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
Sunday, March 27, 2011
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
Subscribe to:
Posts (Atom)