answer=1
Ex2. Derive 5x^2. Answer=10x
While deriving a polynomial just derive each individual part.
Ex.Derive 4x^2+2x+5 Answer=8x+2
Now integrals are just as simple as derivatives, except your doing things backwards.
While integrating always remember to add c to the end in case there was a constant that was turned to 0.**
Some Examples.
Ex.1 Integrate 3x^2.
Just work backwards and you simply get x^3+c
Ex.2 Integrate 4x^3+12x^2+5
Answer would be: x^4+4x^3+5x+c
And thats basically the basics to integrating and derivativing.
First things first:
The derivative of a constant is always 0.
Ex. Derive 1. Answer=0
To derive a variable you take the degree and move it to the front, then subtract one from the original degree.
Ex. Derive x. Answer=1
Ex2. Derive 5x^2. Answer=10x
While deriving a polynomial just derive each individual part.
Ex.Derive 4x^2+2x+5 Answer=8x+2
Now integrals are just as simple as derivatives, except your doing things backwards.
While integrating always remember to add c to the end in case there was a constant that was turned to 0.**
Some Examples.
Ex.1 Integrate 3x^2.
Just work backwards and you simply get x^3+c
Ex.2 Integrate 4x^3+12x^2+5
Answer would be: x^4+4x^3+5x+c
And thats basically the basics to integrating and derivativing.
holiday blog 2
limits at infinity; i remember these as being easy so imma do some of these :D
rules:
if the degree of the top is larger than the degree at the bottom, the limit is infinity or negaative infinity
if the bottom degree is > the top the limit = 0
if the bottom is = to the top degree you take the coefficients
5x^3/x^5 -- the bottom is larger therefore the limit = 0
3x^2/5x^2 -- the bottom is equal to the top therefore the limit = 3/5
x^2/x -- the top is larger than the bottom therefore the limit = infinity
rules:
if the degree of the top is larger than the degree at the bottom, the limit is infinity or negaative infinity
if the bottom degree is > the top the limit = 0
if the bottom is = to the top degree you take the coefficients
5x^3/x^5 -- the bottom is larger therefore the limit = 0
3x^2/5x^2 -- the bottom is equal to the top therefore the limit = 3/5
x^2/x -- the top is larger than the bottom therefore the limit = infinity
Dustin's Holiday Blog 2
For this blog I will review logs. There are many rules with logs and they are all really simple. The first thing i will note is how you take the log notation and rewrite in exponential. If you have an equation such as log10, you rewrite like so:
Your base is understood to be 10 unless stated otherwise**
10^x=10
Now some of the simple rules used for condensing and expanding logs are like so:
When you are adding logs your numbers are multiplied if both have the same base.
Ex. log5+log7
You can condense the equation by multiplying and getting log35 at which point you simplify.
When subtracting logs you do just like addition except you divide instead of multiplying.
Ex.log35-log7
You condense by dividing and get log5 and then you simplify.
Another thing to note about logs is that there are also natural logs, which are notated as ln. Natural logs work the same way as normal logs except for the fact that the base is e instead of 10.
Your base is understood to be 10 unless stated otherwise**
10^x=10
Now some of the simple rules used for condensing and expanding logs are like so:
When you are adding logs your numbers are multiplied if both have the same base.
Ex. log5+log7
You can condense the equation by multiplying and getting log35 at which point you simplify.
When subtracting logs you do just like addition except you divide instead of multiplying.
Ex.log35-log7
You condense by dividing and get log5 and then you simplify.
Another thing to note about logs is that there are also natural logs, which are notated as ln. Natural logs work the same way as normal logs except for the fact that the base is e instead of 10.
Taylor Holiday blog #1
CHAPTER 1 REVIEW
• Remember that estimating limits numerically means to set up and complete the table
The first step to setting up the table is to step up the left side of the graph. To set up the left side of the graph you subtract .1 .01 .001 from the number that X is approaching in the problem and set the subtracted answers to each number into three boxes.
The next box will simply have the number x is approaching in it
The final three boxes will have the results of adding .001 .01 .1 to the number x is approaching respectively.
The first step to estimating the limit is complete
• To complete the table you must enter the given equation into your y=
• After you will hit “2nd” and table and enter in the numbers that you solved for in the seven boxes of the first step.
• now you will fill in the bottom seven boxes of the chart with the results from the table
• (( a quick way to know if you’ve done the steps correctly is to make sure that the box under the box with the number x is approaching should have an error on the table.
Finally to estimate the limit you will read each side toward the number x is approaching and decipher what number each side is headed toward as it approaches the given number x is approaching.
• Remember the three boxes on the left side will read toward the right and the right three boxes will read toward the left
You have the possibility to have two different out comes
• If the numbers on each side of the table match then the number they are headed toward is the estimated limit
• If the numbers on each side of the table do not match the limit does not exist.
I will now give an example of each outcome
EX:1
Lim X-> 0
F(x)= x/ squareroot of (x + 1) -1
The top half of the table will read
[-.1] [-.01] [-.001] [0] [.001] [.01] [.1]
The bottom half of the table will read
[1.9487] [1.995] [1.995] [ERROR] [2.0005] [2.005] [2.0488]
From the left the numbers are approaching 2
From the right the numbers are also approaching 2
Therefore the estimated limit is 2
Ex:2
Lim X-> 0
F(x)= sin 1/x
The top half of the table will read
[2/pi] [2/3pi] [2/5pi] [0] [2/7pi] [2/9pi] [2/11pi]
The bottom half of the table will read
[1] [-1] [1] [ERROR] [-1] [1] [-1]
From the left the numbers are not approaching anything
From the right the numbers are also not approaching anything
Therefore the estimated limit DNE
CHAPTER 2 LESSON 1
SLOPE OF A TANGENT LINE
There are two formulas which need to be memorized
(((& means delta)
• The first formula is
f(x+&x)- f(x)/&x
This is the formula for a derivative. This formula is known as the secant line formula.
• The second formula is only a tiny bit different from the first
Lim f(x+&x)- f(x)/&x
&x -> 0
This formula is known as the slope of a tangent line
Solving for the problems we’ve had thus far in chapter 2 have consisted of plugging into these formulas and solving.
When given an equation you must plug x+&x into all x’s of the given equation and fill in the rest of the formula by placing – f(given equation exactly how its given)/ &x
Ex:
Find the slope of the graph of f(x)= 2x-3 at (2,1)
First you would plug in:
f(x) 2(x+&x)-3-(2x-3)/ &x
&x-> 0
Expand:
2x+2&x – 3 – 2x + 3 / &x
Take out what cancels:
2x + 2&x -3 -2x +3/&x
And you’re left with:
2&x/&x
Simplify:
2 &x/ &x
Therefore the answer is 2
((because there are no x’s left in the equation you can ignore the point (2,1) however if there had been any x’s you would have also plugged in 2 for the x’s before you simplified then solved as normal.))
There are a few helpful hints that need to be remembered
**remember that for each of these formulas any variable can be used for delta x
** remember that slope of the tangent line means to use the derivative formula
** there are many was to ask for a derivative these ways are:
Dy/dx
Y^1
F’(x)
d/dx {f(x)}
Dx[y]
D/dx
CHAPTER 2 LESSON 1
There are two rules to follow when using the derivative shortcut
These rules are : * The Constant Rule
• The Power Rule
The constant rule states that the derivative of a constant is 0
Therefore when asked to find a derivative of an equation with a constant the constant will automatically become 0.
The Power rule states that every time you take a derivative you lose a power.
Therefore if you are not taking the derivative of a constant you bring the exponent to the front of the variable and subtract one from the exponent.
The formula that displays this is
d/dx [x^n ] = nx^n-1
Example: d/dx[x^3] = 3x^2
There are many variations of equations that will call for simplification before you can take the derivative
When taking the derivative of a fraction you must simplify the equation by taking the numerator and placing it in front of the x and the exponent behind the denominator will become a negative
Example: d/dx [1/x^2] = x^-2
Then you would proceed to use the shortcut method to find the derivative.
When taking the derivative of a root you will turn the value of the root into a fraction with the exponent of x over the value of the root.
Example: d/dx [cuberoot x] = X^1/3
Then you would proceed to use the shortcut method to find the derivative.
**do not forget that after finding the derivative of a simplified equation you must convert the derivative back to an unsimplified form
Example: d/dx [ x^1/3] = 1/3x^-2/3 = 1/3cuberoot x^2
• Remember that estimating limits numerically means to set up and complete the table
The first step to setting up the table is to step up the left side of the graph. To set up the left side of the graph you subtract .1 .01 .001 from the number that X is approaching in the problem and set the subtracted answers to each number into three boxes.
The next box will simply have the number x is approaching in it
The final three boxes will have the results of adding .001 .01 .1 to the number x is approaching respectively.
The first step to estimating the limit is complete
• To complete the table you must enter the given equation into your y=
• After you will hit “2nd” and table and enter in the numbers that you solved for in the seven boxes of the first step.
• now you will fill in the bottom seven boxes of the chart with the results from the table
• (( a quick way to know if you’ve done the steps correctly is to make sure that the box under the box with the number x is approaching should have an error on the table.
Finally to estimate the limit you will read each side toward the number x is approaching and decipher what number each side is headed toward as it approaches the given number x is approaching.
• Remember the three boxes on the left side will read toward the right and the right three boxes will read toward the left
You have the possibility to have two different out comes
• If the numbers on each side of the table match then the number they are headed toward is the estimated limit
• If the numbers on each side of the table do not match the limit does not exist.
I will now give an example of each outcome
EX:1
Lim X-> 0
F(x)= x/ squareroot of (x + 1) -1
The top half of the table will read
[-.1] [-.01] [-.001] [0] [.001] [.01] [.1]
The bottom half of the table will read
[1.9487] [1.995] [1.995] [ERROR] [2.0005] [2.005] [2.0488]
From the left the numbers are approaching 2
From the right the numbers are also approaching 2
Therefore the estimated limit is 2
Ex:2
Lim X-> 0
F(x)= sin 1/x
The top half of the table will read
[2/pi] [2/3pi] [2/5pi] [0] [2/7pi] [2/9pi] [2/11pi]
The bottom half of the table will read
[1] [-1] [1] [ERROR] [-1] [1] [-1]
From the left the numbers are not approaching anything
From the right the numbers are also not approaching anything
Therefore the estimated limit DNE
CHAPTER 2 LESSON 1
SLOPE OF A TANGENT LINE
There are two formulas which need to be memorized
(((& means delta)
• The first formula is
f(x+&x)- f(x)/&x
This is the formula for a derivative. This formula is known as the secant line formula.
• The second formula is only a tiny bit different from the first
Lim f(x+&x)- f(x)/&x
&x -> 0
This formula is known as the slope of a tangent line
Solving for the problems we’ve had thus far in chapter 2 have consisted of plugging into these formulas and solving.
When given an equation you must plug x+&x into all x’s of the given equation and fill in the rest of the formula by placing – f(given equation exactly how its given)/ &x
Ex:
Find the slope of the graph of f(x)= 2x-3 at (2,1)
First you would plug in:
f(x) 2(x+&x)-3-(2x-3)/ &x
&x-> 0
Expand:
2x+2&x – 3 – 2x + 3 / &x
Take out what cancels:
2x + 2&x -3 -2x +3/&x
And you’re left with:
2&x/&x
Simplify:
2 &x/ &x
Therefore the answer is 2
((because there are no x’s left in the equation you can ignore the point (2,1) however if there had been any x’s you would have also plugged in 2 for the x’s before you simplified then solved as normal.))
There are a few helpful hints that need to be remembered
**remember that for each of these formulas any variable can be used for delta x
** remember that slope of the tangent line means to use the derivative formula
** there are many was to ask for a derivative these ways are:
Dy/dx
Y^1
F’(x)
d/dx {f(x)}
Dx[y]
D/dx
CHAPTER 2 LESSON 1
There are two rules to follow when using the derivative shortcut
These rules are : * The Constant Rule
• The Power Rule
The constant rule states that the derivative of a constant is 0
Therefore when asked to find a derivative of an equation with a constant the constant will automatically become 0.
The Power rule states that every time you take a derivative you lose a power.
Therefore if you are not taking the derivative of a constant you bring the exponent to the front of the variable and subtract one from the exponent.
The formula that displays this is
d/dx [x^n ] = nx^n-1
Example: d/dx[x^3] = 3x^2
There are many variations of equations that will call for simplification before you can take the derivative
When taking the derivative of a fraction you must simplify the equation by taking the numerator and placing it in front of the x and the exponent behind the denominator will become a negative
Example: d/dx [1/x^2] = x^-2
Then you would proceed to use the shortcut method to find the derivative.
When taking the derivative of a root you will turn the value of the root into a fraction with the exponent of x over the value of the root.
Example: d/dx [cuberoot x] = X^1/3
Then you would proceed to use the shortcut method to find the derivative.
**do not forget that after finding the derivative of a simplified equation you must convert the derivative back to an unsimplified form
Example: d/dx [ x^1/3] = 1/3x^-2/3 = 1/3cuberoot x^2
Dustin's Holiday Blog 1
First thing i'm going to review for the holiday blogs are limits. Limits are the simplest thing in the world. All you do is plug in what the approaching number is into the equation. Like so:
Ex. 1
lim x+2
x->9001
You simply plug in and your answer is 9003.
But Wait! Theres More.
Sometimes when you plug in, it just doesn't work because you gotta divide by zero. Or some other way you get no solution. In this case, you either have to use a trick-a-majigger or you messed up(in my case i have to use the trick-a-majigger because i dont mess up).
So if this happens: lim x+2/x
x->0
in this case you just take a derivative over the top and put it over the derivative of the bottom which gets you one. and that is your limit.
There is one more thing that could happen with limits. You could get a limit approaching infinity. In that case you have to follow a certain set of rules. You need to check the degree of the top and the bottom then use the following rules:
If top>bottom, lim=infinity
If bottom>top, lim=0
If top=bottom, lim=coefficients
Ex. lim x^2+2/4x^2
x->inf
You take your highest degree coefficients and you get 1/4 as your answer.
Ex. 1
lim x+2
x->9001
You simply plug in and your answer is 9003.
But Wait! Theres More.
Sometimes when you plug in, it just doesn't work because you gotta divide by zero. Or some other way you get no solution. In this case, you either have to use a trick-a-majigger or you messed up(in my case i have to use the trick-a-majigger because i dont mess up).
So if this happens: lim x+2/x
x->0
in this case you just take a derivative over the top and put it over the derivative of the bottom which gets you one. and that is your limit.
There is one more thing that could happen with limits. You could get a limit approaching infinity. In that case you have to follow a certain set of rules. You need to check the degree of the top and the bottom then use the following rules:
If top>bottom, lim=infinity
If bottom>top, lim=0
If top=bottom, lim=coefficients
Ex. lim x^2+2/4x^2
x->inf
You take your highest degree coefficients and you get 1/4 as your answer.
Holiday 3
For this final holiday blog (joy :D) I'm going to try to review integrals
Integrals are "backwards derivatives" and are pretty easy to understand, but take practice to work fast with
Most of them remind me of the power rule, which is all i really remember so if we did some with product quotient and chain sorry i forgot! haha
y'= 3x^2 + 4
y = x^3 + 4x
y'= 6x^7 + 4x^2
y = 6/8x^8 + 4/3x^3
since I don't have enough info, I'll talk about limits too
A limit is found at an x value with a given y value, and is usually taken by making a chart or by looking at the graph.
An asymptote is a spot on the graph where the limit stops (discontinuity); it is made of two curves that never touch.
A jump is a skip in the graph, kinda like picking up your pencil and putting it somewhere else if you were graphing by hand.
A removable is literally a hole in the graph, and is also a discontinuity.
Integrals are "backwards derivatives" and are pretty easy to understand, but take practice to work fast with
Most of them remind me of the power rule, which is all i really remember so if we did some with product quotient and chain sorry i forgot! haha
y'= 3x^2 + 4
y = x^3 + 4x
y'= 6x^7 + 4x^2
y = 6/8x^8 + 4/3x^3
since I don't have enough info, I'll talk about limits too
A limit is found at an x value with a given y value, and is usually taken by making a chart or by looking at the graph.
An asymptote is a spot on the graph where the limit stops (discontinuity); it is made of two curves that never touch.
A jump is a skip in the graph, kinda like picking up your pencil and putting it somewhere else if you were graphing by hand.
A removable is literally a hole in the graph, and is also a discontinuity.
taylor holiday blog #2
CHAPTER 2 LESSON 3
For both the product rule and the quotient there is a recognizable format that will allow you tho know which rule you will need to use and for each rule there is a formula to memorize and put into effect to find the derivative.
The product rule:
The product rule is recognized as F(x)G(x)
The formula for solving with the product rule is
D/Dx [F(x)G(x)] = F(x) Gprime(x) + G(x) Fprime(x)
Example: (3x-2x^2) (5+4x)
First I take the derivative e of each so that when I plug in I already know what the derivatives are to be plugged in
D/Dx [3x-2x^2]= 3-4x
Therefore F prime= 3-4x
D/Dx [5+4x]= 4
Therefore G prime= 4
Now you plug into the formula
Therefore
3x-2x^2(4)+ 5+4x (3-4x)
Distribute
12x^2x-8x^2+15-20x+12x-16
Simplify
-24x^2 + 4x +15
Because this equation cannot be simplified any further
Dx= -24x^2+4x+15
The quotient rule:
The quotient rule is recognized as F(x)/G(x)
The formula for solving with the quotient rule is
D/Dx [F(x)/G(x)] = G(x) Fprime(x)- F(x) Gprime(x)/ [G(x)]^2
Example: 5x-2/x^2-1
First I take the derivative of each so that when I plug in I already know what the derivatives are to be plugged in
D/dx [ 5x-2] = 5
Therefore F prime= 5
D/dx [x^2 – 1] = 2x
Therefore G prime = 2x
Now you plug into the formula
Therefore
(x^2+1)(5)-[(5x-2)(2x)]/(x^2+1)^2
Distribute
5x^2+5-[10x^2-4x]/(x^2+1)^2
Distribute the negative
5x^2+5-10x^2+4x/(x^2+1)^2
Simplify
-5x^2 + 4x+5/ (x^2+1)^2
Because this equation cannot be simplified any further
DX= -5x^2 + 4x+5/ (x^2+1)^2
Some things to remember:
**don’t forget that when solving with the quotient rule the negative must be distributed to the entire product of f(x)G prime(X)
** don’t forget which order F(X)G prime(X) and G(X)F prime(X) go in for each equation because it DOES matter.
CHAPTER 2 LESSON 4
The chain rule could loosely be defined as an order of operations that is used when solving composite functions.
A very important key concept to remember when using the chain rule is to work from the outside to the inside.
The “Formula” for the chain rule is
d/dx [fg(x)] = f ’(g(x)) (g’(x))
The most common procedure to using the chain rule is
• First, take the derivative of the outside
• Second, recopy just the inside
• Finally, multiply by the derivative of the inside
Example:
Square root of 3x^2-X+1
= (3x^2-X+1)^1/2
First, take the derivative of the outside:
½(________)^-1/2
Second, recopy just the inside
½(3x^2-X+1)^-1/2
Finally, multiply by the derivative of the inside
½(3x^2-X+1)^-1/2 (6x-1)
Simplify
6x-1/2(3x^2-x+1)^1/2
CHAPTER 2 LESSON 5
The key thing to remember when taking implicit derivatives is that the steps are the same as taking a regular derivative.
“with respect to” is a phrase you will see often when dealing with implicit derivatives.
The phrase “with respect to” gives you a clue as to what you will be solving for
When you see “d/dx” you also must be able to recognize that this is a symbol for “with respect to x”
When you take a derivative of anything besides x you will write it in the asterisk spot of the following example: D*/Dx
In other words, the top letter next to the d is the variable you are taking the derivative of
For example:
With respect to y: dy/dx
With respect to r: dr/dx
With respect to s: ds/dx
To solve a problem you must follow three general steps:
First take the derivative
Second gather all d*/dx
Third solve for d*/dx
Example:::
Sqrtxy= x^2y+1
(xy)1/2=x^2y+1
X^ 1/2 y^ ½ = x^2 y +1
First take the derivative
X^1/2 (1/2y^-1/2 dy/dx) + y^1/2 (1/2x^-1/2)= x^2 (1dy/dx+y(2x))
1/2x^1/2y^-1/2 dy/dx + 1/2x ^-1/2 y^1/2 = x^2 dy/dx + 2xy
Second gather all d*/dx
1/2x^1/2y^-1/2 dy/dx- X^2 dy/dx= 2xy - 1/2x^-1/2y^1y^1/2
Third solve for d*/dx
Dy/dx = 2xy -1/2x^-1/2y^1/2 / ½ x^1/2 y^-1/2 – x^2
Example with trig:
4cosx siny=1
4[cosx(cosy dy/dx)+ siny (-sinx)] = 0
4cosxx cosy dy/dx – 4 siny sin x =0
4 cosx cosy dy/dx= 4siny sinx
Dy/dx= 4 siny sinx / 4 cosx cos y
Dy/dx= tan y tan x
We also learned how to take the second derivative when dealing with implicit derivatives
The steps to solving the second derivative of an implicit derivative are:
Take the first derivative
Solve for dy/dx
Take the derivative of dy/dx (this will be d^2y/dx^2)
Plug in the result of when you solved for dy/dx into the d^2y/dx^2 equation for dy/dx
Solve again
**** sometimes when you solve the numerator will turn out to be your original equation, if so, plug in what the original equation equaled and simplify.
If not then d^2y/dx^2 will be what you found when you “solved again”
Example:
x^2 + y^2 = 25
Take the first derivative
2x + 2y dy/dx = 0
Solve for dy/dx
Dy/dx= -2x/2y therefore -x/y
Take the derivative of dy/dx
- Y(1)- [ x(1dy/dx)]/y^2
- Y-x dy/dx / y^2
Plug in the result of when you solved for dy/dx into the d^2y/dx^2 equation for dy/dx
- Y-x ( -x/y)/ y^2
Solve again
- Y^2 + x^2/y/y^2/1
-y^2 +x^2/ y^3
**** remember that sometimes when you solve the numerator will turn out to be your original equation, if so, plug in what the original equation equaled and simplify.
Therefore:
-25/y^3
CHAPTER 2 LESSON 6
There are a set of steps to follow when solving a related rate problem but first you need to know how to recognize a related rate problem.
A related rate problem is one that involves something per something.
I.E. Gallons per minute
Cubic feet per second
Your key word is PER
Keep in mind that although there are set steps for working related rate problems but each problem is unique in its own method of solution.
To further explain the steps are guidelines but the specific application is dependent upon each problem.
The steps to solving these problems are as follows.
First, identify all given quantities and quantities to be determined. Also make a sketch and be sure to label the quantities.
Second, you must write an equation involving the variables whose rates of change either are given or are to be determined
Third, you must use the chain rule implicitly differentiate both sides of the equation with respect to time (T)
Finally, you must substitute all known variables into the resulting equation and solve for the desired rate of change
Be sure to print and study your Geometric formulas
Example:
A Pebble is dropped into a calm pond causing ripples in the form of concentric circles. The radius of the outer circle is increasing at a rate of 1 foot PER second when the radius is 4 ft at what rate is the total area (A) of the disturbed water changing?
#1 dr/dt= 1ft/sec
R=4 ft
dA/dt= ?
#2 A= pi r ^2
#3 dA/dt= pi{2r dr/dt}
#4 dA/dt= 2pi (4)(1)
Therefore 8 pi ft/sec
For both the product rule and the quotient there is a recognizable format that will allow you tho know which rule you will need to use and for each rule there is a formula to memorize and put into effect to find the derivative.
The product rule:
The product rule is recognized as F(x)G(x)
The formula for solving with the product rule is
D/Dx [F(x)G(x)] = F(x) Gprime(x) + G(x) Fprime(x)
Example: (3x-2x^2) (5+4x)
First I take the derivative e of each so that when I plug in I already know what the derivatives are to be plugged in
D/Dx [3x-2x^2]= 3-4x
Therefore F prime= 3-4x
D/Dx [5+4x]= 4
Therefore G prime= 4
Now you plug into the formula
Therefore
3x-2x^2(4)+ 5+4x (3-4x)
Distribute
12x^2x-8x^2+15-20x+12x-16
Simplify
-24x^2 + 4x +15
Because this equation cannot be simplified any further
Dx= -24x^2+4x+15
The quotient rule:
The quotient rule is recognized as F(x)/G(x)
The formula for solving with the quotient rule is
D/Dx [F(x)/G(x)] = G(x) Fprime(x)- F(x) Gprime(x)/ [G(x)]^2
Example: 5x-2/x^2-1
First I take the derivative of each so that when I plug in I already know what the derivatives are to be plugged in
D/dx [ 5x-2] = 5
Therefore F prime= 5
D/dx [x^2 – 1] = 2x
Therefore G prime = 2x
Now you plug into the formula
Therefore
(x^2+1)(5)-[(5x-2)(2x)]/(x^2+1)^2
Distribute
5x^2+5-[10x^2-4x]/(x^2+1)^2
Distribute the negative
5x^2+5-10x^2+4x/(x^2+1)^2
Simplify
-5x^2 + 4x+5/ (x^2+1)^2
Because this equation cannot be simplified any further
DX= -5x^2 + 4x+5/ (x^2+1)^2
Some things to remember:
**don’t forget that when solving with the quotient rule the negative must be distributed to the entire product of f(x)G prime(X)
** don’t forget which order F(X)G prime(X) and G(X)F prime(X) go in for each equation because it DOES matter.
CHAPTER 2 LESSON 4
The chain rule could loosely be defined as an order of operations that is used when solving composite functions.
A very important key concept to remember when using the chain rule is to work from the outside to the inside.
The “Formula” for the chain rule is
d/dx [fg(x)] = f ’(g(x)) (g’(x))
The most common procedure to using the chain rule is
• First, take the derivative of the outside
• Second, recopy just the inside
• Finally, multiply by the derivative of the inside
Example:
Square root of 3x^2-X+1
= (3x^2-X+1)^1/2
First, take the derivative of the outside:
½(________)^-1/2
Second, recopy just the inside
½(3x^2-X+1)^-1/2
Finally, multiply by the derivative of the inside
½(3x^2-X+1)^-1/2 (6x-1)
Simplify
6x-1/2(3x^2-x+1)^1/2
CHAPTER 2 LESSON 5
The key thing to remember when taking implicit derivatives is that the steps are the same as taking a regular derivative.
“with respect to” is a phrase you will see often when dealing with implicit derivatives.
The phrase “with respect to” gives you a clue as to what you will be solving for
When you see “d/dx” you also must be able to recognize that this is a symbol for “with respect to x”
When you take a derivative of anything besides x you will write it in the asterisk spot of the following example: D*/Dx
In other words, the top letter next to the d is the variable you are taking the derivative of
For example:
With respect to y: dy/dx
With respect to r: dr/dx
With respect to s: ds/dx
To solve a problem you must follow three general steps:
First take the derivative
Second gather all d*/dx
Third solve for d*/dx
Example:::
Sqrtxy= x^2y+1
(xy)1/2=x^2y+1
X^ 1/2 y^ ½ = x^2 y +1
First take the derivative
X^1/2 (1/2y^-1/2 dy/dx) + y^1/2 (1/2x^-1/2)= x^2 (1dy/dx+y(2x))
1/2x^1/2y^-1/2 dy/dx + 1/2x ^-1/2 y^1/2 = x^2 dy/dx + 2xy
Second gather all d*/dx
1/2x^1/2y^-1/2 dy/dx- X^2 dy/dx= 2xy - 1/2x^-1/2y^1y^1/2
Third solve for d*/dx
Dy/dx = 2xy -1/2x^-1/2y^1/2 / ½ x^1/2 y^-1/2 – x^2
Example with trig:
4cosx siny=1
4[cosx(cosy dy/dx)+ siny (-sinx)] = 0
4cosxx cosy dy/dx – 4 siny sin x =0
4 cosx cosy dy/dx= 4siny sinx
Dy/dx= 4 siny sinx / 4 cosx cos y
Dy/dx= tan y tan x
We also learned how to take the second derivative when dealing with implicit derivatives
The steps to solving the second derivative of an implicit derivative are:
Take the first derivative
Solve for dy/dx
Take the derivative of dy/dx (this will be d^2y/dx^2)
Plug in the result of when you solved for dy/dx into the d^2y/dx^2 equation for dy/dx
Solve again
**** sometimes when you solve the numerator will turn out to be your original equation, if so, plug in what the original equation equaled and simplify.
If not then d^2y/dx^2 will be what you found when you “solved again”
Example:
x^2 + y^2 = 25
Take the first derivative
2x + 2y dy/dx = 0
Solve for dy/dx
Dy/dx= -2x/2y therefore -x/y
Take the derivative of dy/dx
- Y(1)- [ x(1dy/dx)]/y^2
- Y-x dy/dx / y^2
Plug in the result of when you solved for dy/dx into the d^2y/dx^2 equation for dy/dx
- Y-x ( -x/y)/ y^2
Solve again
- Y^2 + x^2/y/y^2/1
-y^2 +x^2/ y^3
**** remember that sometimes when you solve the numerator will turn out to be your original equation, if so, plug in what the original equation equaled and simplify.
Therefore:
-25/y^3
CHAPTER 2 LESSON 6
There are a set of steps to follow when solving a related rate problem but first you need to know how to recognize a related rate problem.
A related rate problem is one that involves something per something.
I.E. Gallons per minute
Cubic feet per second
Your key word is PER
Keep in mind that although there are set steps for working related rate problems but each problem is unique in its own method of solution.
To further explain the steps are guidelines but the specific application is dependent upon each problem.
The steps to solving these problems are as follows.
First, identify all given quantities and quantities to be determined. Also make a sketch and be sure to label the quantities.
Second, you must write an equation involving the variables whose rates of change either are given or are to be determined
Third, you must use the chain rule implicitly differentiate both sides of the equation with respect to time (T)
Finally, you must substitute all known variables into the resulting equation and solve for the desired rate of change
Be sure to print and study your Geometric formulas
Example:
A Pebble is dropped into a calm pond causing ripples in the form of concentric circles. The radius of the outer circle is increasing at a rate of 1 foot PER second when the radius is 4 ft at what rate is the total area (A) of the disturbed water changing?
#1 dr/dt= 1ft/sec
R=4 ft
dA/dt= ?
#2 A= pi r ^2
#3 dA/dt= pi{2r dr/dt}
#4 dA/dt= 2pi (4)(1)
Therefore 8 pi ft/sec
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