These are explanations that I had emailed to various students while we were working on this chapter. Use this information as you review old problems, or while you're studying for the final exam.
Chp 5 Review "T" Answers:
Chp 5 Review (pg. 241):
R9ai) Make the exponent "one bigger", then divide by the new exponent. So, your integrated equation (which is the anti-derivative) is:
R9d) I found a picture online that illustrates the statement in this problem (see below). It's basically saying that if your entire integral goes from a to b, that you could split that whole area at some "c" location. This would make it two areas, where one goes from a to c, and the other goes from c to b.
R10a) Check the answer in the back of the book for this picture.
R10b) This says that displacement (dy) is equal to the velocity (v) times the time (t). Write an equation saying this:
dy = v * dt
To “write dy in terms of t and dt”, substitute the velocity equation they gave you in for “v”.
dy = 150t^0.5 * dt
R10c) Remember how displacement, velocity, and acceleration are related to each other:
displacement —derivative—> velocity —derivative—> acceleration
acceleration —integral—> velocity —integral—> displacement
So in this problem, if they want you to use an integral to find displacement, you’ll need to integrate the velocity equation. The integration method they’re asking you to use here is a Riemann Sum. The answer in the back of the book is a different form than the way we’ve been writing them in class, but you could write a Riemann Sum however you’d like. Here’s one example:
Mid-Riemann Sum of 3 rectangles. You’re integrating velocity, so you’ll use the velocity equation to find the height of each rectangle. The width of each rectangle would be 3:
3 * v(1.5) + 3 * v(4.5) + 3 * v(7.5)
…or could be written as:
3 * ( v(1.5) + v(4.5) + v(7.5) )
R10d) The limit of ANY Riemann Sum as dt (the width of each rectangle) approaches zero will be equal to the exact definite integral. This is because as the width of the rectangles approaches zero, there must be more and more rectangles being used.
So, to find this limit, you could just find the integral by using the Fundamental Theorem of Calculus (FTC).
(integral sign from 0 to 9) 150t^0.5
Make the exponent “one bigger”, then divide by the new exponent:
100t^1.5 (bounded 0 to 9) =
100(9)^1.5 - 100(0)^1.5 =
2700 feet
R10e) This problem is trying to help you see that two smaller integrals added together will equal the larger integral. Find the integral from 0 to 4, and find the integral from 4 to 9. It should equal the same answer you got in Part D.
Sec 5.10 worksheet
These links are Educreations videos I made for Sec 5.10 homework (worksheet). Make sure your sound is on!
Sec 5.10 worksheet #1 (this is an explanation of the problem we already did in class)
Sec 5.10 homework #1 & #3
Sec 5.9 help: Fundamental Theorem of Calculus
This link is an Educreations video I made for the Sec 5.9 homework.
Examples of how to calculate using FTC (using problems from Sec 5.9 in the book)
Winter Break Packet:
2c) An easy way to do this is to plug in a number really close to x = 2. I would try x = 1.99 and x = 2.01 (do both, just to make sure that you’re getting the same answer on both sides of the limit).
You could try factoring the numerator and denominator as the “difference of two squares”, but I don’t think it will help too much. The numerator becomes: (2 + x)(2 - x). The denominator could be factored the same way (like using the square root of 3 in each factor, etc.), but it wouldn’t create anything to cancel out factors in the numerator.
2d) The “trick” here is to recognize that this problem is actually the forward difference quotient:
lim f(x + h) - f(x)
h—>0 h
See how the two functions in the numerator are the same, but the first one has a “+h" in it? And the denominator is just an “h”, and the limit is h—>0.
So, the original function that the forward difference quotient is being applied to is: 3 / (x - 2)
This means that this question is just asking you to find the derivative of f(x) = 3 / (x - 2). You could use the quotient rule to do this, or you could roll the denominator up and then use the chain rule:
f(x) = 3(x - 2)^-1
f’(x) = -3(x - 2)^-2
f’(x) = -3 / (x - 2)^2
3f) I haven’t taught you this yet! Stay tuned for class tomorrow.
3g) An integral is an anti-derivative, so we need to find the anti-derivative of f(x) = x^2 + 1. Remember, “add 1 to the exponent, then divide by the new exponent”.
g(x) = 1/3x^3 + x + c
3h) This tells you that when x = 3, the g(x) equation equals 1. Plug these in to find c:
1 = 1/3(3)^3 + 3 + c
1 = 1/3(27) + 3 + c
1 = 9 + 3 + c
1 = 12 + c
-11 = c
g(x) = 1/3x^3 + x - 11
6) Factor the numerator as the difference of two squares. It’s like doing this:
x^2 - 25 = (x + 5)(x - 5)
But here, it’s like this:
x - 1 = (sqrt(x) + 1)(sqrt(x) - 1)
Now your function looks like:
(sqrt(x) + 1)(sqrt(x) - 1) / (sqrt(x) - 1)
You can cancel out a factor from the top and bottom, which will then let you plug in x = 1 to find the limit. Here’s your equation after canceling:
lim sqrt(x) + 1
x—>1
Then plug in x = 1:
sqrt(1) + 1
1 + 1
= 2
17) Finding the area between a curve and the x-axis is the same as finding its integral. Here, you need to find the points where the function intersects the x-axis so you'll know which bounds to use in your integral.
y = x^2 - x - 6
0 = x^2 - x - 6
0 = (x - 3)(x + 2)
x = 3 and -2
So, you need to integrate from -2 to 3:
(integral sign -2 to 3) x^2 - x - 6 * dx
= 1/3x^3 - 1/2x^2 - 6x
= [ 1/3(3)^3 - 1/2(3)^2 - 6(3) ] - [ 1/3(-2)^3 - 1/2(-2)^2 - 6(-2) ]
= [ 9 - 4.5 - 18 ] - [ -8/3 - 2 + 12 ]
= -13.5 - 22/3
= -20.833
Because the question is asking us about the area under the curve, we'll answer in terms of the positive value, 20.833.
18a) This question is basically giving you the answer, and asking you to show how you find it. Find the implicit derivative of both sides of the equation to answer this question.
x^2 - sin y = y + 4
Remember that every time you find a derivative involving "y", you need to multiply by dy/dx Derivative is:
2x - cos y * dy/dx = dy/dx
Get all the dy/dx terms on one side, and all the other terms on the other side:
2x = dy/dx + cos y * dy/dx
Factor the "dy/dx" out of both terms, then divide to get the "dy/dx" to find the answer.
2x = dy/dx(1 + cos y)
2x / (1 + cos y) = dy/dx
18b) To find a tangent line equation, you need a slope and a point.
Find the point at y = 0 by plugging in y = 0 to the original equation:
x^2 - sin 0 = 0 + 4
x^2 = 4
x = 2 and -2
Find the slope at both (x,y) points by plugging in (2,0) and (-2,0) to the dy/dx equation:
dy/dx = 2(2) / (1 + cos 0) = 4 / 2 = 2
dy/dx = 2(-2) / (1 + cos 0) = -4 / 2 = -2
You can write two tangent line equations. One has a point at (2,0) and dy/dx = 2. The other has a point at (-2,0) and dy/dx = -2.
Answers:
y - 0 = 2(x - 2)
y = 2x - 4
y - 0 = -2(x + 2)
y = -2x - 4
AP Q #6) WHOOPS! We haven't learned this one yet. Skip it!
AP Q #7) Start by finding the tangent line equation. You need a slope and a point. They gave you the point, but you need to find the derivative equation to find the slope.
Find the derivative f'(x) by using the quotient rule:
f(x) = sec x / (x + 2)
f'(x) = [sec x tan x * (x + 2) - sec x] / (x + 2)^2
f'(x) = sec x * [tan x * (x + 2) - 1] / (x + 2)^2
f'(x) = [tan x * (x + 2) - 1] / [cos x * (x + 2)^2]
Plug in x = 0 to find f'(0):
f'(0) = [tan 0 * (0 + 2) - 1] / [cos 0 * (0 + 2)^2]
f'(0) = [0 * (2) - 1] / [1 * 4]
f'(0) = -1/4
Your tangent line equation is:
y - 1/2 = -1/4(x - 0)
y = -1/4x + 1/2
Now use this to approximate f(0.1) by plugging in x = 0.1 for x in the tangent line equation:
y = -1/4(0.1) + 1/2 = -0.025 + 0.5 = 0.475
Sec 5.6 (pg. 208):
General help: Mean Value Theorem
These links are Educreations videos I made for Sec 5.6 Exploration Wkst. last year. They should mostly line up with the worksheet we did in class this year.
Sec 5.6 Exploration Problem #1
Sec 5.6 Exploration Problem #2
Sec 5.6 Exploration Problem #3
Sec 5.6 Exploration Problem #4
Sec 5.6 Exploration - BACK OF WORKSHEET (lots of important information about Mean Value Theorem in this video especially)
3) The Mean Value Theorem (MVT) says that if a graph is differentiable, then there is some point in the interval that has the same slope as the average slope between the two endpoints.
Sec 5.5: Riemann Sum
Note: This is a Khan Academy/YouTube video that I didn't create, but it's a good explanation. How do I find a "Lower Riemann Sum"?
Sec 5.4 (pg. 193):
33) You can't find the integral (anti-derivative) the way it's written in the book, because there isn't a "derivative of the inside" anywhere in the integral, which would have shown up from doing the chain rule.
Chp 5 Review (pg. 241):
R9ai) Make the exponent "one bigger", then divide by the new exponent. So, your integrated equation (which is the anti-derivative) is:
-x^-1
Then, plug in 5, then plug in 1, then subtract:
(-5^-1) - (-1^-1)
Remember when you're figuring these out: a negative exponent can be changed to a positive by using the reciprocal.
(-1/5) - (-1/1)
-1/5 + 1
4/5
R9aii) Because there's an "inside" function, and there appears to be another function multiplied, you could use U-substitution. Make u equal to the inside function:
u = x^2 + 3
Then find du by finding the derivative of u:
du = 2x dx
Your given integral only has x dx, so divide both sides of the du equation by 2 to make it look the same:
1/2 du = x dx
Now substitute back into the integral. You should have only u (no x).
1/2 (integral sign) u^5 du
Now use the power rule to integrate (make the exponent one bigger, then divide by the new exponent).
1/2 * 1/6 u^6
1/12 u^6
Substitute u back out:
1/12(x^2 + 3)^6
Plug in 4, then plug in 3, then subtract:
1/12(4^2 + 3)^6 - 1/12(3^2 + 3)^6
1/12(19)^6 - 1/12(12)^6
= 3,671,658.083
R9aiii) Integrate each term separately. The derivative of what gives you sin x? --> -cos x
The derivative of what gives you 5? --> 5x
So, your integrated equation is:
(integral sign) -cos x - 5x
Plug in pi, then plug in 0, then subtract:
(-cos pi - 5pi) - (-cos 0 - 5*0)
(1 - 5pi) - (-1)
2 - 5pi
Picture from:
http://www.drcruzan.com/Images/Mathematics/DefiniteIntegrals/DecompositionGraph.png
R10a) Check the answer in the back of the book for this picture.
R10b) This says that displacement (dy) is equal to the velocity (v) times the time (t). Write an equation saying this:
dy = v * dt
To “write dy in terms of t and dt”, substitute the velocity equation they gave you in for “v”.
dy = 150t^0.5 * dt
R10c) Remember how displacement, velocity, and acceleration are related to each other:
displacement —derivative—> velocity —derivative—> acceleration
acceleration —integral—> velocity —integral—> displacement
So in this problem, if they want you to use an integral to find displacement, you’ll need to integrate the velocity equation. The integration method they’re asking you to use here is a Riemann Sum. The answer in the back of the book is a different form than the way we’ve been writing them in class, but you could write a Riemann Sum however you’d like. Here’s one example:
Mid-Riemann Sum of 3 rectangles. You’re integrating velocity, so you’ll use the velocity equation to find the height of each rectangle. The width of each rectangle would be 3:
3 * v(1.5) + 3 * v(4.5) + 3 * v(7.5)
…or could be written as:
3 * ( v(1.5) + v(4.5) + v(7.5) )
R10d) The limit of ANY Riemann Sum as dt (the width of each rectangle) approaches zero will be equal to the exact definite integral. This is because as the width of the rectangles approaches zero, there must be more and more rectangles being used.
So, to find this limit, you could just find the integral by using the Fundamental Theorem of Calculus (FTC).
(integral sign from 0 to 9) 150t^0.5
Make the exponent “one bigger”, then divide by the new exponent:
100t^1.5 (bounded 0 to 9) =
100(9)^1.5 - 100(0)^1.5 =
2700 feet
R10e) This problem is trying to help you see that two smaller integrals added together will equal the larger integral. Find the integral from 0 to 4, and find the integral from 4 to 9. It should equal the same answer you got in Part D.
Sec 5.10 worksheet
These links are Educreations videos I made for Sec 5.10 homework (worksheet). Make sure your sound is on!
Sec 5.10 worksheet #1 (this is an explanation of the problem we already did in class)
Sec 5.10 homework #1 & #3
Sec 5.9 help: Fundamental Theorem of Calculus
This link is an Educreations video I made for the Sec 5.9 homework.
Examples of how to calculate using FTC (using problems from Sec 5.9 in the book)
Winter Break Packet:
2c) An easy way to do this is to plug in a number really close to x = 2. I would try x = 1.99 and x = 2.01 (do both, just to make sure that you’re getting the same answer on both sides of the limit).
You could try factoring the numerator and denominator as the “difference of two squares”, but I don’t think it will help too much. The numerator becomes: (2 + x)(2 - x). The denominator could be factored the same way (like using the square root of 3 in each factor, etc.), but it wouldn’t create anything to cancel out factors in the numerator.
2d) The “trick” here is to recognize that this problem is actually the forward difference quotient:
lim f(x + h) - f(x)
h—>0 h
See how the two functions in the numerator are the same, but the first one has a “+h" in it? And the denominator is just an “h”, and the limit is h—>0.
So, the original function that the forward difference quotient is being applied to is: 3 / (x - 2)
This means that this question is just asking you to find the derivative of f(x) = 3 / (x - 2). You could use the quotient rule to do this, or you could roll the denominator up and then use the chain rule:
f(x) = 3(x - 2)^-1
f’(x) = -3(x - 2)^-2
f’(x) = -3 / (x - 2)^2
3f) I haven’t taught you this yet! Stay tuned for class tomorrow.
3g) An integral is an anti-derivative, so we need to find the anti-derivative of f(x) = x^2 + 1. Remember, “add 1 to the exponent, then divide by the new exponent”.
g(x) = 1/3x^3 + x + c
3h) This tells you that when x = 3, the g(x) equation equals 1. Plug these in to find c:
1 = 1/3(3)^3 + 3 + c
1 = 1/3(27) + 3 + c
1 = 9 + 3 + c
1 = 12 + c
-11 = c
g(x) = 1/3x^3 + x - 11
6) Factor the numerator as the difference of two squares. It’s like doing this:
x^2 - 25 = (x + 5)(x - 5)
But here, it’s like this:
x - 1 = (sqrt(x) + 1)(sqrt(x) - 1)
Now your function looks like:
(sqrt(x) + 1)(sqrt(x) - 1) / (sqrt(x) - 1)
You can cancel out a factor from the top and bottom, which will then let you plug in x = 1 to find the limit. Here’s your equation after canceling:
lim sqrt(x) + 1
x—>1
Then plug in x = 1:
sqrt(1) + 1
1 + 1
= 2
17) Finding the area between a curve and the x-axis is the same as finding its integral. Here, you need to find the points where the function intersects the x-axis so you'll know which bounds to use in your integral.
y = x^2 - x - 6
0 = x^2 - x - 6
0 = (x - 3)(x + 2)
x = 3 and -2
So, you need to integrate from -2 to 3:
(integral sign -2 to 3) x^2 - x - 6 * dx
= 1/3x^3 - 1/2x^2 - 6x
= [ 1/3(3)^3 - 1/2(3)^2 - 6(3) ] - [ 1/3(-2)^3 - 1/2(-2)^2 - 6(-2) ]
= [ 9 - 4.5 - 18 ] - [ -8/3 - 2 + 12 ]
= -13.5 - 22/3
= -20.833
Because the question is asking us about the area under the curve, we'll answer in terms of the positive value, 20.833.
18a) This question is basically giving you the answer, and asking you to show how you find it. Find the implicit derivative of both sides of the equation to answer this question.
x^2 - sin y = y + 4
Remember that every time you find a derivative involving "y", you need to multiply by dy/dx Derivative is:
2x - cos y * dy/dx = dy/dx
Get all the dy/dx terms on one side, and all the other terms on the other side:
2x = dy/dx + cos y * dy/dx
Factor the "dy/dx" out of both terms, then divide to get the "dy/dx" to find the answer.
2x = dy/dx(1 + cos y)
2x / (1 + cos y) = dy/dx
18b) To find a tangent line equation, you need a slope and a point.
Find the point at y = 0 by plugging in y = 0 to the original equation:
x^2 - sin 0 = 0 + 4
x^2 = 4
x = 2 and -2
Find the slope at both (x,y) points by plugging in (2,0) and (-2,0) to the dy/dx equation:
dy/dx = 2(2) / (1 + cos 0) = 4 / 2 = 2
dy/dx = 2(-2) / (1 + cos 0) = -4 / 2 = -2
You can write two tangent line equations. One has a point at (2,0) and dy/dx = 2. The other has a point at (-2,0) and dy/dx = -2.
Answers:
y - 0 = 2(x - 2)
y = 2x - 4
y - 0 = -2(x + 2)
y = -2x - 4
AP Q #6) WHOOPS! We haven't learned this one yet. Skip it!
AP Q #7) Start by finding the tangent line equation. You need a slope and a point. They gave you the point, but you need to find the derivative equation to find the slope.
Find the derivative f'(x) by using the quotient rule:
f(x) = sec x / (x + 2)
f'(x) = [sec x tan x * (x + 2) - sec x] / (x + 2)^2
f'(x) = sec x * [tan x * (x + 2) - 1] / (x + 2)^2
f'(x) = [tan x * (x + 2) - 1] / [cos x * (x + 2)^2]
Plug in x = 0 to find f'(0):
f'(0) = [tan 0 * (0 + 2) - 1] / [cos 0 * (0 + 2)^2]
f'(0) = [0 * (2) - 1] / [1 * 4]
f'(0) = -1/4
Your tangent line equation is:
y - 1/2 = -1/4(x - 0)
y = -1/4x + 1/2
Now use this to approximate f(0.1) by plugging in x = 0.1 for x in the tangent line equation:
y = -1/4(0.1) + 1/2 = -0.025 + 0.5 = 0.475
Sec 5.6 (pg. 208):
General help: Mean Value Theorem
These links are Educreations videos I made for Sec 5.6 Exploration Wkst. last year. They should mostly line up with the worksheet we did in class this year.
Sec 5.6 Exploration Problem #1
Sec 5.6 Exploration Problem #2
Sec 5.6 Exploration Problem #3
Sec 5.6 Exploration Problem #4
Sec 5.6 Exploration - BACK OF WORKSHEET (lots of important information about Mean Value Theorem in this video especially)
3) The Mean Value Theorem (MVT) says that if a graph is differentiable, then there is some point in the interval that has the same slope as the average slope between the two endpoints.
So…
This function, 6/x, isn't continuous and differentiable everywhere, but it is for the interval we're looking at: [1, 4]. So, the hypotheses are met!
1st step: Find the average slope between the endpoints of the interval:
g(4) - g(1) / 4 - 1
3/2 - 6 / 3
-4.5 / 3
-3/2
2nd step: Set the derivative equal to the average slope. (Reason: We are trying to find the x = c point where the slope is the same as the average slope).
g'(x) = -6/x^2
-6/x^2 = -3/2
3x^2 = 12
x^2 = 4
Final answer: x = 2
4) Same ideas!
1st step: Find the average slope between the endpoints.
f(2) - f(-1) / 2 - -1
16 - 1 / 3
15 / 3
5
2nd step: Set the derivative equal to the average slope.
f'(x) = 4x^3
4x^3 = 5
x^3 = 1.25
Final answer: x = 1.077Sec 5.5: Riemann Sum
Note: This is a Khan Academy/YouTube video that I didn't create, but it's a good explanation. How do I find a "Lower Riemann Sum"?
Sec 5.4 (pg. 193):
33) You can't find the integral (anti-derivative) the way it's written in the book, because there isn't a "derivative of the inside" anywhere in the integral, which would have shown up from doing the chain rule.
Instead, you need to multiply it out, then find the integral by doing reverse power rule.
So…
(x^2 + 5)^3 multiplied out becomes…
[ Remember Pascal's Triangle? (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 ]
x^6 + 15x^4 + 75x^2 + 125
Now you can find the integral! "Make the exponent one bigger, then divide by the new exponent."
1/7x^7 + 3x^5 + 25x^3 + 125x + c
