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.
Remember, my Google Drive folder contains copies of worksheets and ANSWERS (on the last page of each file). CLICK HERE TO ACCESS.
Chp 3 Help: Finding the equation of a tangent line
This link is an Educreations video I made for a specific tangent line problem a couple years ago. Make sure your sound is on!
How to find a tangent line equation? (AP practice)
Because this is saying that f'(2) = 5, that means the derivative of f(x) at x = 2 is 5. In order for a function to have a derivative, it MUST be continuous. So then we can conclude...
I. --> f is definitely continuous at x = 2.
II. --> f is definitely differentiable at x = 2.
III. --> They don't say anything else about the graph, so we have no idea if the f'(x) graph is continuous at x = 2 or not. Maybe the f'(x) graph ends at x = 2, or maybe it doesn't. We can't make any conclusions about it!
ANSWER: C
12) They are giving you a(t), and asking you about v(t). We know that the derivative of v(t) gives you a(t), so to go backwards: the anti-derivative of a(t) will give you v(t).
We need to find an anti-derivative of a(t):
a(t) = t + sin t
Make the exponent of "t" one bigger, then divide by the new exponent. Then say, "the derivative of what would give me sin? -cos!"
v(t) = 1/2*t^2 - cos t + c
They gave you a specific value of v(t) in order to take your general anti-derivative to a particular anti-derivative. They said v(0) = -2, so plug that in to solve for c:
-2 = 1/2*0^2 - cos 0 + c
-2 = 0 - 1 + c
-1 = c
So, v(t) = 1/2*t^2 - cos t - 1. They want to know when v(t) = 0, so set v(t) equal to 0, then solve for t:
0 = 1/2*t^2 - cos t -1
To solve for t, there are several methods (all involving your graphing calculator). I would graph this on your calculator, then find the x-intercepts by using 2ND, TRACE, "Zero".
ANSWER: B (t = 1.4781703 sec)
Chp 3 Review (pg. 123)
R2b) "Why do I get a different derivative from using the power rule than using the derivative formula?"
21) This is a polynomial function, so we can use (the opposite of) the power rule. With the regular power rule, you can find a derivative by multiplying the coefficient by the old exponent, and making your new exponent "one smaller". We are going to find the anti-derivative by making each exponent "one bigger", then dividing by the new exponent.
Final answer: f(x) = f(x) = 1/3x^3 - 4x^2 + 3x + 113/3
24a) This is not a polynomial function, so you are not using (the opposite of) the power rule. No exponents! If you want to find the anti-derivative of sin, say to yourself: "The derivative of what would give me sin?" The derivative of -cos would give you sin. So, your anti-derivative function will have -cos in it.
Sec 3.8 "Waterwheel Problem" worksheet:
7) It will be easiest to find the derivative of d(t) if you distribute the inside parentheses first (so that you only have one set of parentheses in the whole equation). So, you should be starting with this d(t) function:
d(t) = 7 cos (pi/5 * t - 2pi/5) + 6
To do the chain rule, find the derivative of the “outside” function first, while keeping the “inside” the same. The outside function here is cos, so the derivative of it is -sin:
d’(t) = -7 sin (pi/5 * t - 2pi/5) + 6 [this answer is not done yet!]
Then, multiply by the derivative of the inside. The “inside” is (pi/5 * t - 2pi/5), where the derivative of pi/5 * t is pi/5, and the derivative of -2pi/5 is 0:
d’(t) = -7pi/5 sin (pi/5 * t - 2pi/5) + 6 [this answer is not done yet!]
Then, the derivative of the +6 on the end is 0. Here is the FINAL DERIVATIVE EQUATION:
d’(t) = -7pi/5 sin (pi/5 * t - 2pi/5)
Sec 3.7 help: Chain Rule
This link is an Educreations video I made. Make sure your sound is on!
Examples (notes) on how to use the Chain Rule
Sketching Graphs practice wkst.
Sec 3.4 (pg. 95):
Q5) "I tried to use the difference quotient, but it did not work."
Think about what the graph of (x-5) / (x+3) looks like. It's a rational function with a vertical asymptote at x = -3 and a horizontal asymptote at y = 1. So, as x approaches -3 (where the vertical asymptote is), the limit is infinite.
3) Use the power rule to find the derivative. Roll down the exponent (which means multiply the coefficient of 0.007 by the exponent -83), then make the new exponent one less (which means subtract 1 from -83). Your answer should be:
v’ = -0.581t^-84
or using the other notation for derivatives, the answer is:
dv/dt = -0.581t^-84
Sec 3.3 worksheet: "The Derivative is a Function, Too"
Below, I have included thoughts, answers, math lessons, etc. for each of the questions on this worksheet for #1 - 6. You need to answer each of these questions in your own words (sometimes your answer and my answer will be very similar or identical, but not all the time), and then complete the rest of the worksheet. You can find the answers to this homework assignment online as the last page of this file in my Google Drive folder.
1) The apostrophe on "f" means that we are now referring to the derivative of f. If we just wrote f'(x), then we're referring to the derivative function (an entire equation). If we write f'(c), then we're referring to the derivative value at a specific "c" location.
2) Here, use nDeriv (MATH, 8, then type the rest in) that I showed you in class today. Look at your "Calculus Skills Sheet" if you've forgotten how to do it.
3) When it asks you to find f'(c), it's actually asking you to find the exact derivative at x = c. We have been using the "exact definition of a derivative" (also called the "difference quotient", also called the "forward difference quotient") to do this. Write down the formula we've been using that involves a limit as x approaches c.
4) "the derivative"
5) "the slope"
These two problems are trying to get you to remember that the derivative is the same as the slope, and that f'(x) is referring to the derivative of f(x).
6) TO GRAPH THE DERIVATIVE FUNCTION f'(x):
In Y1, type the f(x) equation you've started with.
In Y2, type: nDeriv(Y1, X, X) (older calculators) or d/dx (Y1) | X = X (newer calculators)
The "Y1" tells your calculator to find the derivative of Y1. The first "X" tells your calculator to find the derivative in terms of "X" (it will always be this). The second "X" tells your calculator to find the derivative value at every single value of x, not just at one specific value.
Then, press "Graph". It will graph your derivative, Y2, very very slowly, because your calculator is actually calculating every single derivative value individually.
Here's what your screen should look like (both in "Mathprint" mode and "Classic" mode).
6a - d) I am looking for answers involving < or > signs, and using words like "increasing" or "decreasing" or "positive values" or "negative values".
6h) HINT: In your explanation of how the derivative graph has or has not changed, talk about the slope of f(x), because this is the same thing as talking about the derivative of f(x). This all the help you get for now, but please feel free to email me with any questions you may have! Good luck! :)
Sec 3.2 (pg. 81):
3) Did you try factoring out a common factor first? You have 0.6x^2 - 5.4 in the numerator. Start by factoring out 0.6:
0.6(x^2 - 9)
Now you can factor the x^2 - 9 part like normal:
0.6(x + 3)(x - 3)
Now you can cancel with the denominator! Hope this helps get you started!
Remember, my Google Drive folder contains copies of worksheets and ANSWERS (on the last page of each file). CLICK HERE TO ACCESS.
Chp 3 Help: Finding the equation of a tangent line
This link is an Educreations video I made for a specific tangent line problem a couple years ago. Make sure your sound is on!
How to find a tangent line equation? (AP practice)
1997 AP Exam Problems worksheet
5) This link is an Educreations video I made for this specific problem. Make sure your sound is on!
#5 How to find a tangent line equation? (AP practice)
9) Problem #9 is tough! The best way to go about this is to picture what the graph COULD look like, then eliminate answers that don't work. It says that the graph is "continuous" and "differentiable" between a and b. Differentiable just means that you can find the derivative at any point (which in other words means every point has a slope!).
5) This link is an Educreations video I made for this specific problem. Make sure your sound is on!
#5 How to find a tangent line equation? (AP practice)
9) Problem #9 is tough! The best way to go about this is to picture what the graph COULD look like, then eliminate answers that don't work. It says that the graph is "continuous" and "differentiable" between a and b. Differentiable just means that you can find the derivative at any point (which in other words means every point has a slope!).
An example of a graph that fits these conditions would be y = 3x, or y = sin x.
An example of a graph that DOES NOT fit these conditions would be a vertical line, like x = 10, or a non-continuous graph with holes or asymptotes.
To decide which answer is correct, let's envision a graph that is a straight line, like y = 3x. Remember, the question is asking us which one could be false.
Choice A) This is the equation for slope! This just says that there is some point c on the interval that we could find the slope of. This is not our answer.
Choice B) This says that some point c has a slope of 0. Remember, the derivative at c is referring to the slope of the line at c. This is NOT necessarily true, especially for the y = 3x graph we are envisioning. This is our answer.
Choice C) Every interval has a minimum. It's just the lowest point on the graph in that interval.
Choice D) Every interval has a maximum. It's just the highest point on the graph in that interval.
Choice E) Remember that the "integral" is the area under the curve? As long as the graph is continuous (which it is), you can always find the area under the curve.
10) The equation given to you here is the forward difference quotient for an f(x) function at x = 2. Notice that it's the limit as h approaches 0, and the numerator has the "x+h" part (here it's 2+h) and the "x" part (here it's 2). In other words, the equation given to you here is the derivative of f(x) at x = 2.
10) The equation given to you here is the forward difference quotient for an f(x) function at x = 2. Notice that it's the limit as h approaches 0, and the numerator has the "x+h" part (here it's 2+h) and the "x" part (here it's 2). In other words, the equation given to you here is the derivative of f(x) at x = 2.
Because this is saying that f'(2) = 5, that means the derivative of f(x) at x = 2 is 5. In order for a function to have a derivative, it MUST be continuous. So then we can conclude...
I. --> f is definitely continuous at x = 2.
II. --> f is definitely differentiable at x = 2.
III. --> They don't say anything else about the graph, so we have no idea if the f'(x) graph is continuous at x = 2 or not. Maybe the f'(x) graph ends at x = 2, or maybe it doesn't. We can't make any conclusions about it!
ANSWER: C
12) They are giving you a(t), and asking you about v(t). We know that the derivative of v(t) gives you a(t), so to go backwards: the anti-derivative of a(t) will give you v(t).
We need to find an anti-derivative of a(t):
a(t) = t + sin t
Make the exponent of "t" one bigger, then divide by the new exponent. Then say, "the derivative of what would give me sin? -cos!"
v(t) = 1/2*t^2 - cos t + c
They gave you a specific value of v(t) in order to take your general anti-derivative to a particular anti-derivative. They said v(0) = -2, so plug that in to solve for c:
-2 = 1/2*0^2 - cos 0 + c
-2 = 0 - 1 + c
-1 = c
So, v(t) = 1/2*t^2 - cos t - 1. They want to know when v(t) = 0, so set v(t) equal to 0, then solve for t:
0 = 1/2*t^2 - cos t -1
To solve for t, there are several methods (all involving your graphing calculator). I would graph this on your calculator, then find the x-intercepts by using 2ND, TRACE, "Zero".
ANSWER: B (t = 1.4781703 sec)
Chp 3 Review (pg. 123)
R2b) "Why do I get a different derivative from using the power rule than using the derivative formula?"
Keep in mind that the definition of a derivative formula (as x approaches c) is NOT the same as the actual derivative formula. When you use f(x) - f(x) / x - c, you are setting up a derivative specifically at the value of x = c. In this problem, your formula will give you the correct derivative, but only for x = 3, because that’s how you set up the formula.
On the other hand, the forward difference quotient f(x + h) - f(x) / h WILL give you the correct derivative formula, because it’s not set up dependent on a certain x-value.
R5bi) The "y" equation given is displacement. To find the velocity, you need the derivative of displacement, which is y'. Use the power rule:
velocity = y' = -0.03t^2 + 1.8t - 25
To find acceleration, you need the derivative of velocity, which is y''. Use the power rule:
acceleration = y'' = -0.06t + 1.8
R5bii) To find the acceleration at t = 15 sec, plug t = 15 into the acceleration equation y'':
y'' = -0.06(15) + 1.8 = 0.9 km/sec^2
When you're figuring out the units for these types of equations, remember that derivatives always divide the units you already had. So, if the displacement units are in feet, then velocity will be feet/sec, and acceleration will be feet/sec^2.
R5biii) "Direct calculation" is their way of saying "don't use your graphing calculator, use calculus stuff". If we want to know when the spaceship is stopped, then we are looking for when velocity is equal to 0. Set the y' equation equal to 0:
0 = -0.03t^2 + 1.8t - 25
and use the quadratic formula to solve. You should get t = 21.835 sec and t = 38.165 sec.
R5biv) The spaceship will touch Mars when its displacement is equal to 0, so set y equal to 0. Solve by using your graphing calculator (this means you could graph the cubic function and find all its x-intercerpts by using the "zero" function under 2nd, TRACE).
0 = -0.01t^3 + 0.9t^2 - 25t + 250
t = 50 sec
To find the velocity at t = 50 sec, plug t = 50 into the velocity equation y':
y' = -0.03(50)^2 + 1.8(50) - 25
y' = -10 km/sec
This would be way too fast for a landing! -10 km/sec is about -22,000 mph, so the spaceship would be coming in fast!
R7ai) Chain rule definition in dy/dx form:
dy/dx = dy/du * du/dx
In plain words, this says that you first find the derivative of the outside function (y) with respect to the inside function (u). "With respect to" here basically means that we're not touching the inside part yet.
Then, you multiply by the derivative of the inside (u) with respect to x (the actual variable). This is saying that NOW we can find the derivative of the inside.
The definition in the next part of the problem usually makes more sense to people... :)
R7aii) Chain rule definition in f(x) form:
When f(x) = g( h(x) ) , then the derivative is given by f'(x) = g'( h(x) ) * h'(x)
In plain words, this says that when you have an outside function (g) and an inside function (h), you can find the derivative by using the chain rule. You start by finding the derivative of the outside (g') while keeping the inside the same (h), then multiplying by the derivative of the inside (h').
R7ci) To find the derivative by using the chain rule, you find the derivative of the outside by rolling the exponent down and making the new exponent one less:
f'(x) = 3(x^2 - 4)^2 ...
...then multiplying by the derivative of the inside:
f'(x) = 3(x^2 - 4)^2 * 2x ...
...then simplifying by multiplying the 3 and 2x together at the front. Notice that the inside function itself stayed the same the whole time:
f'(x) = 6x(x^2 - 4)^2
R7cii) To expand the binomial, write it out:
f(x) = (x^2 - 4)(x^2 - 4)(x^2 - 4)
then start distributing. I'll leave the first factor alone for now, and distribute the second and third factors together:
f(x) = (x^2 - 4)(x^4 - 4x^2 - 4x^2 + 16)
f(x) = (x^2 - 4)(x^4 - 8x^2 + 16)
Now distribute the first factor into the second big factor, then combine like terms:
f(x) = x^6 - 8x^4 + 16x^2 - 4x^4 + 32x^2 - 64
f(x) = x^6 - 12x^4 + 48x^2 - 64
Now find f'(x) by using the power rule:
f'(x) = 6x^5 - 48x^3 + 96x
To show that the two answers are equivalent:
You could expand the binomial from your chain rule answer in R7ci:
f'(x) = 6x(x^2 - 4)(x^2 - 4)
f'(x) = 6x(x^4 - 8x^2 + 16)
f'(x) = 6x^5 - 48x^3 + 96x
and see that your two derivatives are actually the same.
R7e) Use the dy/dx definition of the chain rule. Here, you are given the equation W that is using the variable x. But, you are being asked about an answer in pounds per day, which would be the derivative of W with respect to time t. So, they're actually asking you for dW/dt. Let's set up the big idea here:
dW/dt = dW/dx * dx/dt
[See my explanation in R7ai above regarding this chain rule definition.]
We can find dW/dx, because this is the derivative of the W equation with respect to x:
W = 0.6x^3
dW/dx = 1.8x^2
They told us that dx/dt = 0.4 ft/day, so we can just plug that constant into the formula for dx/dt. Now, we have:
dW/dt = 1.8x^2 * 0.4
or simplified:
dW/dt = 0.72x^2
"How heavy is the shark when it is 2 ft long? 10 ft long?" This is giving you x and asking for W, so use your original equation involving W and x:
W = 0.6(2)^3 = 4.8 lbs
W = 0.6(10)^3 = 600 lbs
"At what rate is it gaining weight (pounds per day) when it is these lengths?" This is using the same x values and asking you for dW/dt, because it's asking about the rate of change of W (pounds per day). Use your dW/dt equation:
dW/dt = 0.72(2)^2 = 2.88 lbs/day
dW/dt = 0.72(10)^2 = 72 lbs/day
R8c) This is the same equation I made you guys look at on the overhead when we talked about Sec 3.8. (sin x)/x can be squeezed in between y = cos x and y = 1, because (sin x)/x in always in between cos x and 1 when we're looking at the neighborhood of x = 0. y = cos x and y = 1 both have a limit of 1 as x approaches 0, which means (sin x)/x will also have a limit of 1 as x approaches 0, per the squeeze theorem.
R9b) This is asking you to find a particular anti-derivative. It's giving you the derivative, and giving you a value of the original y function. Start by finding the general anti-derivative:
dy/dx = sin 0.2x
[The derivative of what will give you sin? ---> -cos . Then, divide the current coefficient by the derivative of the inside function.]
y = -5 cos 0.2x + c
You can solve for the +c by plugging in the original values given of y = 3 and x = 0:
3 = -5 cos 0.2(0) + c
3 = -5(1) + c
3 = -5 + c
8 = c
Particular anti-derivative answer: y = -5 cos 0.2x + 8
R9c) Because the derivative of displacement will give you velocity, you'll need to find the anti-derivative of velocity to give you displacement.
v(t) = 6t^(1/2)
Add one to the exponent, then divide by your new exponent:
y(t) = 4t^(3/2) + c
To solve for +c, notice that it said she starts 100 ft from her house at t = 0 sec. Use this to find +c.
100 = 4(0)^(3/2) + c
100 = c
Particular anti-derivative: y(t) = 4t^(3/2) + 100
"How far from the house is she 1 min after she starts?" This is asking for y(1).
"How fast is she going at that time?" This is asking for v(1).
Sec 3.9 (pg. 121):
On the other hand, the forward difference quotient f(x + h) - f(x) / h WILL give you the correct derivative formula, because it’s not set up dependent on a certain x-value.
R5bi) The "y" equation given is displacement. To find the velocity, you need the derivative of displacement, which is y'. Use the power rule:
velocity = y' = -0.03t^2 + 1.8t - 25
To find acceleration, you need the derivative of velocity, which is y''. Use the power rule:
acceleration = y'' = -0.06t + 1.8
R5bii) To find the acceleration at t = 15 sec, plug t = 15 into the acceleration equation y'':
y'' = -0.06(15) + 1.8 = 0.9 km/sec^2
When you're figuring out the units for these types of equations, remember that derivatives always divide the units you already had. So, if the displacement units are in feet, then velocity will be feet/sec, and acceleration will be feet/sec^2.
R5biii) "Direct calculation" is their way of saying "don't use your graphing calculator, use calculus stuff". If we want to know when the spaceship is stopped, then we are looking for when velocity is equal to 0. Set the y' equation equal to 0:
0 = -0.03t^2 + 1.8t - 25
and use the quadratic formula to solve. You should get t = 21.835 sec and t = 38.165 sec.
R5biv) The spaceship will touch Mars when its displacement is equal to 0, so set y equal to 0. Solve by using your graphing calculator (this means you could graph the cubic function and find all its x-intercerpts by using the "zero" function under 2nd, TRACE).
0 = -0.01t^3 + 0.9t^2 - 25t + 250
t = 50 sec
To find the velocity at t = 50 sec, plug t = 50 into the velocity equation y':
y' = -0.03(50)^2 + 1.8(50) - 25
y' = -10 km/sec
This would be way too fast for a landing! -10 km/sec is about -22,000 mph, so the spaceship would be coming in fast!
R7ai) Chain rule definition in dy/dx form:
dy/dx = dy/du * du/dx
In plain words, this says that you first find the derivative of the outside function (y) with respect to the inside function (u). "With respect to" here basically means that we're not touching the inside part yet.
Then, you multiply by the derivative of the inside (u) with respect to x (the actual variable). This is saying that NOW we can find the derivative of the inside.
The definition in the next part of the problem usually makes more sense to people... :)
R7aii) Chain rule definition in f(x) form:
When f(x) = g( h(x) ) , then the derivative is given by f'(x) = g'( h(x) ) * h'(x)
In plain words, this says that when you have an outside function (g) and an inside function (h), you can find the derivative by using the chain rule. You start by finding the derivative of the outside (g') while keeping the inside the same (h), then multiplying by the derivative of the inside (h').
R7ci) To find the derivative by using the chain rule, you find the derivative of the outside by rolling the exponent down and making the new exponent one less:
f'(x) = 3(x^2 - 4)^2 ...
...then multiplying by the derivative of the inside:
f'(x) = 3(x^2 - 4)^2 * 2x ...
...then simplifying by multiplying the 3 and 2x together at the front. Notice that the inside function itself stayed the same the whole time:
f'(x) = 6x(x^2 - 4)^2
R7cii) To expand the binomial, write it out:
f(x) = (x^2 - 4)(x^2 - 4)(x^2 - 4)
then start distributing. I'll leave the first factor alone for now, and distribute the second and third factors together:
f(x) = (x^2 - 4)(x^4 - 4x^2 - 4x^2 + 16)
f(x) = (x^2 - 4)(x^4 - 8x^2 + 16)
Now distribute the first factor into the second big factor, then combine like terms:
f(x) = x^6 - 8x^4 + 16x^2 - 4x^4 + 32x^2 - 64
f(x) = x^6 - 12x^4 + 48x^2 - 64
Now find f'(x) by using the power rule:
f'(x) = 6x^5 - 48x^3 + 96x
To show that the two answers are equivalent:
You could expand the binomial from your chain rule answer in R7ci:
f'(x) = 6x(x^2 - 4)(x^2 - 4)
f'(x) = 6x(x^4 - 8x^2 + 16)
f'(x) = 6x^5 - 48x^3 + 96x
and see that your two derivatives are actually the same.
R7e) Use the dy/dx definition of the chain rule. Here, you are given the equation W that is using the variable x. But, you are being asked about an answer in pounds per day, which would be the derivative of W with respect to time t. So, they're actually asking you for dW/dt. Let's set up the big idea here:
dW/dt = dW/dx * dx/dt
[See my explanation in R7ai above regarding this chain rule definition.]
We can find dW/dx, because this is the derivative of the W equation with respect to x:
W = 0.6x^3
dW/dx = 1.8x^2
They told us that dx/dt = 0.4 ft/day, so we can just plug that constant into the formula for dx/dt. Now, we have:
dW/dt = 1.8x^2 * 0.4
or simplified:
dW/dt = 0.72x^2
"How heavy is the shark when it is 2 ft long? 10 ft long?" This is giving you x and asking for W, so use your original equation involving W and x:
W = 0.6(2)^3 = 4.8 lbs
W = 0.6(10)^3 = 600 lbs
"At what rate is it gaining weight (pounds per day) when it is these lengths?" This is using the same x values and asking you for dW/dt, because it's asking about the rate of change of W (pounds per day). Use your dW/dt equation:
dW/dt = 0.72(2)^2 = 2.88 lbs/day
dW/dt = 0.72(10)^2 = 72 lbs/day
R8c) This is the same equation I made you guys look at on the overhead when we talked about Sec 3.8. (sin x)/x can be squeezed in between y = cos x and y = 1, because (sin x)/x in always in between cos x and 1 when we're looking at the neighborhood of x = 0. y = cos x and y = 1 both have a limit of 1 as x approaches 0, which means (sin x)/x will also have a limit of 1 as x approaches 0, per the squeeze theorem.
R9b) This is asking you to find a particular anti-derivative. It's giving you the derivative, and giving you a value of the original y function. Start by finding the general anti-derivative:
dy/dx = sin 0.2x
[The derivative of what will give you sin? ---> -cos . Then, divide the current coefficient by the derivative of the inside function.]
y = -5 cos 0.2x + c
You can solve for the +c by plugging in the original values given of y = 3 and x = 0:
3 = -5 cos 0.2(0) + c
3 = -5(1) + c
3 = -5 + c
8 = c
Particular anti-derivative answer: y = -5 cos 0.2x + 8
R9c) Because the derivative of displacement will give you velocity, you'll need to find the anti-derivative of velocity to give you displacement.
v(t) = 6t^(1/2)
Add one to the exponent, then divide by your new exponent:
y(t) = 4t^(3/2) + c
To solve for +c, notice that it said she starts 100 ft from her house at t = 0 sec. Use this to find +c.
100 = 4(0)^(3/2) + c
100 = c
Particular anti-derivative: y(t) = 4t^(3/2) + 100
"How far from the house is she 1 min after she starts?" This is asking for y(1).
"How fast is she going at that time?" This is asking for v(1).
Sec 3.9 (pg. 121):
21) This is a polynomial function, so we can use (the opposite of) the power rule. With the regular power rule, you can find a derivative by multiplying the coefficient by the old exponent, and making your new exponent "one smaller". We are going to find the anti-derivative by making each exponent "one bigger", then dividing by the new exponent.
f'(x) = x^2 - 8x + 3
f(x) = 1/3x^3 - 4x^2 + 3x + c
The first term was degree 2, and is now degree 3. We divided the coefficient by 3.
The second term was degree 1, and is now degree 2. We divided the coefficient by 2.
The third term was degree 0, and is now degree 1.
Then +c :)
Now, use the given information f(-2) = 13. Plug in x = -2 and y = 13 to your new f(x) equation in order to solve for c:
13 = 1/3(-2)^3 - 4(-2)^2 + 3(-2) + c
13 = 1/3(-8) - 4(4) - 6 + c
13 = -8/3 - 16 - 6 + c
Making a common denominator everywhere:
39/3 = -8/3 - 48/3 - 18/3 + c
39/3 = -74/3 + c
113/3 = c
24a) This is not a polynomial function, so you are not using (the opposite of) the power rule. No exponents! If you want to find the anti-derivative of sin, say to yourself: "The derivative of what would give me sin?" The derivative of -cos would give you sin. So, your anti-derivative function will have -cos in it.
So far, we have v(t) = -18 cos 3t. The inside always stays the same, and the coefficient at the front will still be there. Lastly, we need to account for the derivative of the inside function. The derivative of 3t is 3, so we need to balance that out in the anti-derivative by multiplying by 1/3.
Our final anti-derivative answer is: v(t) = -6 cos 3t + c
Checking your answer:
You can check this by finding the derivative of this answer. Its derivative should give you the equation you started with in the book:
v(t) = -6 cos 3t + c
Do the derivative of the outside, keep the inside the same, then multiply by the derivative of the inside:
v'(t) = 6 sin 3t * 3
v'(t) = 18 sin 3t
24b) Velocity shows which direction your speed is traveling. So, negative velocity would be when the swing is traveling backward.
24d) To find the maximum velocity, you would set v'(t) = 0, because that is the vertex of v(t) (where the slope of v(t) is zero). So, you'd start with:
v'(t) = 18 sin 3t = 0
then solve from there.
sin 3t = 0
3t = 0 and 3t = pi
t = 0 and t = pi/3
t = 0 is irrelevant, because that's when he starts. We find that he reaches maximum velocity when t = pi/3 seconds, and that his acceleration at this time is 0 ft/sec^2.
7) It will be easiest to find the derivative of d(t) if you distribute the inside parentheses first (so that you only have one set of parentheses in the whole equation). So, you should be starting with this d(t) function:
d(t) = 7 cos (pi/5 * t - 2pi/5) + 6
To do the chain rule, find the derivative of the “outside” function first, while keeping the “inside” the same. The outside function here is cos, so the derivative of it is -sin:
d’(t) = -7 sin (pi/5 * t - 2pi/5) + 6 [this answer is not done yet!]
Then, multiply by the derivative of the inside. The “inside” is (pi/5 * t - 2pi/5), where the derivative of pi/5 * t is pi/5, and the derivative of -2pi/5 is 0:
d’(t) = -7pi/5 sin (pi/5 * t - 2pi/5) + 6 [this answer is not done yet!]
Then, the derivative of the +6 on the end is 0. Here is the FINAL DERIVATIVE EQUATION:
d’(t) = -7pi/5 sin (pi/5 * t - 2pi/5)
Sec 3.7 help: Chain Rule
This link is an Educreations video I made. Make sure your sound is on!
Examples (notes) on how to use the Chain Rule
Sketching Graphs practice wkst.
What can you try while you're learning to draw an antiderivative ("going backwards") graph?
It would be pretty easy to make up your own practice problems. Type whatever you want into Y1= on your calculator to see a starting graph, then try to draw the antiderivative graph on your own paper. Then, in Y2= you should type fnInt(Y1, x, 0, x). It will graph it REALLY slowly, but it should work. You’ll find “fnInt” under the MATH button, then it’s #9 on the list.
It would be pretty easy to make up your own practice problems. Type whatever you want into Y1= on your calculator to see a starting graph, then try to draw the antiderivative graph on your own paper. Then, in Y2= you should type fnInt(Y1, x, 0, x). It will graph it REALLY slowly, but it should work. You’ll find “fnInt” under the MATH button, then it’s #9 on the list.
Here are a couple examples of what it could look like. The bold line on each graph is the original graph you had - that we were going to call f’(x). The thinner line would be its anti-derivative, f(x).
Q5) "I tried to use the difference quotient, but it did not work."
3) Use the power rule to find the derivative. Roll down the exponent (which means multiply the coefficient of 0.007 by the exponent -83), then make the new exponent one less (which means subtract 1 from -83). Your answer should be:
v’ = -0.581t^-84
or using the other notation for derivatives, the answer is:
dv/dt = -0.581t^-84
Below, I have included thoughts, answers, math lessons, etc. for each of the questions on this worksheet for #1 - 6. You need to answer each of these questions in your own words (sometimes your answer and my answer will be very similar or identical, but not all the time), and then complete the rest of the worksheet. You can find the answers to this homework assignment online as the last page of this file in my Google Drive folder.
1) The apostrophe on "f" means that we are now referring to the derivative of f. If we just wrote f'(x), then we're referring to the derivative function (an entire equation). If we write f'(c), then we're referring to the derivative value at a specific "c" location.
2) Here, use nDeriv (MATH, 8, then type the rest in) that I showed you in class today. Look at your "Calculus Skills Sheet" if you've forgotten how to do it.
3) When it asks you to find f'(c), it's actually asking you to find the exact derivative at x = c. We have been using the "exact definition of a derivative" (also called the "difference quotient", also called the "forward difference quotient") to do this. Write down the formula we've been using that involves a limit as x approaches c.
4) "the derivative"
5) "the slope"
These two problems are trying to get you to remember that the derivative is the same as the slope, and that f'(x) is referring to the derivative of f(x).
6) TO GRAPH THE DERIVATIVE FUNCTION f'(x):
In Y1, type the f(x) equation you've started with.
In Y2, type: nDeriv(Y1, X, X) (older calculators) or d/dx (Y1) | X = X (newer calculators)
The "Y1" tells your calculator to find the derivative of Y1. The first "X" tells your calculator to find the derivative in terms of "X" (it will always be this). The second "X" tells your calculator to find the derivative value at every single value of x, not just at one specific value.
Then, press "Graph". It will graph your derivative, Y2, very very slowly, because your calculator is actually calculating every single derivative value individually.
Here's what your screen should look like (both in "Mathprint" mode and "Classic" mode).
6a - d) I am looking for answers involving < or > signs, and using words like "increasing" or "decreasing" or "positive values" or "negative values".
6h) HINT: In your explanation of how the derivative graph has or has not changed, talk about the slope of f(x), because this is the same thing as talking about the derivative of f(x). This all the help you get for now, but please feel free to email me with any questions you may have! Good luck! :)
3) Did you try factoring out a common factor first? You have 0.6x^2 - 5.4 in the numerator. Start by factoring out 0.6:
0.6(x^2 - 9)
Now you can factor the x^2 - 9 part like normal:
0.6(x + 3)(x - 3)
Now you can cancel with the denominator! Hope this helps get you started!
