Well, I feel as though I have reviewed just about everything we’ve learned several times over…except word problems. The free response questions on the practice AP tests that we’ve been doing are essentially just word problems. I’ll go over how to work through a simple related rate problem.
There are steps that you can take to decipher related rate word problems and make them easier to understand. First, you need to identify all of the information the word problem gives, such as variables and what you’re actually trying to find. Once you do that, then you need to find a formula that connects all of your variables and your shape together (such as the area of a triangle or the volume of a cube). After this, solve for your desired rate with respect to t (such as dA/dt for the rate of area changing or dV/dt for the rate of volume changing). And lastly, plug in all of your variables to get your answer.
*Don’t forget to include your units (such as cm/min or ft^3/sec).
Some key words you need to know are:
“rate of change” = derivative
“speed” = derivative
“with respect to” = what’s the bottom variable of the derivative (ex. rate of changing area A with respect to time t : dA/dt)
“area” or “volume” = formula
“cube” or “sphere” or “square” or “triangle” or “circle” = shape
“radius” or “diameter” or “height” or “depth” = possibly what you need to take the derivate of or important components of the formula
“increasing” or “decreasing” = whether the derivative is positive or negative
and any form of units (especially squared or cubic units) = squared means area; cubic means volume
Ex. The radius r of a circle is increasing at a rate of 4 centimeters per minute. Find the rate of change of the area when r = 8 cm.
First of all, we know the following info: r = 8 & dr/dt = 4 cm/min
We also know we are looking for: dA/dt = ?
Second, because we are looking for the rate of change of the AREA, we know where using the area of a circle formula: A = π(r^2)
Now, solve for dA/dt by taking the derivative of the area formula: dA/dt = 2π r dr/dt
**Remember to put dr/dt after you take the derivative of r because it is an implicit derivative.
Finally, plug in your variables: dA/dt = 2π (8)(4) = 64π cm^2/min
***cm is squared because area is always squared, just like volume is always cubed.
Sunday, April 3, 2011
Alaina's blog, 3 April 2011
Something some of us may have forgotten how to do: 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.
4/3/11
Here are some of the rules we learned a reallllly long time ago. I guess a lot of them can be applied to the AP test so it does come in handy :)
THE CONSTANT RULE:
The constant rule: the derivative of a constant is always 0
Ex. 1) d/dx 7 = 0
THE POWER RULE:
The power rule: whenever you take the derivative of something with an exponent, you lose a power. The formula for this shortcut is d/dx (x^n) = nx^(n-1)
Ex. 2) d/dx 3x^2
First, bring the exponent to the front (and if there is a constant in front of the x, multiply the two). Then, subtract one from the exponent.
= 2(3)x^(2-1) = 6x
Ex. 3) d/dx (x^3 + 9x)
When there are multiple terms in an equation, take the derivative of each term individually.
= 3x^2 + 9
THE PRODUCT RULE:
The product rule formula: f(x)g’(x) + g(x)f’(x)
When using the product rule, the problem must state that you take the derivative of two separate equations multiplied by each other.
*note: g’(x) or f’(x) means take the derivative of g(x) or f(x)
Ex. 4) (x+3)(x^2+6x)
(x+3)(2x+6) + (x^2+6x)(1)
2x^2 + 6x + 6x + 18 + x^2 + 6x
3x^2 + 18x + 18
= 3(x^2 + 6x +6)
**note: Make sure the answer is as fully simplified as possible.
THE QUOTIENT RULE:
The quotient rule formula: (g(x)f’(x) – f(x)g’(x)) / (g(x))^2
When using the quotient rule, the problem must state that you take the derivative of two equations divided by each other.
***note: if the denominator is just a single term, it is not considered an equation and you do not use the quotient rule; you simply rewrite the problem and take the derivative normally
Ex. 5) (x^3+4) / (x-2)
((x-2)(3x^2) – (x^3+4)(1)) / ((x-2)^2)
((3x^3-6x^2) – (x^3+4)) / ((x-2)^2)
****note: to keep things in simplest terms, do not factor out the bottom
(3x^3 – 6x^2 – x^3 – 4) / ((x-2)^2)
*****note: be sure to distribute the negative throughout the second equation
(2x^3 – 6x^2 – 4) / ((x – 2)^2)
(2(x^3 – 3x^2 – 2)) / ((x – 2)^2)
THE CHAIN RULE:
You use the chain rule when you have a function inside of a function, f(g(x)).
Ex. 6) (x^2 + 3)^4
First, take the exponent of the entire function, bring it to the front, then subtract one from the exponent, and then just copy the inside.
4(x^2 + 3)^3
Now, multiply that by the derivative of the inside equation.
4(x^2 + 3)^3 (2x)
Put in simplest terms.
= 8x(x^2 + 3)^3
You can also use the chain rule with the product rule or quotient rule. If this is the case, then you first need to determine whether the chain rule is on the outside or the inside; this is to figure out what order you do things in. If the chain rule is on the outside, you do that first. If not, you do the product rule or quotient rule first.
Ex. 7) (x + 3)(x^2 – 1)^2
In this problem, the product rule comes first. When taking the derivative of the second equation, use the chain rule.
(x + 3)(2(x^2 – 1)(2x)) + (x^2 – 1)^2 (1)
= 4x(x + 3)(x^2 – 1) + (x^2 – 1)^2
Ex. 8) ((x + 2)/(x – 1))^2
For this problem, the chain rule is on the outside, therefore it goes first.
2((x + 2)/(x – 1)) (((x – 1)(1) – (x + 2)(1)) / (x – 1)^2))
2((x + 2)/(x – 1)) ((x – 1 – x – 2) / (x – 1)^2)
2((x + 2)/(x – 1)) ((-3)/(x – 1)^2)
= -3(2x + 4) / (x – 1) (x – 1)^2
THE CONSTANT RULE:
The constant rule: the derivative of a constant is always 0
Ex. 1) d/dx 7 = 0
THE POWER RULE:
The power rule: whenever you take the derivative of something with an exponent, you lose a power. The formula for this shortcut is d/dx (x^n) = nx^(n-1)
Ex. 2) d/dx 3x^2
First, bring the exponent to the front (and if there is a constant in front of the x, multiply the two). Then, subtract one from the exponent.
= 2(3)x^(2-1) = 6x
Ex. 3) d/dx (x^3 + 9x)
When there are multiple terms in an equation, take the derivative of each term individually.
= 3x^2 + 9
THE PRODUCT RULE:
The product rule formula: f(x)g’(x) + g(x)f’(x)
When using the product rule, the problem must state that you take the derivative of two separate equations multiplied by each other.
*note: g’(x) or f’(x) means take the derivative of g(x) or f(x)
Ex. 4) (x+3)(x^2+6x)
(x+3)(2x+6) + (x^2+6x)(1)
2x^2 + 6x + 6x + 18 + x^2 + 6x
3x^2 + 18x + 18
= 3(x^2 + 6x +6)
**note: Make sure the answer is as fully simplified as possible.
THE QUOTIENT RULE:
The quotient rule formula: (g(x)f’(x) – f(x)g’(x)) / (g(x))^2
When using the quotient rule, the problem must state that you take the derivative of two equations divided by each other.
***note: if the denominator is just a single term, it is not considered an equation and you do not use the quotient rule; you simply rewrite the problem and take the derivative normally
Ex. 5) (x^3+4) / (x-2)
((x-2)(3x^2) – (x^3+4)(1)) / ((x-2)^2)
((3x^3-6x^2) – (x^3+4)) / ((x-2)^2)
****note: to keep things in simplest terms, do not factor out the bottom
(3x^3 – 6x^2 – x^3 – 4) / ((x-2)^2)
*****note: be sure to distribute the negative throughout the second equation
(2x^3 – 6x^2 – 4) / ((x – 2)^2)
(2(x^3 – 3x^2 – 2)) / ((x – 2)^2)
THE CHAIN RULE:
You use the chain rule when you have a function inside of a function, f(g(x)).
Ex. 6) (x^2 + 3)^4
First, take the exponent of the entire function, bring it to the front, then subtract one from the exponent, and then just copy the inside.
4(x^2 + 3)^3
Now, multiply that by the derivative of the inside equation.
4(x^2 + 3)^3 (2x)
Put in simplest terms.
= 8x(x^2 + 3)^3
You can also use the chain rule with the product rule or quotient rule. If this is the case, then you first need to determine whether the chain rule is on the outside or the inside; this is to figure out what order you do things in. If the chain rule is on the outside, you do that first. If not, you do the product rule or quotient rule first.
Ex. 7) (x + 3)(x^2 – 1)^2
In this problem, the product rule comes first. When taking the derivative of the second equation, use the chain rule.
(x + 3)(2(x^2 – 1)(2x)) + (x^2 – 1)^2 (1)
= 4x(x + 3)(x^2 – 1) + (x^2 – 1)^2
Ex. 8) ((x + 2)/(x – 1))^2
For this problem, the chain rule is on the outside, therefore it goes first.
2((x + 2)/(x – 1)) (((x – 1)(1) – (x + 2)(1)) / (x – 1)^2))
2((x + 2)/(x – 1)) ((x – 1 – x – 2) / (x – 1)^2)
2((x + 2)/(x – 1)) ((-3)/(x – 1)^2)
= -3(2x + 4) / (x – 1) (x – 1)^2
Friday, April 1, 2011
4/1/11
This week I'm going to go over PARTICULAR SOLUTIONS. It's a really simple concept, but yet it's something that everyone forgets the steps for and cannot do. So to explain the concept...For the problem given to you, you're typically given a derivative already solved, in the form dy/dx = 'something'..And they'll tell you something along the lines of, "Find the particular solution (or they might just say "solution", but particular solution is implied) with initial condition f(#)=#..(which means they give you an x and y value that they want you to plug in after). Okay so here's a simple example explaining all of the steps thoroughly:
Ex. 1) Find the particular solution for 3y^2 + xy + 2x = 6 given that x=0 when y=1.
*Alright, so first you should notice the obvious--that you're not given what the derivative is. So they expect you to derive that equation before you start the particular solution
*So using implicit differentiation, you should get this as your first step when deriving the equation:
6ydy/dx + (x)(dy/dx) + (y)(1) + 2 = 0
which is
6ydy/dx + xdy/dx + y + 2 = 0
Now to simplify this you're going to have to get all the dy/dx's on one side and everything else on the other side:
6dy/dx + xdy/dx = -y-2
Now factor out a dy/dx
dy/dx(6+x) = -y-2
And divide by 6+x
dy/dx = (-y-2)/(6+x)
**Now we can use that to find our particular solution
*So the first step in finding the particular solution is to cross multiply..as in, multiply dy by 6+x and multiply dx by -y-2, and set them equal like this:
(6+x)dy = (-y-2)dx
*Alright now for the next step you're supposed to get the x's with the dx's and the y's with the dy's (by dividing usually)..butttttt, I don't see how that's possible with this equation..I mean I did make up this problem so that could be why, but this usually wouldn't happen on an AP. Soooo, for me to explain how to work the rest of this problem let's just pretend for our derivative we got this instead:
dy/dx = (6+x)/(-y-2)
*Now cross multiply and you should get this:
(-y-2)dy = (6+x)dx
*Now you're going to integrate both sides of the equation and put the "+C" on the right hand side. So after integrating you should have this:
-1/2y^2 - 2y = 6x + 1/2x^2 + C
*Now you're going to solve for C..And to do this you have to plug in the x and y value they gave you (x=0; y=1)
*So plugging in you should have:
-1/2(1)^2 - 2(1) = 6(0) + 1/2(0)^2 + C
C = 5/2
*Now plug your C into your integrated equation and solve for y to get your final answer
-1/2y^2 - 2y = 6x + 1/2x^2 + 5/2
*To solve this for y, you can first get rid of the fractions by multiplying every term by 2..After doing that you should have:
-y^2 - 4y = 12x + x^2 - 5
Rearranging that you have
-y^2 - 4y = x^2 + 12x - 5
Now you can factor out a -y..then you should get this:
-y(y+4) = x^2+12x-5
And divide by y+4
-y = (x^2+12x-5)/(y+4)
Divide by -1
y = - (x^2+12x-5)/(y+4) ..And that's your particular solution!
Ex. 1) Find the particular solution for 3y^2 + xy + 2x = 6 given that x=0 when y=1.
*Alright, so first you should notice the obvious--that you're not given what the derivative is. So they expect you to derive that equation before you start the particular solution
*So using implicit differentiation, you should get this as your first step when deriving the equation:
6ydy/dx + (x)(dy/dx) + (y)(1) + 2 = 0
which is
6ydy/dx + xdy/dx + y + 2 = 0
Now to simplify this you're going to have to get all the dy/dx's on one side and everything else on the other side:
6dy/dx + xdy/dx = -y-2
Now factor out a dy/dx
dy/dx(6+x) = -y-2
And divide by 6+x
dy/dx = (-y-2)/(6+x)
**Now we can use that to find our particular solution
*So the first step in finding the particular solution is to cross multiply..as in, multiply dy by 6+x and multiply dx by -y-2, and set them equal like this:
(6+x)dy = (-y-2)dx
*Alright now for the next step you're supposed to get the x's with the dx's and the y's with the dy's (by dividing usually)..butttttt, I don't see how that's possible with this equation..I mean I did make up this problem so that could be why, but this usually wouldn't happen on an AP. Soooo, for me to explain how to work the rest of this problem let's just pretend for our derivative we got this instead:
dy/dx = (6+x)/(-y-2)
*Now cross multiply and you should get this:
(-y-2)dy = (6+x)dx
*Now you're going to integrate both sides of the equation and put the "+C" on the right hand side. So after integrating you should have this:
-1/2y^2 - 2y = 6x + 1/2x^2 + C
*Now you're going to solve for C..And to do this you have to plug in the x and y value they gave you (x=0; y=1)
*So plugging in you should have:
-1/2(1)^2 - 2(1) = 6(0) + 1/2(0)^2 + C
C = 5/2
*Now plug your C into your integrated equation and solve for y to get your final answer
-1/2y^2 - 2y = 6x + 1/2x^2 + 5/2
*To solve this for y, you can first get rid of the fractions by multiplying every term by 2..After doing that you should have:
-y^2 - 4y = 12x + x^2 - 5
Rearranging that you have
-y^2 - 4y = x^2 + 12x - 5
Now you can factor out a -y..then you should get this:
-y(y+4) = x^2+12x-5
And divide by y+4
-y = (x^2+12x-5)/(y+4)
Divide by -1
y = - (x^2+12x-5)/(y+4) ..And that's your particular solution!
Sunday, March 27, 2011
Blog #31
I’m going to review how to integrate using substitution. This is used when you have a product or quotient rule. This can’t always be used, but these are the instances in which it can:
S (u)(v)
…or…
S u/v
where v is the derivative of u
Here’s an example:
Given…S (x²+1)(2x) dx
First, identify u and v. Which is the derivative and which is the original?
u = x²+1
v = 2x
…because 2x is the derivative of x²+1…
Okay, next you remove v (2x dx) from the equation and substitute “u” for x²+1.
S u du
*You insert du because v is the derivative of u, hence du.
Now, integrate u.
½ u² + C
Finally, replace “u” with the original.
½ (x²+1)² + C
I’m also going to review the Second Fundamental Theorem of Calculus:
This theorem state that the derivative of an integral results in the original equation.
The formula: d/dx vSa f(x) dx = (f(v)) (v’) where “a” is a number (usually equaling 0) and v is an equation
An example:
d/dx x²S0 (x+1)² dx
= (((x²) +1)²) (2x)
= 2x(x²+1)²
S (u)(v)
…or…
S u/v
where v is the derivative of u
Here’s an example:
Given…S (x²+1)(2x) dx
First, identify u and v. Which is the derivative and which is the original?
u = x²+1
v = 2x
…because 2x is the derivative of x²+1…
Okay, next you remove v (2x dx) from the equation and substitute “u” for x²+1.
S u du
*You insert du because v is the derivative of u, hence du.
Now, integrate u.
½ u² + C
Finally, replace “u” with the original.
½ (x²+1)² + C
I’m also going to review the Second Fundamental Theorem of Calculus:
This theorem state that the derivative of an integral results in the original equation.
The formula: d/dx vSa f(x) dx = (f(v)) (v’) where “a” is a number (usually equaling 0) and v is an equation
An example:
d/dx x²S0 (x+1)² dx
= (((x²) +1)²) (2x)
= 2x(x²+1)²
blog blog blog blog blOg :)
For this blog I'm going to review Chapter 5..taking derivatives of logs, natural logs, e, exponents and whatnot. *first let's start with natural logs. The rule for taking the derivative of those is to first write the inside as a fraction over 1 and then multiply it by its derivative. So it would be like this: 1/# x d/dx of # So here are some examples like that: 1.) ln(5x+4) =1/5x+4 x 5 =5/5x+4 2.) ln(5x^2+2x+4) =1/(5x^2+2x+4) x 10x+2 =(10x+2)/(5x^2+2x+4) 3.) ln(x^2-16)^1/2 =1/(x^2-16)^1/2 x 1/2(x^2-16)^-1/2 x 2x =x/(x^2-16) *Now let's go over how to derive exponential functions dealing with e. The rule says that all you do to take the derivative is first recopy the function they give you, then multiply it by the exponent's derivative So here are some examples: 1.) f(x)=(4e)^-3x^2 -First instead of recopying the whole thing, you leave the 4 out in front just like you would do when deriving any other equation. Then you recopy the e^-3x^2 -So you should have 4(e^-3x^2 -Now multiply that by the derivative of e^-3x^2 which is -6x -So now you have 4(e^-3x^2 x -6x) = -24xe^-3x^2 2.) f(x)=x^4e^x -Okay first thing you should realize is that this is a product rule. The same rules apply even though there's an e involved..and to save time, the derivative of e^x is e^x -So using the product rule you get: (x^4)(e^x)+(e^x)(4x^3) =x^4e^x + 4x^3e^x -Then you can simplify that^ further by taking out an x^3 and an e^x -So your final answer is: x^3e^x(x+4) *Last, let's go over how to derive logs. You take the inside of the function, put it as a fraction over 1 times ln(of whatever base) times the derivative of the inside..I'm not sure if I explained that right haha, but anyway here's an example of that: 1.) log (base 3) (6x^2+5x+2) =1/(6x^2=5x+2)ln3 x 12x+5 = (12x+5)/(6x^2+5x+2)ln3
Soooo... differentiability. There are four ways to determine whether a function is differentiable. If it is differentiable it pretty much means it isn't continuous at some point in the graph. *Using these steps to figure out if a function is differentiable or not you can: look at the graph, factor out the problem, or simply just by looking at the problem. So here are the four ways to determine if a function is differentiable:
1) If there is a corner
Corners occur when absolute values and/or piecewises are found in the function that you are trying to take the derivative.
An example of a function with a corner 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
**When graphing this function you will see what a corner is(it will be the point where the graph makes a v).
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
**Vertical asymptotes can also be graphed
3) If it's not continuous
If a function is not continuous, then it's not differentiable. Discontinuities can occur if theres an vertical asymptote, removable, or jump.
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
**When graphing: if there is a point in there graph where there is a dip then it goes back straight, then it would be a cusp.
1) If there is a corner
Corners occur when absolute values and/or piecewises are found in the function that you are trying to take the derivative.
An example of a function with a corner 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
**When graphing this function you will see what a corner is(it will be the point where the graph makes a v).
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
**Vertical asymptotes can also be graphed
3) If it's not continuous
If a function is not continuous, then it's not differentiable. Discontinuities can occur if theres an vertical asymptote, removable, or jump.
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
**When graphing: if there is a point in there graph where there is a dip then it goes back straight, then it would be a cusp.
3/27/11
Soooo, time for another blog. This time I'm going to go over PVA, which we all know stands for Position, Velocity, and Acceleration. So, to go over the basics: If you're given a position function and they're asking for velocity, you're going to take the derivative of that equation (i.e...for "PVA", DERIVATIVE TO THE RIGHT; INTEGRAL TO THE LEFT)..If you're given position and they ask you for acceleration, you take the derivative twice. If you're given velocity and they're asking for the position (or distance), then you're going to integrate the equation they gave you and add your +c to that. With that, here are some examples:
Ex. 1) At t=0 a particle starts at rest and moves along a line in such a way that at time t its acceleration is 24t^2 feet per second per second. Through how many feet does the particle move during thye first 2 seconds?
*Okay first let's identify what we're given to work with in the problem:
They tell you 24t^2 is the acceleration; they're asking for how many feet the particle moves during the first 2 seconds, so that means you're going to integrate the equation TWICE because "the FIRST 2 seconds" implies that they want you to use the position function
*So integrating 24t^2 for the first time you get 8t^3; then integrating it a 2nd time you get 2t^4
*Now you plug in you time value (2):
2(2)^4
2(16)
=32
Ex. 2) A particle moves along the x-axis so that its position at time t is given by x(t)=t^2-6t+5. For what value of t is the velocity of the particle zero?
*So for this problem we are given the position function and we're asked to find the velocity at zero; All that means is to derive the position function to get velocity and then set it equal to zero and solve for x
*So deriving the function you should get: 2t-6
*Set it equal to zero and solve for x:
2t-6=0
2t=6
t=3
Ex. 3) The maximum acceleration attained on the interval 0
*Doing that you get: 3t^2-6t+12
*Since they say MAXIMUM, you're going to also have to do the 1st derivative test
*So deriving the equation you just got, you should now have: 6t-6
*1st derivative test says to set the derivative = 0: 6t-6=0...and solve for t
t = 1
*Set up intervals: (*they gave you the interval [0,3] so you MUST use that!)
(0,1)u(1,3)
*plug in values on both intervals (1st interval is negative; 2nd is positive, which means there is a max at x=1)
*Alsoooo, since you were given an interval in the problem, you MUST check your endpoints:
f(0)=3(0)^2-6(0)+12 = 12
f(1)=9
f(3)=21
*And whichever number you get once you've plugged in is your maximum value
*So your maximum acceleration is 21
Ex. 1) At t=0 a particle starts at rest and moves along a line in such a way that at time t its acceleration is 24t^2 feet per second per second. Through how many feet does the particle move during thye first 2 seconds?
*Okay first let's identify what we're given to work with in the problem:
They tell you 24t^2 is the acceleration; they're asking for how many feet the particle moves during the first 2 seconds, so that means you're going to integrate the equation TWICE because "the FIRST 2 seconds" implies that they want you to use the position function
*So integrating 24t^2 for the first time you get 8t^3; then integrating it a 2nd time you get 2t^4
*Now you plug in you time value (2):
2(2)^4
2(16)
=32
Ex. 2) A particle moves along the x-axis so that its position at time t is given by x(t)=t^2-6t+5. For what value of t is the velocity of the particle zero?
*So for this problem we are given the position function and we're asked to find the velocity at zero; All that means is to derive the position function to get velocity and then set it equal to zero and solve for x
*So deriving the function you should get: 2t-6
*Set it equal to zero and solve for x:
2t-6=0
2t=6
t=3
Ex. 3) The maximum acceleration attained on the interval 0
*Doing that you get: 3t^2-6t+12
*Since they say MAXIMUM, you're going to also have to do the 1st derivative test
*So deriving the equation you just got, you should now have: 6t-6
*1st derivative test says to set the derivative = 0: 6t-6=0...and solve for t
t = 1
*Set up intervals: (*they gave you the interval [0,3] so you MUST use that!)
(0,1)u(1,3)
*plug in values on both intervals (1st interval is negative; 2nd is positive, which means there is a max at x=1)
*Alsoooo, since you were given an interval in the problem, you MUST check your endpoints:
f(0)=3(0)^2-6(0)+12 = 12
f(1)=9
f(3)=21
*And whichever number you get once you've plugged in is your maximum value
*So your maximum acceleration is 21
Monday, March 21, 2011
3/20/11
Sooo.. curve sketching! Remember: this is on every single AP pretty much.
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 horizontal asymptotes (limits approaching infinity) and set up intervals
y=2; (-infinity,2)u(2,infinity)
2. vertical asymptotes: x=+/-2
horizontal asymptotes: y=2
x intercepts: (3,0)(-3,0)
y intercepts: (0,9/2)
3.a) use quotient rule
f'(x)=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; decreasing; increasing; increasing
f) x=0 is a min
4. a) second derivative(quotient rule)
f''(x)=-20(3x^2+4)/(x^2-4)^3=0
x=+/- squareroot (-4/3)
b)x=+/-2
c)(-infinity,-2)u(-2,2)u(2,infinity)
d)f''(-3)=-ve; f''(0)=+ve; f''(3)=-v3
e)decreasing; increasing; decreasing
f)concave down; concave up; concave down
x=-2,2 are points of inflection
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 horizontal asymptotes (limits approaching infinity) and set up intervals
y=2; (-infinity,2)u(2,infinity)
2. vertical asymptotes: x=+/-2
horizontal asymptotes: y=2
x intercepts: (3,0)(-3,0)
y intercepts: (0,9/2)
3.a) use quotient rule
f'(x)=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; decreasing; increasing; increasing
f) x=0 is a min
4. a) second derivative(quotient rule)
f''(x)=-20(3x^2+4)/(x^2-4)^3=0
x=+/- squareroot (-4/3)
b)x=+/-2
c)(-infinity,-2)u(-2,2)u(2,infinity)
d)f''(-3)=-ve; f''(0)=+ve; f''(3)=-v3
e)decreasing; increasing; decreasing
f)concave down; concave up; concave down
x=-2,2 are points of inflection
Sunday, March 20, 2011
Blog #30
Well, more review I guess…let’s go over concavity and the 2nd Derivative Test.
Okay, concavity. To test concavity, you use the same steps as the first derivative test, but with the 2nd derivative. To review, here are the steps:
1. Take the 2nd derivative.
2. Set it equal to 0.
3. Solve for x.
4. Check differentiability.
5. Set up intervals.
6. Pick numbers on the intervals and plug them into the 2nd derivative.
7. If you get a positive number, it is concave up. If you get a negative number it is concave down.
8. If there is a change (like from positive to negative or vice versa), it is a point of inflection.
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: At x = 2 there is a point of inflection.
Now, the 2nd Derivative Test. It is simply a shortcut for the 1st Derivative Test. Also, this test does not always work. If for step 7 you get a 0, the test fails. If that ever happens, you must use the 1st Derivative Test. Anyway, the steps are:
1. Take the 1st derivative.
2. Set it equal to 0.
3. Solve for x.
4. Check differentiability.
5. Take the 2nd derivative.
6. Plug in your critical points (your x-values) into the 2nd derivative.
7. If you get a positive number, it is a min. If you get a negative number, it is a max.
Ex. 2) Use the 2nd Derivative Test to find the relative extrema of f(x) = x^3 – 3x^2 + 3.
f’(x) = 3x^2 – 6x
3x^2 – 6x = 0
3x(x – 2) = 0
x = 0, 2
f’’(x) = 6x – 6
6(0) – 6 = negative #
6(2) – 6 = positive #
x = 0 is a max
x = 2 is a min
Okay, concavity. To test concavity, you use the same steps as the first derivative test, but with the 2nd derivative. To review, here are the steps:
1. Take the 2nd derivative.
2. Set it equal to 0.
3. Solve for x.
4. Check differentiability.
5. Set up intervals.
6. Pick numbers on the intervals and plug them into the 2nd derivative.
7. If you get a positive number, it is concave up. If you get a negative number it is concave down.
8. If there is a change (like from positive to negative or vice versa), it is a point of inflection.
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: At x = 2 there is a point of inflection.
Now, the 2nd Derivative Test. It is simply a shortcut for the 1st Derivative Test. Also, this test does not always work. If for step 7 you get a 0, the test fails. If that ever happens, you must use the 1st Derivative Test. Anyway, the steps are:
1. Take the 1st derivative.
2. Set it equal to 0.
3. Solve for x.
4. Check differentiability.
5. Take the 2nd derivative.
6. Plug in your critical points (your x-values) into the 2nd derivative.
7. If you get a positive number, it is a min. If you get a negative number, it is a max.
Ex. 2) Use the 2nd Derivative Test to find the relative extrema of f(x) = x^3 – 3x^2 + 3.
f’(x) = 3x^2 – 6x
3x^2 – 6x = 0
3x(x – 2) = 0
x = 0, 2
f’’(x) = 6x – 6
6(0) – 6 = negative #
6(2) – 6 = positive #
x = 0 is a max
x = 2 is a min
Bloggeroonie
Okay for this blog I'm going to review implicit differentiation-and all that means is to take the derivative of an equation with respect to other variables (like y) besides x. So that would make it dy/dx, dt/dx, etc...
So here are some examples:
Ex 1.) Find dy/dx......x^2+y^2=36
*For these you're going to take the derivative all the way across the equation so for the derivative of 36 you'd get zero
*So once you take the derivative you should get:
2x+2ydy/dx=0
*Now you have to move the dy/dx to the other side of the equal sign so you can solve for it
*So now you have:
2ydy/dx = -2x
*Now divide by 2y and should should get this:
dy/dx = -2x/2y which simplifies to -x/y
Ex. 2) Find dy/dx.....x^8+9x+2xy-y^5=4
*First take the derivative of each term all the way across the equation and you should get:
8x^7+9+2(x)(dy/dx)+(y)(1)-5y^4dy/dx = 0
*Now simplify that and you should have:
8x^7+9+2xdy/dx+2y-5y^4dy/dx = 0
*Now take the dy/dx's and move them to one side of the equal sign and the other terms to the other side
*So you should now have:
2xdy/dx-5ydy/dx = -8x^7-2y-9
*Now divide -8x^7-2y-9 by 2x-5y^4
*And your answer is (-8x^7-2y-9)/(2x-5y^4)
So here are some examples:
Ex 1.) Find dy/dx......x^2+y^2=36
*For these you're going to take the derivative all the way across the equation so for the derivative of 36 you'd get zero
*So once you take the derivative you should get:
2x+2ydy/dx=0
*Now you have to move the dy/dx to the other side of the equal sign so you can solve for it
*So now you have:
2ydy/dx = -2x
*Now divide by 2y and should should get this:
dy/dx = -2x/2y which simplifies to -x/y
Ex. 2) Find dy/dx.....x^8+9x+2xy-y^5=4
*First take the derivative of each term all the way across the equation and you should get:
8x^7+9+2(x)(dy/dx)+(y)(1)-5y^4dy/dx = 0
*Now simplify that and you should have:
8x^7+9+2xdy/dx+2y-5y^4dy/dx = 0
*Now take the dy/dx's and move them to one side of the equal sign and the other terms to the other side
*So you should now have:
2xdy/dx-5ydy/dx = -8x^7-2y-9
*Now divide -8x^7-2y-9 by 2x-5y^4
*And your answer is (-8x^7-2y-9)/(2x-5y^4)
Friday, March 18, 2011
3/18/11
Okay, this week I'll go over the concept of LIMITS: how to find them, when to find them, blah blah blah...
1. The first type of limit you could have is a FINITE limit, which means all they're going to do is give you some type of equation, and they'll say lim x># (meaning, find the limit as x approaches whatever # they give you as x). And all you have to do is plug in that x value into the equation they gave you.
Ex: lim x>2 (4/3x^2 + 6x + 1)
*So all you have to do for this is plug in 2 for x:
4/3(2)^2 + 6(2) + 1
= 55/3
*Now not all finite limit problems will be this easy. Typically, you'll be given some form of a fraction. In some cases, you can just plug in like usual and you'll get an answer, but in other cases, if you plug in first, you'll get "undefined"...which means you'll probably have to factor the equation and cancel and then plug in your x-value.
Ex: lim x>2 (3x^2 - 8x + 4)/(x-2)
*Alright since it says lim x>2, you're going to plug in 2 right?..Well, that doesn't really work for us because we'll get 0 in the denominator, which is undefined. So let's see if we can manipulate this function and then plug in.
*So looking at the numerator, it is factorable. (factors of 12 that add to give you 8 are 6 and 2)..So it'll look like this: (3x^2-6x)-(2x+4)..then once you factor that out you should end up with (x-2)(3x-2) as your simplification
*So let's put that^ as the new numerator of our function:
(x-2)(3x-2)/(x-2)
Now we can cancel x-2 from the top and bottom, and we're left with 3x-2 in the numerator. Now we can plug in 2 for x: 3(2)-2 = 4
2. The next type of limits you'll see are INFINITE limits...where you're given some type of function (pretty much always in fractional form) and it says take the limit as x goes to infinity. Now obviously it's not like finite limits, because you can't just plug in 'infinity' into the equation for x...I mean you can try, but it won't exactly work out for you. There's actually specific guidelines for finding infinite limits: 1. If the degree (exponent) of the top is greater than the degree of the bottom (referring to fractions, duh), then the limit is infinity. 2. If the degree of the top is less than the degree of the bottom, then the limit is 0. 3. If the degree of the top is equal to the degree of the bottom, then the limit is the coefficients of the top and bottom number
Ex: lim x>infinity (3x^2 - 8x + 4)/(x-2)
*Now this is the same problem as before but you do NOT solve it in the same way. All you have to do is look at the degree of the exponents. There's an x^2 in the top and only an x in the bottom; therefore the degree of the top>bottom, so the limit is infinity
Ex: lim x>infinity (3x^2 - 8x + 4)/(7x^2 + 1)
*Again, when taking the limit as x goes to infinity, just look at the exponents. For this one, the highest degree of the top is the same as the highest of the bottom, so the limit will be the coefficients of those variables. So your limit is 3/7
3. Lastly, you could be asked to find the horizontal asymptotes of a particular equation. You do NOT do this in the same way that you find vertical asymptotes, though. All you do is take the limit as x goes to infinity (again, always given a fraction to work with). The only thing that's different is that for your answer, horizontal asymptotes are written in the form "y = # is a horizontal asymptote"..And also, if you get infinity/-infinity for the limit, there are NO horizontal asymptotes.
Ex: The horizontal asymptote(s) of the equation f(x) = (5x + 7)/(2x^2 + 3x) are what?
*So all you do for this problem is take the limit as x approaches infinity (just look for the highest degree of the exponents)
*So x^2 is the highest degree, and it's in the denominator...which means the limit as x approaches infinity is 0
*And y = 0 is a horizontal asymptote
1. The first type of limit you could have is a FINITE limit, which means all they're going to do is give you some type of equation, and they'll say lim x># (meaning, find the limit as x approaches whatever # they give you as x). And all you have to do is plug in that x value into the equation they gave you.
Ex: lim x>2 (4/3x^2 + 6x + 1)
*So all you have to do for this is plug in 2 for x:
4/3(2)^2 + 6(2) + 1
= 55/3
*Now not all finite limit problems will be this easy. Typically, you'll be given some form of a fraction. In some cases, you can just plug in like usual and you'll get an answer, but in other cases, if you plug in first, you'll get "undefined"...which means you'll probably have to factor the equation and cancel and then plug in your x-value.
Ex: lim x>2 (3x^2 - 8x + 4)/(x-2)
*Alright since it says lim x>2, you're going to plug in 2 right?..Well, that doesn't really work for us because we'll get 0 in the denominator, which is undefined. So let's see if we can manipulate this function and then plug in.
*So looking at the numerator, it is factorable. (factors of 12 that add to give you 8 are 6 and 2)..So it'll look like this: (3x^2-6x)-(2x+4)..then once you factor that out you should end up with (x-2)(3x-2) as your simplification
*So let's put that^ as the new numerator of our function:
(x-2)(3x-2)/(x-2)
Now we can cancel x-2 from the top and bottom, and we're left with 3x-2 in the numerator. Now we can plug in 2 for x: 3(2)-2 = 4
2. The next type of limits you'll see are INFINITE limits...where you're given some type of function (pretty much always in fractional form) and it says take the limit as x goes to infinity. Now obviously it's not like finite limits, because you can't just plug in 'infinity' into the equation for x...I mean you can try, but it won't exactly work out for you. There's actually specific guidelines for finding infinite limits: 1. If the degree (exponent) of the top is greater than the degree of the bottom (referring to fractions, duh), then the limit is infinity. 2. If the degree of the top is less than the degree of the bottom, then the limit is 0. 3. If the degree of the top is equal to the degree of the bottom, then the limit is the coefficients of the top and bottom number
Ex: lim x>infinity (3x^2 - 8x + 4)/(x-2)
*Now this is the same problem as before but you do NOT solve it in the same way. All you have to do is look at the degree of the exponents. There's an x^2 in the top and only an x in the bottom; therefore the degree of the top>bottom, so the limit is infinity
Ex: lim x>infinity (3x^2 - 8x + 4)/(7x^2 + 1)
*Again, when taking the limit as x goes to infinity, just look at the exponents. For this one, the highest degree of the top is the same as the highest of the bottom, so the limit will be the coefficients of those variables. So your limit is 3/7
3. Lastly, you could be asked to find the horizontal asymptotes of a particular equation. You do NOT do this in the same way that you find vertical asymptotes, though. All you do is take the limit as x goes to infinity (again, always given a fraction to work with). The only thing that's different is that for your answer, horizontal asymptotes are written in the form "y = # is a horizontal asymptote"..And also, if you get infinity/-infinity for the limit, there are NO horizontal asymptotes.
Ex: The horizontal asymptote(s) of the equation f(x) = (5x + 7)/(2x^2 + 3x) are what?
*So all you do for this problem is take the limit as x approaches infinity (just look for the highest degree of the exponents)
*So x^2 is the highest degree, and it's in the denominator...which means the limit as x approaches infinity is 0
*And y = 0 is a horizontal asymptote
Monday, March 14, 2011
3/13/11
Soo.. still reviewing for AP's and i'll review the 2 theorem's. They have showed up a lot lately on the AP test, so here is the Rolle's theorem and Mean Value theorem. I'll also review the first derivative test since you pretty much need the first derivative test to do both of the theorems.
-Rolle’s Theorem tells you if there’s AT LEAST one max or min, but it doesn’t tell you how many or what they are.
-The Mean Value Theorem tells you that the slope of a secant line equals the slope of a tangent line between a and b.
-The First Derivative Test tells you were on the graph is there a max or min by telling you were the graph is decreasing or increasing. It also finds the max or min.
Next, you need to know when to use them.
But, in order to use Rolle’s Theorem or the Mean Value Theorem, you first need to make sure your graph is continuous and differentiable.
1) If a problem asks “Find all values of c in the given interval such f’(c) = 0”, use Rolle’s Theorem.
*c = x
**Rolle’s Theorem generally uses x-intercepts.
2) If a problem asks “Find all values of c in the open interval [a, b] such that
f’(c) = f(b) – f(a) / b – a”, use the Mean Value Theorem.
***This theorem always uses the formula above.
3) If a problem asks “Find the relative extrema of…” or “Find the open intervals on which the graph is increasing or decreasing”, use the First Derivative Test.
Ex. 1) Let f(x) = x^4 – 2x^2. Find all values of c in the interval [-2, 2] such f’(c) = 0.
1. Is it continuous? Yes
2. Is it differentiable? Yes
The next step for Rolle’s Theorem is to check if the y-values are the same by plugging in the x-values from the interval into the original equation.
3. f(-2) = (-2)^4 – 2(-2)^2 = 8
f(2) = (2)^4 – 2(2)^2 = 8
Are the y-values the same? Yes
Next, take the derivative and set it equal to 0, then solve for x.
4. f’(x) = 4x^3 – 4x
4x^3 – 4x = 0
4x(x^2 – 1) = 0
x = 0, 1, -1
Therefore, c = 0, 1, -1
Finally, justify your answer.
By Rolle’s theorem, the function is continuous, differentiable, and f(-2) = f(2), therefore there is at least one max or min on [-2, 2].
Ex. 2) Given f(x) = 5 – (4/x), find all values of c in the open interval [1, 4] such that f’(c) = f(4) – f(1) / 4 – 1.
1. Is it continuous? Yes
2. Is it differentiable? Yes
The next step of the Mean Value Theorem is to find the slope by plugging into the formula.
3. 5 – (4/4) = 4
5 – (4/1) = 1
4 – 1 = 3
4 – 1 / 3 = 1
Next, take the derivative and set it equal to the slope you just found, then solve for x.
4. f’(x) = 4/x^2
4/x^2 = 1
x = 2, -2
Therefore, c = 2, -2
By the Mean Value Theorem, the function is continuous and differentiable on the interval, therefore there is some value (c = 2) where the derivative equals the slope between 1 and 4.
sorrrrry if this was kind of confusing.. I tried to explain the best I can :)
-Rolle’s Theorem tells you if there’s AT LEAST one max or min, but it doesn’t tell you how many or what they are.
-The Mean Value Theorem tells you that the slope of a secant line equals the slope of a tangent line between a and b.
-The First Derivative Test tells you were on the graph is there a max or min by telling you were the graph is decreasing or increasing. It also finds the max or min.
Next, you need to know when to use them.
But, in order to use Rolle’s Theorem or the Mean Value Theorem, you first need to make sure your graph is continuous and differentiable.
1) If a problem asks “Find all values of c in the given interval such f’(c) = 0”, use Rolle’s Theorem.
*c = x
**Rolle’s Theorem generally uses x-intercepts.
2) If a problem asks “Find all values of c in the open interval [a, b] such that
f’(c) = f(b) – f(a) / b – a”, use the Mean Value Theorem.
***This theorem always uses the formula above.
3) If a problem asks “Find the relative extrema of…” or “Find the open intervals on which the graph is increasing or decreasing”, use the First Derivative Test.
Ex. 1) Let f(x) = x^4 – 2x^2. Find all values of c in the interval [-2, 2] such f’(c) = 0.
1. Is it continuous? Yes
2. Is it differentiable? Yes
The next step for Rolle’s Theorem is to check if the y-values are the same by plugging in the x-values from the interval into the original equation.
3. f(-2) = (-2)^4 – 2(-2)^2 = 8
f(2) = (2)^4 – 2(2)^2 = 8
Are the y-values the same? Yes
Next, take the derivative and set it equal to 0, then solve for x.
4. f’(x) = 4x^3 – 4x
4x^3 – 4x = 0
4x(x^2 – 1) = 0
x = 0, 1, -1
Therefore, c = 0, 1, -1
Finally, justify your answer.
By Rolle’s theorem, the function is continuous, differentiable, and f(-2) = f(2), therefore there is at least one max or min on [-2, 2].
Ex. 2) Given f(x) = 5 – (4/x), find all values of c in the open interval [1, 4] such that f’(c) = f(4) – f(1) / 4 – 1.
1. Is it continuous? Yes
2. Is it differentiable? Yes
The next step of the Mean Value Theorem is to find the slope by plugging into the formula.
3. 5 – (4/4) = 4
5 – (4/1) = 1
4 – 1 = 3
4 – 1 / 3 = 1
Next, take the derivative and set it equal to the slope you just found, then solve for x.
4. f’(x) = 4/x^2
4/x^2 = 1
x = 2, -2
Therefore, c = 2, -2
By the Mean Value Theorem, the function is continuous and differentiable on the interval, therefore there is some value (c = 2) where the derivative equals the slope between 1 and 4.
sorrrrry if this was kind of confusing.. I tried to explain the best I can :)
Saturday, March 12, 2011
Blog #29
Once again, more review.
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.
The Second Fundamental Theorem of Calculus:
This theorem states that the derivative of an integral results in the original equation.
The formula: d/dx vSa f(x) dx = (f(v)) (v’) where “a” is a number (usually equaling 0) and v is an equation
An example:
d/dx x²S0 (x+1)² dx
= (((x²) +1)²) (2x)
= 2x(x²+1)²
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.
The Second Fundamental Theorem of Calculus:
This theorem states that the derivative of an integral results in the original equation.
The formula: d/dx vSa f(x) dx = (f(v)) (v’) where “a” is a number (usually equaling 0) and v is an equation
An example:
d/dx x²S0 (x+1)² dx
= (((x²) +1)²) (2x)
= 2x(x²+1)²
Friday, March 11, 2011
3/11/11
Alright, so for this blog I decided that I'm going to go over the concept of SLOPES and how to find them, when they are needed, etc.
1. So first, and example would be if you are asked to find the "slope of the normal line" and they give you a simple equation like this:
4x^3 + 12x^2 + 5x
and they'll say "at the point x=2"..
*Now in calculus, whenever you hear the word "slope", you're typically going to have to take the derivative of something. And in this case you will have to derive the equation they gave you and then plug in 2 for x to get the slope.
*So once you derive the equation you should get 12x^2 + 24x + 5
*Now plug in 2: 12(2)^2 + 24(2) + 5
*And once you simplify that, you should get 101 as your slope.
***Buttttttt!..since they asked for the slope of the NORMAL line, that means you have to take the negative reciprocal of that number you just got
*So the slope of the NORMAL line is -1/101
2. The next thing you could be asked to find is "the slope of the tangent line" for a certain equation...and you'll be given an x-value to plug in again here.
*You find this the same way as you did above^ except you keep the answer the way it was...As in, 101 would be the slope of the tangent line, while -1/101 is the slope of the normal line
3. Another time you'll have to find the slope is if you're asked to find "the average rate of change" or "average velocity"..Those are key words just to find the slope using this formula: f(b)-f(a)/b-a...And typically you'll just be given an interval [#,#] to work with.
Ex. Find the average rate of change of a particle for f(x)=3x^2+5x during the time interval [1,4]
*All you have to do for this is simply plug into the formula:
f(4)-f(1)/4-1
(68-8)/3
*slope = 20
4. Another time you'll need to find the slope is when in a problem you're asked to apply the Mean Value Theorem for an equation and a time interval..Remember for the Mean Value Theorem, you have to find the slope and then set it equal to the derivative of the equation and solve to find a value of c
5. You may also be asked to "find the slope of the horizontal tangent" for a certain equation...but, do not be fooled by the "slope" in the question! It does NOT mean you have to find the slope! "Slope of the horizontal tangent" simply means to take the derivative of the equation they gave you, set it equal to zero, and solve for x.
Ex. Find the slope of the horizontal tangent for f(x) = (6x^2+3x)/(5x+1)
*Oh, a fraction...the same rules above still apply here! Just use the quotient rule to differentiate. And you should get this:
(30x^2+12x+3)/(5x+1)^2
Now to find the slope of the horizontal tangent, all you have to do is set the top of the fraction equal to zero and solve for x
30x^2+12x+3 = 0
..And it doesn't factor..but you get the point on what to do haha
1. So first, and example would be if you are asked to find the "slope of the normal line" and they give you a simple equation like this:
4x^3 + 12x^2 + 5x
and they'll say "at the point x=2"..
*Now in calculus, whenever you hear the word "slope", you're typically going to have to take the derivative of something. And in this case you will have to derive the equation they gave you and then plug in 2 for x to get the slope.
*So once you derive the equation you should get 12x^2 + 24x + 5
*Now plug in 2: 12(2)^2 + 24(2) + 5
*And once you simplify that, you should get 101 as your slope.
***Buttttttt!..since they asked for the slope of the NORMAL line, that means you have to take the negative reciprocal of that number you just got
*So the slope of the NORMAL line is -1/101
2. The next thing you could be asked to find is "the slope of the tangent line" for a certain equation...and you'll be given an x-value to plug in again here.
*You find this the same way as you did above^ except you keep the answer the way it was...As in, 101 would be the slope of the tangent line, while -1/101 is the slope of the normal line
3. Another time you'll have to find the slope is if you're asked to find "the average rate of change" or "average velocity"..Those are key words just to find the slope using this formula: f(b)-f(a)/b-a...And typically you'll just be given an interval [#,#] to work with.
Ex. Find the average rate of change of a particle for f(x)=3x^2+5x during the time interval [1,4]
*All you have to do for this is simply plug into the formula:
f(4)-f(1)/4-1
(68-8)/3
*slope = 20
4. Another time you'll need to find the slope is when in a problem you're asked to apply the Mean Value Theorem for an equation and a time interval..Remember for the Mean Value Theorem, you have to find the slope and then set it equal to the derivative of the equation and solve to find a value of c
5. You may also be asked to "find the slope of the horizontal tangent" for a certain equation...but, do not be fooled by the "slope" in the question! It does NOT mean you have to find the slope! "Slope of the horizontal tangent" simply means to take the derivative of the equation they gave you, set it equal to zero, and solve for x.
Ex. Find the slope of the horizontal tangent for f(x) = (6x^2+3x)/(5x+1)
*Oh, a fraction...the same rules above still apply here! Just use the quotient rule to differentiate. And you should get this:
(30x^2+12x+3)/(5x+1)^2
Now to find the slope of the horizontal tangent, all you have to do is set the top of the fraction equal to zero and solve for x
30x^2+12x+3 = 0
..And it doesn't factor..but you get the point on what to do haha
Sunday, March 6, 2011
Blog #28
Well, this week we had exams. (The Calc exam was actually really easy!) But anyway, I’m just going to review cross-sections and substitution since those two things seem to be what I trip up on the most while doing the practice AP’s.
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
Next, Substitution:
This is used when you have a product or quotient rule. This can’t always be used, but these are the instances in which it can:
S u x v
…or…
S u/v
where v is the derivative of u
Ex. 2) S (x²+1)(2x) dx
First, identify u and v. Which is the derivative and which is the original?
u = x²+1
v = 2x
…because 2x is the derivative of x²+1…
Okay, next you remove v (2x dx) from the equation and substitute “u” for x²+1.
S u du
*You insert du because v is the derivative of u, hence du.
Now, integrate u.
½ u² + C
Finally, replace “u” with the original.
½ (x²+1)² + C
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
Next, Substitution:
This is used when you have a product or quotient rule. This can’t always be used, but these are the instances in which it can:
S u x v
…or…
S u/v
where v is the derivative of u
Ex. 2) S (x²+1)(2x) dx
First, identify u and v. Which is the derivative and which is the original?
u = x²+1
v = 2x
…because 2x is the derivative of x²+1…
Okay, next you remove v (2x dx) from the equation and substitute “u” for x²+1.
S u du
*You insert du because v is the derivative of u, hence du.
Now, integrate u.
½ u² + C
Finally, replace “u” with the original.
½ (x²+1)² + C
3/6/11
What to doooo, what to do...Guess I'll just continue giving example problems.
Ex. 1) What is the x-coordinate of the point of inflection on the graph of y=1/3x^3+5x^2+24 ?
*Okay, first let's identify our key words in this problem and see what we actually have to work with
*POINT OF INFLECTION is the key word..which means 2nd derivative! So all you have to do for this problem is follow the guidelines for the 2nd derivative test which says you take the 2nd dervative, set it equal to zero, solve for x, set up intervals, and check numbers on each interval to see where the interval is negative/positive. If the signs change, that means there IS in fact a point of inflection. If the signs DO NOT change, guess what, there's NO point of inflection
*So when you take the 1st derivative of the equation they gave you, you should get:
x^2 + 10x
*And when you take the 2nd derivative you should get 2x+10
*Set it equal to zero and solve for x:
2x+10=0
2x=-10
x=-5
*Set up intervals:
(-infinity, -5)u(-5, infinity)
*for the 1st interval you can plug in -6 and you should get a negative number; for the 2nd interval you can plug in 0 and you should get a positive number; therforeeeeee, there IS a point of inflection at x = -5, so that's your answer!
Ex. 2) If x^2 + xy = 10, then when x = 2, dy/dx = ?
*Again, let's first find our key word(s)...which is DY/DX!!!! And all that means is they want you to take the derivative of the equation they gave you (using implicit differentiation, by the way, because there is a y in the equation) And thennnnn, they want you to plug in 2 for x afterwards
*So first let's differentiate this problem..You should have this before you simplify:
2x + (x)(dy/dx)+(y)(1) = 0
*Then simplifying that all the way you should wind up with this:
dy/dx = (-2x-y)/x
*Now in order to plug back into that equation you have to get a y-value also..because you were only given an x-value. So to do that, you plug your x, 2, into the original equation to get the y-value:
(2)^2 + 2y = 10
4 + 2y = 10
y = 3
*Now plug your x and y-values into the dy/dx equation above:
[-2(2)-3]/(2)
=(-4-3)/2
= -7/2 and that's the answer they wanted
Ex. 1) What is the x-coordinate of the point of inflection on the graph of y=1/3x^3+5x^2+24 ?
*Okay, first let's identify our key words in this problem and see what we actually have to work with
*POINT OF INFLECTION is the key word..which means 2nd derivative! So all you have to do for this problem is follow the guidelines for the 2nd derivative test which says you take the 2nd dervative, set it equal to zero, solve for x, set up intervals, and check numbers on each interval to see where the interval is negative/positive. If the signs change, that means there IS in fact a point of inflection. If the signs DO NOT change, guess what, there's NO point of inflection
*So when you take the 1st derivative of the equation they gave you, you should get:
x^2 + 10x
*And when you take the 2nd derivative you should get 2x+10
*Set it equal to zero and solve for x:
2x+10=0
2x=-10
x=-5
*Set up intervals:
(-infinity, -5)u(-5, infinity)
*for the 1st interval you can plug in -6 and you should get a negative number; for the 2nd interval you can plug in 0 and you should get a positive number; therforeeeeee, there IS a point of inflection at x = -5, so that's your answer!
Ex. 2) If x^2 + xy = 10, then when x = 2, dy/dx = ?
*Again, let's first find our key word(s)...which is DY/DX!!!! And all that means is they want you to take the derivative of the equation they gave you (using implicit differentiation, by the way, because there is a y in the equation) And thennnnn, they want you to plug in 2 for x afterwards
*So first let's differentiate this problem..You should have this before you simplify:
2x + (x)(dy/dx)+(y)(1) = 0
*Then simplifying that all the way you should wind up with this:
dy/dx = (-2x-y)/x
*Now in order to plug back into that equation you have to get a y-value also..because you were only given an x-value. So to do that, you plug your x, 2, into the original equation to get the y-value:
(2)^2 + 2y = 10
4 + 2y = 10
y = 3
*Now plug your x and y-values into the dy/dx equation above:
[-2(2)-3]/(2)
=(-4-3)/2
= -7/2 and that's the answer they wanted
3/6/11
Okay for this blog I'm going to review implicit differentiation-and all that means is to take the derivative of an equation with respect to other variables (like y) besides x. So that would make it dy/dx, dt/dx, etc...
So here are some examples:
Ex 1.) Find dy/dx......x^2+y^2=36
*For these you're going to take the derivative all the way across the equation so for the derivative of 36 you'd get zero
*So once you take the derivative you should get:
2x+2ydy/dx=0
*Now you have to move the dy/dx to the other side of the equal sign so you can solve for it
*So now you have:
2ydy/dx = -2x
*Now divide by 2y and should should get this:
dy/dx = -2x/2y which simplifies to -x/y
Ex. 2) Find dy/dx.....x^8+9x+2xy-y^5=4
*First take the derivative of each term all the way across the equation and you should get:
8x^7+9+2(x)(dy/dx)+(y)(1)-5y^4dy/dx = 0
*Now simplify that and you should have:
8x^7+9+2xdy/dx+2y-5y^4dy/dx = 0
*Now take the dy/dx's and move them to one side of the equal sign and the other terms to the other side
*So you should now have:
2xdy/dx-5ydy/dx = -8x^7-2y-9
*Now divide -8x^7-2y-9 by 2x-5y^4
*And your answer is (-8x^7-2y-9)/(2x-5y^4)
That's ittt, oh and HAPPY MARDI GRAS EVERYONE:D
So here are some examples:
Ex 1.) Find dy/dx......x^2+y^2=36
*For these you're going to take the derivative all the way across the equation so for the derivative of 36 you'd get zero
*So once you take the derivative you should get:
2x+2ydy/dx=0
*Now you have to move the dy/dx to the other side of the equal sign so you can solve for it
*So now you have:
2ydy/dx = -2x
*Now divide by 2y and should should get this:
dy/dx = -2x/2y which simplifies to -x/y
Ex. 2) Find dy/dx.....x^8+9x+2xy-y^5=4
*First take the derivative of each term all the way across the equation and you should get:
8x^7+9+2(x)(dy/dx)+(y)(1)-5y^4dy/dx = 0
*Now simplify that and you should have:
8x^7+9+2xdy/dx+2y-5y^4dy/dx = 0
*Now take the dy/dx's and move them to one side of the equal sign and the other terms to the other side
*So you should now have:
2xdy/dx-5ydy/dx = -8x^7-2y-9
*Now divide -8x^7-2y-9 by 2x-5y^4
*And your answer is (-8x^7-2y-9)/(2x-5y^4)
That's ittt, oh and HAPPY MARDI GRAS EVERYONE:D
Monday, February 28, 2011
blogety
ok so this week we reviewed for the ap like the bosses we are so i get to talk about quotient rule. yay me! so you use quotient rule wen finding the derivative of a fraction. you start by taking the bottom and multiplying it by the derivative of the top and then subtracting that from the derivative of the bottom multiplied by the top. all that is then over the bottom squared.
Sunday, February 27, 2011
2/27
So we did APs again this week as well as for the rest of the year to prep for THE BIG ONE.
Basically, the free response consists of certain questions:
1. Derivative work
2. Integral work
3. "Physics" (PVA)
Key words to look for!
Antiderivative, area, and volume all involve integration
If the "graph of f'" is shown in the problem, you may need to predict where f or f'' is
make sure if youre integrating a fraction to see if its a natural log
Its all a review of the first few chapters we covered
keep reviewing and you'll get a 3 and up!
I got a 2 last time and im pretty proud of that
Basically, the free response consists of certain questions:
1. Derivative work
2. Integral work
3. "Physics" (PVA)
Key words to look for!
Antiderivative, area, and volume all involve integration
If the "graph of f'" is shown in the problem, you may need to predict where f or f'' is
make sure if youre integrating a fraction to see if its a natural log
Its all a review of the first few chapters we covered
keep reviewing and you'll get a 3 and up!
I got a 2 last time and im pretty proud of that
Blog #27
Well, more AP tests…so, more review! This time, I’m reviewing Antiderivatives (or 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.
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 sinx dx = -cosx
Ex. 5) S x^1/2 dx = (1/2) + 1 = 3/2 = 2x^(3/2)/3 + c
Ex. 6) S 3x^2 dx = x^3 + c
Ex. 7) Find the integral of f’(x) = 1/x^2 that satisfies the initial f(1) = 0.
S x^-2 dx = (1/-1)(x^-1) = -1/x + c
-1/1 + c = 0
-1 + c = 0
c = 1
-1/x + 1
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.
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 sinx dx = -cosx
Ex. 5) S x^1/2 dx = (1/2) + 1 = 3/2 = 2x^(3/2)/3 + c
Ex. 6) S 3x^2 dx = x^3 + c
Ex. 7) Find the integral of f’(x) = 1/x^2 that satisfies the initial f(1) = 0.
S x^-2 dx = (1/-1)(x^-1) = -1/x + c
-1/1 + c = 0
-1 + c = 0
c = 1
-1/x + 1
2/27/11
Okay, sooo for some reason for the most recent practice AP's I forgot how to do chain rule.. really weird I know. So I went back and looked and how to do it and I guess i'll explain it now.
CHAIN RULE:
You use the chain rule when you have a function inside of a function, f(g(x)).
Chain rule is way different that the product and quotient rule, however it can be used in both of the rules.
Heres an example of just the chain rule by itself:
Ex. (x^2 + 3)^4
First, take the exponent of the entire function, bring it to the front, then subtract one from the exponent, and then just copy the inside.
4(x^2 + 3)^3
Now, multiply that by the derivative of the inside equation.
4(x^2 + 3)^3 (2x)
Put in simplest terms.
= 8x(x^2 + 3)^3
You can also use the chain rule with the product rule or quotient rule. If this is the case, then you first need to determine whether the chain rule is on the outside or the inside; this is to figure out what order you do things in. If the chain rule is on the outside, you do that first. If not, you do the product rule or quotient rule first.
Example with product rule and chain rule:
Ex. (x + 3)(x^2 – 1)^2
In this problem, the product rule comes first. When taking the derivative of the second equation, use the chain rule.
(x + 3)(2(x^2 – 1)(2x)) + (x^2 – 1)^2 (1)
= 4x(x + 3)(x^2 – 1) + (x^2 – 1)^2
Example with quotient rule and chain rule:
Ex. ((x + 2)/(x – 1))^2
For this problem, the chain rule is on the outside, therefore it goes first.
2((x + 2)/(x – 1)) (((x – 1)(1) – (x + 2)(1)) / (x – 1)^2))
2((x + 2)/(x – 1)) ((x – 1 – x – 2) / (x – 1)^2)
2((x + 2)/(x – 1)) ((-3)/(x – 1)^2)
= -3(2x + 4) / (x – 1)^3
CHAIN RULE:
You use the chain rule when you have a function inside of a function, f(g(x)).
Chain rule is way different that the product and quotient rule, however it can be used in both of the rules.
Heres an example of just the chain rule by itself:
Ex. (x^2 + 3)^4
First, take the exponent of the entire function, bring it to the front, then subtract one from the exponent, and then just copy the inside.
4(x^2 + 3)^3
Now, multiply that by the derivative of the inside equation.
4(x^2 + 3)^3 (2x)
Put in simplest terms.
= 8x(x^2 + 3)^3
You can also use the chain rule with the product rule or quotient rule. If this is the case, then you first need to determine whether the chain rule is on the outside or the inside; this is to figure out what order you do things in. If the chain rule is on the outside, you do that first. If not, you do the product rule or quotient rule first.
Example with product rule and chain rule:
Ex. (x + 3)(x^2 – 1)^2
In this problem, the product rule comes first. When taking the derivative of the second equation, use the chain rule.
(x + 3)(2(x^2 – 1)(2x)) + (x^2 – 1)^2 (1)
= 4x(x + 3)(x^2 – 1) + (x^2 – 1)^2
Example with quotient rule and chain rule:
Ex. ((x + 2)/(x – 1))^2
For this problem, the chain rule is on the outside, therefore it goes first.
2((x + 2)/(x – 1)) (((x – 1)(1) – (x + 2)(1)) / (x – 1)^2))
2((x + 2)/(x – 1)) ((x – 1 – x – 2) / (x – 1)^2)
2((x + 2)/(x – 1)) ((-3)/(x – 1)^2)
= -3(2x + 4) / (x – 1)^3
Alaina's blog, 27 February 2011
So most of this week, we focused on our AP tests, which I like because its good practice for the real thing. But I'm having some problems with the APs.
First, sometimes the way the questions on the free response are worded, I don't exactly know what to do. So, for example, when a question tells you that g(x)=f '(x) and h(x)=g(x), and says to find g(x). What are you supposed to do?
So I guess I'll explain explain area if only one equation is given (area between the graph and the x/y axis) and the area between two equations.
a) area under the graph: you integrate the equation and use the Fundamental Theorem of Calculus, plugging in the two given x-values. F(b)-F(a) where b is the larger number and a is the smaller one.
B) area between two graphs: 1st: if the bounds are not given, set the equations equal and solve for the variable.
If you are not allowed a calculator, plug into the formula integral from a to b of top-bottom. Once you integrate, use the Fundamental Theorem of Calculus and plug in end points.
If you are allowed a calculator, graph the two equations in the calculator and use the intersect function.
These concepts are fairly simple, yet I struggle with them when it comes down to working problems involving them on the AP.
So I'm studying my key words sheets for the exammm! Good luck everyone!
First, sometimes the way the questions on the free response are worded, I don't exactly know what to do. So, for example, when a question tells you that g(x)=f '(x) and h(x)=g(x), and says to find g(x). What are you supposed to do?
So I guess I'll explain explain area if only one equation is given (area between the graph and the x/y axis) and the area between two equations.
a) area under the graph: you integrate the equation and use the Fundamental Theorem of Calculus, plugging in the two given x-values. F(b)-F(a) where b is the larger number and a is the smaller one.
B) area between two graphs: 1st: if the bounds are not given, set the equations equal and solve for the variable.
If you are not allowed a calculator, plug into the formula integral from a to b of top-bottom. Once you integrate, use the Fundamental Theorem of Calculus and plug in end points.
If you are allowed a calculator, graph the two equations in the calculator and use the intersect function.
These concepts are fairly simple, yet I struggle with them when it comes down to working problems involving them on the AP.
So I'm studying my key words sheets for the exammm! Good luck everyone!
Stephen Ledbetter Blog Post 2/27/2011
This week went by pretty fast but the days seemed to just drag on and on and on. I'm going to explain how to work on integration by substitution.
If you have the integral of u * v or the integral of u/v, where v is the derivative of u, v will disappear.
Example:
the integral of (x^2 + 1)^2 (2x) dx
u = x^2 + 1
du = 2x dx
plug in to...
the integral of u^2 du
1/3u^3 + c
1/3(x^2 + 1)^3 + c
the integral of 5 cos 5x dx
u = 5x
du = 5 dx
plug in to...
the integral of cos u du
-sin u + c
-sin 5x + c
e^x
d/dx e^(x^3) = e^(x^3) * 3x^2
the integral of e^2x = e^2x/2 = 1/2 e^2x + c
the integral of x^2 e^(x^3) dx
u = x^3
du = 3x^2
1/3 * the integral of 3x^2 e^(x^3) dx
1/3 * the integral of e^u du
1/3 e^u/1 + c
1/3 e^(x^3) + c
blog of chaos
well, let's try to avoid talking about last week for now.....
anyways, almost didn't get this blog done, RTC has faulty hardware, and my internet's out for a while
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.
anyways, almost didn't get this blog done, RTC has faulty hardware, and my internet's out for a while
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.
Saturday, February 26, 2011
BLOGGG
soooo RAM RAM RAM RAM RAM RAM A BAMMER BAM BAM
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.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.
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.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.
2/26/11
Okayyyy, so for this blog I'll just go over some problems that we normally see on the multiple choice no calculator section..
Ex. 1) 1/2 S e^t/2 dt =
*(1/2 times the integral of e^t/2)
*Okay so to solve this integral we must first integrate e..but remember to leave the number, 1/2, out in front until the end
*So to integrate e^t/2, it's going to be e^t/2 / 1/2...where 1/2 is the derivative of the exponent of e
*So you should have this 1/2[e^t/2 / 1/2], which simplifies to:
1/2[2e^t/2]
*And distributing the 1/2 through you get e^t/2 + c...*don't forget the +c because you weren't given bounds for this integral
Ex. 2) At what point on the graph of y=1/2x^2 is the tangent line parallel to the line 2x-4y=3?
*First, since they say "tangent line" you're going to have to derive the 1st equation they gave you, and you should get that y'=x
*And in order to find where the graph of that^ is parallel to the line (2nd equ.) you must find the slope of the 2nd equation and then set that slope and the derivative you just found equal to each other
*So let's find the slope of 2x-4y=3
-first subtract the 2x over to get -4y=-2x+3
-now divide 4 by everything and you should get y=1/2x-3/4, which means your slope is 1/2
*Now set your derivative equal to your slope:
x=1/2 (so that's also your x-value...to find your y-value you must plug the x-value into the original equation y=1/2x^2..so you get that y=1/8
*Since you found your x and y, now you have a point!
(1/2, 1/8) is your final answer
Ex. 3) The average value of cos x on the interval [-3,5] is
*AVERAGE VALUE is the key word in this problem! All you do for this problem is simply plug into a formula..which is this:
1/b-a Sa,b f(x)...where -3 is a and 5 is b
*So plugging in you get: 1/5-(-3) S-3,5 cos x dx
= 1/8 [sinx] l-3,5 ...where the "l" is the bar
Fundamental Theorem of Calculus:
1/8 sin(5) - [1/8 sin(-3)]
(*Since the negative for -3 would become -sin(3) that means the subtraction sign would become a + sign)
*So now you have 1/8 sin(5) + 1/8 sin(3) and that's your final answer..or you could put it as a fraction like this:
(sin5 + sin3)/8
Ex. 1) 1/2 S e^t/2 dt =
*(1/2 times the integral of e^t/2)
*Okay so to solve this integral we must first integrate e..but remember to leave the number, 1/2, out in front until the end
*So to integrate e^t/2, it's going to be e^t/2 / 1/2...where 1/2 is the derivative of the exponent of e
*So you should have this 1/2[e^t/2 / 1/2], which simplifies to:
1/2[2e^t/2]
*And distributing the 1/2 through you get e^t/2 + c...*don't forget the +c because you weren't given bounds for this integral
Ex. 2) At what point on the graph of y=1/2x^2 is the tangent line parallel to the line 2x-4y=3?
*First, since they say "tangent line" you're going to have to derive the 1st equation they gave you, and you should get that y'=x
*And in order to find where the graph of that^ is parallel to the line (2nd equ.) you must find the slope of the 2nd equation and then set that slope and the derivative you just found equal to each other
*So let's find the slope of 2x-4y=3
-first subtract the 2x over to get -4y=-2x+3
-now divide 4 by everything and you should get y=1/2x-3/4, which means your slope is 1/2
*Now set your derivative equal to your slope:
x=1/2 (so that's also your x-value...to find your y-value you must plug the x-value into the original equation y=1/2x^2..so you get that y=1/8
*Since you found your x and y, now you have a point!
(1/2, 1/8) is your final answer
Ex. 3) The average value of cos x on the interval [-3,5] is
*AVERAGE VALUE is the key word in this problem! All you do for this problem is simply plug into a formula..which is this:
1/b-a Sa,b f(x)...where -3 is a and 5 is b
*So plugging in you get: 1/5-(-3) S-3,5 cos x dx
= 1/8 [sinx] l-3,5 ...where the "l" is the bar
Fundamental Theorem of Calculus:
1/8 sin(5) - [1/8 sin(-3)]
(*Since the negative for -3 would become -sin(3) that means the subtraction sign would become a + sign)
*So now you have 1/8 sin(5) + 1/8 sin(3) and that's your final answer..or you could put it as a fraction like this:
(sin5 + sin3)/8
Sunday, February 20, 2011
aps are ca-razy
well, all we've done this week were ap reviews and more practice aps, but at least the last one wasn't as bad as the others
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.
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.
blogtiem
so this week we did a bunch of stuff for the AP which would mean that its review tiem. i believe ill go with this weeks popular choice LRAM and RRAM. quick sidenote the average of the two comes to TRAM. anyram here we go. first of all you usually use LRAM and RRAM when you are given several coordinate points and have to find an integral or a missing coordinate point. you start off by dividing the area into sections (usually using the coordinate points given). the y-values become your height and the length between the x-values become the width of each subsection. for LRAM you add all of the y-values together excluding the one furthest to the right and for RRAM you add up all the y-values excluding the one furthest left. you then multiply your answer by the width of your subsections. and shazzam the rabbit was in the hat the whole tiem.
Blog #26
This week, we continued with the practice AP tests, so 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
alaina's blog, 20 Feb 2011
I'm just going to review LRAM, RRAM, MRAM, and TRAP and 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.
2/20/11
I think i'll go over definite integrals since I saw a couple of problems like that on the practice AP's. Sooo, 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. 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.
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.
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.
2/20
So we've been doing practice AP's lately and I got a 2 on the last one which is good by my standards. I learned that the free response is a point fest when it comes to just showing that you know how to set it up.
for example, if youre confused by a certain problem: find the key words in it (most likely a derivative or an integral) then set it up (free point). If you know what to do after setting up, carry it out for 1 to 2 more points.
The calculator portion is where I'm sort of weak at at the moment, along with the free response, but after some binder assignments and extra practice I can easily get a 3 :D
for example, if youre confused by a certain problem: find the key words in it (most likely a derivative or an integral) then set it up (free point). If you know what to do after setting up, carry it out for 1 to 2 more points.
The calculator portion is where I'm sort of weak at at the moment, along with the free response, but after some binder assignments and extra practice I can easily get a 3 :D
Saturday, February 19, 2011
2/19/11
So this week we continued with practice AP's; I'll just give a few example problems like those on the multiple choice section..
Ex. 1) If f(x) = -x^3 + x + 1/x, then f'(-1) =
*So first let's determine what we're given in this problem.
*They give you the original equation, and they're asking for f'(-1)..which means they want you to take the derivative and then plug in -1 for x after.
*So first let's take the derivative and you should have this:
-3x^2 + 1 - 1/x^2
*Now all you do is plug in -1 for x:
-3(-1)^2 + 1 - 1/(-1)^2
= -3
Ex. 2) d/dx cos^2(x^3)
*So when you first take a look at this problem, you see d/dx, which means all you have to do is differentiate it
*So since this is a chain rule, first you have to derive the outside, cos^2, so you should first get this:
2(cos(x^3))
*Buttt, you're not done yet; next you have to derive cos and then recopy the rest of the equation. So now you should have this:
2(cos(x^3)) * sin(x^3)
*Last, multiply that by the derivative of the inside, x^3, which is 3x^2
*So now you have 2(cos(x^3)) * sin(x^3) * 3x^2
*Now simplifying all of that, your final answer should be:
6x^2cos(x^3)sin(x^3)
Ex. 3) The area of the region enclosed by the graph of y = x^2 + 1 and the line y = 5 is....
*Alright they're asking for AREA so that means you're going to use some kind of INTEGRAL..
*First, let's figure out what our graphs look like...y=x^2+1 is a parabola with vertex (0,1) and y=5 is a horizontal line at the point (0,5)
*So looking at the two graphs on the same coordinate plane, the line y=5 is on the top and the parabola is on the bottom, and you're looking for the area in between.
So you're going to have to use the formula: S top-bottom
*Plugging into the formula you get:
S 5-x^2-1
= S 4-x^2
Now integrating that^ you get:
4x - 1/3x^3....but we still have to find the bounds!
So to do that all you do is take the two equations and set them equal and solve for x:
x^2 + 1 = 5
x^2 = 4
x = -2,2 ...and those are your bounds-- -2 being the lower bound and 2 being the upper bound
*So now you have this:
[4x - 1/3x^3] I-2,2 ...the "I" being the bar (meaning you didn't plug in yet)
*Now Fundamental Theorem of Calculus!
4(2)-1/3(2)^3-[4(-2)-1/3(-2)^3]
Simplifying that you should end up with 32/3 and that's your area
Ex. 1) If f(x) = -x^3 + x + 1/x, then f'(-1) =
*So first let's determine what we're given in this problem.
*They give you the original equation, and they're asking for f'(-1)..which means they want you to take the derivative and then plug in -1 for x after.
*So first let's take the derivative and you should have this:
-3x^2 + 1 - 1/x^2
*Now all you do is plug in -1 for x:
-3(-1)^2 + 1 - 1/(-1)^2
= -3
Ex. 2) d/dx cos^2(x^3)
*So when you first take a look at this problem, you see d/dx, which means all you have to do is differentiate it
*So since this is a chain rule, first you have to derive the outside, cos^2, so you should first get this:
2(cos(x^3))
*Buttt, you're not done yet; next you have to derive cos and then recopy the rest of the equation. So now you should have this:
2(cos(x^3)) * sin(x^3)
*Last, multiply that by the derivative of the inside, x^3, which is 3x^2
*So now you have 2(cos(x^3)) * sin(x^3) * 3x^2
*Now simplifying all of that, your final answer should be:
6x^2cos(x^3)sin(x^3)
Ex. 3) The area of the region enclosed by the graph of y = x^2 + 1 and the line y = 5 is....
*Alright they're asking for AREA so that means you're going to use some kind of INTEGRAL..
*First, let's figure out what our graphs look like...y=x^2+1 is a parabola with vertex (0,1) and y=5 is a horizontal line at the point (0,5)
*So looking at the two graphs on the same coordinate plane, the line y=5 is on the top and the parabola is on the bottom, and you're looking for the area in between.
So you're going to have to use the formula: S top-bottom
*Plugging into the formula you get:
S 5-x^2-1
= S 4-x^2
Now integrating that^ you get:
4x - 1/3x^3....but we still have to find the bounds!
So to do that all you do is take the two equations and set them equal and solve for x:
x^2 + 1 = 5
x^2 = 4
x = -2,2 ...and those are your bounds-- -2 being the lower bound and 2 being the upper bound
*So now you have this:
[4x - 1/3x^3] I-2,2 ...the "I" being the bar (meaning you didn't plug in yet)
*Now Fundamental Theorem of Calculus!
4(2)-1/3(2)^3-[4(-2)-1/3(-2)^3]
Simplifying that you should end up with 32/3 and that's your area
Sunday, February 13, 2011
blog? blog. *serious face* BLOG.
duuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuude, we've learned nothing new lately, which is awesome
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
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
Blog #25
Well, this week we just went over the AP practice test, so I’m just going to review a few things that were on the AP tests.
First, 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”.
Ex. 1) Find the area of the shape between the boundaries given.
2S0 (x + 1) dx
F(x) = ½ x² + x
Area = (½ (2)² + (2)) – (½ (0)² + (0)) = 4 – 0
= 4
Next, Cross-Sections:
Ex. 2) 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
Finally, Substitution:
This is used when you have a product or quotient rule. This can’t always be used, but these are the instances in which it can:
S u x v
…or…
S u/v
where v is the derivative of u
Ex. 3) S (x²+1)(2x) dx
First, identify u and v. Which is the derivative and which is the original?
u = x²+1
v = 2x
…because 2x is the derivative of x²+1…
Okay, next you remove v (2x dx) from the equation and substitute “u” for x²+1.
S u du
*You insert du because v is the derivative of u, hence du.
Now, integrate u.
½ u² + C
Finally, replace “u” with the original.
½ (x²+1)² + C
First, 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”.
Ex. 1) Find the area of the shape between the boundaries given.
2S0 (x + 1) dx
F(x) = ½ x² + x
Area = (½ (2)² + (2)) – (½ (0)² + (0)) = 4 – 0
= 4
Next, Cross-Sections:
Ex. 2) 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
Finally, Substitution:
This is used when you have a product or quotient rule. This can’t always be used, but these are the instances in which it can:
S u x v
…or…
S u/v
where v is the derivative of u
Ex. 3) S (x²+1)(2x) dx
First, identify u and v. Which is the derivative and which is the original?
u = x²+1
v = 2x
…because 2x is the derivative of x²+1…
Okay, next you remove v (2x dx) from the equation and substitute “u” for x²+1.
S u du
*You insert du because v is the derivative of u, hence du.
Now, integrate u.
½ u² + C
Finally, replace “u” with the original.
½ (x²+1)² + C
2/13/11
Alright, sooo were still doing AP stuff.. with nothing new learned I think i'm just gonna review some old topics. I know questions similar to these showed up on the practice AP's we took, and I forgot how to do them. So i'll explain it and give some examples.
EXAMPLES OF SLOPE:
1. Find the slope m of the line tangent to the graph of the function g(x) = 5 - x^2 at the point (2,1).
*To do this, you must find the derivative, then plug in your x-value.
Derivative = -2x
Plug in: -2 (2) = -4
The slope is -4
2. Find the slope of the graph of the function at the given value.
f(x) = 2 (4x + 6)^2 when x = 5
*Find the derivative and then plug in.
Derivative:
2 * (2 (4x + 6)) * 4
16(4x + 6)
16(4(5) + 6)
16(20 + 6)
16 (26) = 416
416 is the slope of the graph.
EXAMPLE: Mean Value Theorem
1. Check if the function is continuous.
2. Check for differentiability or non-differentiability.
3. Slope
f ' (c) = f(b) - f(a) / b - a
* It means on an interval [a,b]. There must be some x-value where the derivative = the slope between a & b.
slope of a secant line = slope of a tangent line between a & b
1. Given f(x) = 5 - 4/x, find all values of c in the open interval (1,4) such that
f ' (c) = f(4) - f(1) / 4-1
yes it is both continuous and differentiable
= 4-1/4-1 = 1
5 - 4x^-1
4/x^2 = 1
4 = x^2
x = +/- 2
c = +/- 2
*By the mean value theorem, the function is continuous and differentiable, therefore some value c=2 where f ' (x) = the slope between (1,4).
EXAMPLES OF SLOPE:
1. Find the slope m of the line tangent to the graph of the function g(x) = 5 - x^2 at the point (2,1).
*To do this, you must find the derivative, then plug in your x-value.
Derivative = -2x
Plug in: -2 (2) = -4
The slope is -4
2. Find the slope of the graph of the function at the given value.
f(x) = 2 (4x + 6)^2 when x = 5
*Find the derivative and then plug in.
Derivative:
2 * (2 (4x + 6)) * 4
16(4x + 6)
16(4(5) + 6)
16(20 + 6)
16 (26) = 416
416 is the slope of the graph.
EXAMPLE: Mean Value Theorem
1. Check if the function is continuous.
2. Check for differentiability or non-differentiability.
3. Slope
f ' (c) = f(b) - f(a) / b - a
* It means on an interval [a,b]. There must be some x-value where the derivative = the slope between a & b.
slope of a secant line = slope of a tangent line between a & b
1. Given f(x) = 5 - 4/x, find all values of c in the open interval (1,4) such that
f ' (c) = f(4) - f(1) / 4-1
yes it is both continuous and differentiable
= 4-1/4-1 = 1
5 - 4x^-1
4/x^2 = 1
4 = x^2
x = +/- 2
c = +/- 2
*By the mean value theorem, the function is continuous and differentiable, therefore some value c=2 where f ' (x) = the slope between (1,4).
Saturday, February 12, 2011
2/12/11
Soooooo...time for another blog I suppose. About what? Nothing specific really..All this week we just took practice AP's and did work with that. But anyway, I think I'll just give a few examples of some recent things we learned..
Ex. 1) S sin(3x+5) dx ...(integrate)
*Okay first you should notice that with this problem there's two things going on at once..kind of like a chain rule; so, that tells you that you're going to have to use substitution to solve this
*So your u is going to be what's inside the function, which is 3x+5 and your du will be 3 dx (because 3 is the derivative of your u)
*Now you're going to replace 3x+5 with u in your equation to solve...butttttt, you're going to have to add a 1/3 on the outside to even it out (since your derivative isn't actually in the original problem) So you should get this:
1/3 S sin(u) du
*Now you're going to integrate sin(u)....and you should get -cos(u)
So now you have this:
1/3[-cos(u)] + c
*Now all you have left to do is plug back in for u like this:
= 1/3[-cos(3x+5)] + c
Ex. 2) For y = 3x^2+2x+5 at the point (4,0), what is the slope of the normal line?
*First, since they're asking for slope you know that you're going to have to take the derivative of the equation they gave you.
*So once you take the derivative you should get 6x+2
*Since they didn't ask for an equation, all they want is the slope (just the number)..and they gave you an x-value so you're going to plug that (4) into your derivative (6x+2) like this:
6(4)+2
= 26...so that's your slope
*Buttttt, since they're asking for the slope of the NORMAL LINE, that means to get your final answer you'll have to take the negative reciprocal of the slope you just found
*So the slope of the normal line is -1/26
Ex. 3) Find the average value for f(x)=5x^2+6x+2 on the interval [0,2].
*Sooooo when you first look at this problem I bet the word "average" stands out to you..it probably makes you think like average rate of change or average velocity or something which means that you'll have to take the derivative....wellllll, not in this case. Average VALUE is totally different. It involves an integral, but it's super easy because all you have to do is use this simple formula:
1/b-a Sa,b f(x) ....(where its the integral of f(x) from bounds a-b multiplied by 1/b-a
*So plugging in the information they gave us into the formula you should get this:
1/2 S0,2 5x^2+6x+2 dx
*Now all you do is integrate it! And you should end up with this:
1/2[5/3x^3 + 3x^2 +2x] I2,0 ...(pretend that "I" is the bar that means you haven't plugged in yet, haha)
*So now do the Fundamental Theorem of Calculus:
1/2 [5/3(2)^3 + 3(2)^2 + 2(2)] - [0]
**^don't forget, the 1/2 is still out front because that's in the formula
*So once you simplify everything you should get this for your final answer:
44/3
Ex. 1) S sin(3x+5) dx ...(integrate)
*Okay first you should notice that with this problem there's two things going on at once..kind of like a chain rule; so, that tells you that you're going to have to use substitution to solve this
*So your u is going to be what's inside the function, which is 3x+5 and your du will be 3 dx (because 3 is the derivative of your u)
*Now you're going to replace 3x+5 with u in your equation to solve...butttttt, you're going to have to add a 1/3 on the outside to even it out (since your derivative isn't actually in the original problem) So you should get this:
1/3 S sin(u) du
*Now you're going to integrate sin(u)....and you should get -cos(u)
So now you have this:
1/3[-cos(u)] + c
*Now all you have left to do is plug back in for u like this:
= 1/3[-cos(3x+5)] + c
Ex. 2) For y = 3x^2+2x+5 at the point (4,0), what is the slope of the normal line?
*First, since they're asking for slope you know that you're going to have to take the derivative of the equation they gave you.
*So once you take the derivative you should get 6x+2
*Since they didn't ask for an equation, all they want is the slope (just the number)..and they gave you an x-value so you're going to plug that (4) into your derivative (6x+2) like this:
6(4)+2
= 26...so that's your slope
*Buttttt, since they're asking for the slope of the NORMAL LINE, that means to get your final answer you'll have to take the negative reciprocal of the slope you just found
*So the slope of the normal line is -1/26
Ex. 3) Find the average value for f(x)=5x^2+6x+2 on the interval [0,2].
*Sooooo when you first look at this problem I bet the word "average" stands out to you..it probably makes you think like average rate of change or average velocity or something which means that you'll have to take the derivative....wellllll, not in this case. Average VALUE is totally different. It involves an integral, but it's super easy because all you have to do is use this simple formula:
1/b-a Sa,b f(x) ....(where its the integral of f(x) from bounds a-b multiplied by 1/b-a
*So plugging in the information they gave us into the formula you should get this:
1/2 S0,2 5x^2+6x+2 dx
*Now all you do is integrate it! And you should end up with this:
1/2[5/3x^3 + 3x^2 +2x] I2,0 ...(pretend that "I" is the bar that means you haven't plugged in yet, haha)
*So now do the Fundamental Theorem of Calculus:
1/2 [5/3(2)^3 + 3(2)^2 + 2(2)] - [0]
**^don't forget, the 1/2 is still out front because that's in the formula
*So once you simplify everything you should get this for your final answer:
44/3
Sunday, February 6, 2011
umad bloggity bloggers?
duuuuuuuuuuuuuuuuuuuuuuuuude, the super bowl was boring
anyway, we didn't learn a single new thing this week, so what am I supposed to blog about?
something: !
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
anyway, we didn't learn a single new thing this week, so what am I supposed to blog about?
something: !
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
BLOGbro
This week we started taking practice AP testing and we also went over PVA too. PVA: position, velocity, and acceleration. Displacement or position is the integral of velocity; and distance is the absolute value of the integral of velocity. Most of the PVA questions will have more than one part to them. It's good to realize what the question is asking for (as far as distance/displacement).
Here's an example:
Chad accelerates his car, giving it a velocity of v = t^1/2 - 2, in feet per second, at time t, in seconds after he started accelerating.
A: Find the time(s) at which v = 0
Ask yourself, what is this problem looking for? They're asking for "t", so all you do is plug in 0 for v <--We know to plug in this because it tells us that at t v=0.
-So set the equation they gave you equal to zero and solve for t.
t^1/2 - 2 = 0
t^1/2 = 2 ..(t^1/2 is the same thing as the squareroot of t)
(t^1/2)^2 = 2^2
t=4
In reality these problems are really easy, you just have to follow the formulas that we already learned about distance and displacement, and also finding integrals.
Here's an example:
Chad accelerates his car, giving it a velocity of v = t^1/2 - 2, in feet per second, at time t, in seconds after he started accelerating.
A: Find the time(s) at which v = 0
Ask yourself, what is this problem looking for? They're asking for "t", so all you do is plug in 0 for v <--We know to plug in this because it tells us that at t v=0.
-So set the equation they gave you equal to zero and solve for t.
t^1/2 - 2 = 0
t^1/2 = 2 ..(t^1/2 is the same thing as the squareroot of t)
(t^1/2)^2 = 2^2
t=4
In reality these problems are really easy, you just have to follow the formulas that we already learned about distance and displacement, and also finding integrals.
Blog #24
This week, we mainly just did AP practice tests. Besides that, we learned how to integrate using substitution. This is used when you have a product or quotient rule. This can’t always be used, but these are the instances in which it can:
S u x v
…or…
S u/v
where v is the derivative of u
Here’s an example:
Given…S (x²+1)(2x) dx
First, identify u and v. Which is the derivative and which is the original?
u = x²+1
v = 2x
…because 2x is the derivative of x²+1…
Okay, next you remove v (2x dx) from the equation and substitute “u” for x²+1.
S u du
*You insert du because v is the derivative of u, hence du.
Now, integrate u.
½ u² + C
Finally, replace “u” with the original.
½ (x²+1)² + C
This week, we also learned the Second Fundamental Theorem of Calculus:
This theorem state that the derivative of an integral results in the original equation.
The formula: d/dx vSa f(x) dx = (f(v)) (v’) where “a” is a number (usually equaling 0) and v is an equation
An example:
d/dx x²S0 (x+1)² dx
= (((x²) +1)²) (2x)
= 2x(x²+1)²
S u x v
…or…
S u/v
where v is the derivative of u
Here’s an example:
Given…S (x²+1)(2x) dx
First, identify u and v. Which is the derivative and which is the original?
u = x²+1
v = 2x
…because 2x is the derivative of x²+1…
Okay, next you remove v (2x dx) from the equation and substitute “u” for x²+1.
S u du
*You insert du because v is the derivative of u, hence du.
Now, integrate u.
½ u² + C
Finally, replace “u” with the original.
½ (x²+1)² + C
This week, we also learned the Second Fundamental Theorem of Calculus:
This theorem state that the derivative of an integral results in the original equation.
The formula: d/dx vSa f(x) dx = (f(v)) (v’) where “a” is a number (usually equaling 0) and v is an equation
An example:
d/dx x²S0 (x+1)² dx
= (((x²) +1)²) (2x)
= 2x(x²+1)²
2/6/11
Soooo this week we started taking practice AP testing..well just the multipule choice part, and actually it is really hard. I know the information, but for some reason I just can't work the problems out..hopefully with practice I will get better at taking the test. Anyways, since we started taking the AP test we haven't learned anything new. I pretty much went over everything besides PVA, so i'll explain how to do that. PVA: position, velocity, and acceleration. The week before we took the AP test we learned about these functions and how to apply integrals with them. For instance, displacement or position is the integral of velocity; and distance is the absolute value of the integral of velocity. Most of the PVA questions will have more than one part to them. It's good to realize what the question is asking for (as far as distance/displacement).
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
*Ask yourself, what is this problem looking for? They're asking for "t", so all you do is plug in 0 for v <--We know to plug in this because it tells us that at t v=0.
-So set the equation they gave you equal to zero and solve for t.
t^1/2 - 2 = 0
t^1/2 = 2 ..(t^1/2 is the same thing as the squareroot of t)
(t^1/2)^2 = 2^2
t=4
B: Find her net displacement for the time interval [1,9].
*Notice they give you the time interval [1,9]..which is going to be your bounds for your integral
*Also notice there asking for the displacement, you're going to 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:
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
These problems are reallllly easy, you just have to follow the formulas that we already learned about distance and displacement, and also finding integrals. It's pretty much everything rolled into one. And that's alllll :)
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
*Ask yourself, what is this problem looking for? They're asking for "t", so all you do is plug in 0 for v <--We know to plug in this because it tells us that at t v=0.
-So set the equation they gave you equal to zero and solve for t.
t^1/2 - 2 = 0
t^1/2 = 2 ..(t^1/2 is the same thing as the squareroot of t)
(t^1/2)^2 = 2^2
t=4
B: Find her net displacement for the time interval [1,9].
*Notice they give you the time interval [1,9]..which is going to be your bounds for your integral
*Also notice there asking for the displacement, you're going to 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:
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
These problems are reallllly easy, you just have to follow the formulas that we already learned about distance and displacement, and also finding integrals. It's pretty much everything rolled into one. And that's alllll :)
2/6/11
Sooo..starting off this week we learned the Second Fundamental Theorem of Calculus. All that is is the derivative of an integral. So it would be written in the form: d/dx S1,2x x^2 dx...where d/dx infers derivative; 1,2x are the bounds; and x^2 is the equation. In order to solve a problem like this, all you do is take f(x) (which would be the equation), plug in the top bound for x and multiply that by the derivative of the top bound. *Also, in order for this method to work, the bottom bound has to be a number. Nexxttttt, we learned integration by Substitution..you use this method when you have an equation that looks like it would be a product rule or a quotient rule. To use substitution, one part of the equation is going to be the derivative of the other part..or close to it. It's a weird process to explain so I'll just give an example haha...And also this week we learned how to integrate e..and all you do for that is basically the opposite that you would do to take its derivative. First you recopy the e part, then instead of multiplying by the exponent's derivative, you divide by it. So here are a few examples of what I just talked about:
Ex. 1) d/dx S6,2x^4 2x+1 dx......where 6,2x^4 are the bounds and 2x+1 is the equation
*So first you're going to take the equation (f) and plug in the top bound for x..so you should get this:
2(2x^4)+1
*Now multiply that by the derivative of 2x^4...which is 8x^3
*So you should have 8x^3(4x^4+1)
Ex. 2) S(6x)(3x^2+1) dx
*Now since this looks like it would be a product rule, you have to check to see if you can use substitution to solve it
*6x is the derivative of 3x^2+1..so yes you can use substitution
*To use this method, you're going to replace one of the terms with u to make it simpler to solve
*Since 6x is the derivative of 3x^2+1, u becomes 3x^2+1 And du, the derivative of u, would be 6x dx..(<
S u du (and the 6x is not included because it's going to disappear)
*So once you integrate u, you should end up with this:
1/2u^2 + c ...*don't forget the +c because there are no bounds
*Now all you do is plug back in for u. So you're final equation should look like this:
1/2(3x^2+1) + c
Ex. 3) S e^4x^2
*To integrate this, you first recopy the equation:
= e^4x^2
*Now all you do is divide by the derivative of e's exponent...which is 8x
*So you should get this for your answer:
e^4x^2/8x
**Well this week was good overall. I understood everything; I just think I could use more practice on the substitution problems because those are kinda tricky.
Ex. 1) d/dx S6,2x^4 2x+1 dx......where 6,2x^4 are the bounds and 2x+1 is the equation
*So first you're going to take the equation (f) and plug in the top bound for x..so you should get this:
2(2x^4)+1
*Now multiply that by the derivative of 2x^4...which is 8x^3
*So you should have 8x^3(4x^4+1)
Ex. 2) S(6x)(3x^2+1) dx
*Now since this looks like it would be a product rule, you have to check to see if you can use substitution to solve it
*6x is the derivative of 3x^2+1..so yes you can use substitution
*To use this method, you're going to replace one of the terms with u to make it simpler to solve
*Since 6x is the derivative of 3x^2+1, u becomes 3x^2+1 And du, the derivative of u, would be 6x dx..(<
S u du (and the 6x is not included because it's going to disappear)
*So once you integrate u, you should end up with this:
1/2u^2 + c ...*don't forget the +c because there are no bounds
*Now all you do is plug back in for u. So you're final equation should look like this:
1/2(3x^2+1) + c
Ex. 3) S e^4x^2
*To integrate this, you first recopy the equation:
= e^4x^2
*Now all you do is divide by the derivative of e's exponent...which is 8x
*So you should get this for your answer:
e^4x^2/8x
**Well this week was good overall. I understood everything; I just think I could use more practice on the substitution problems because those are kinda tricky.
Sunday, January 30, 2011
blabbity blabbity blog
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)
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)
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
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