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.
Extra topics we've covered in class:
Educreations Video:
YouTube Video:
Chapter 4 Review Packet:
These links are Educreations videos I made. Make sure your sound is on! (I might refer to the book problem number instead of the packet problem number in the video, since I made the videos last year).
The actual goal is to find the derivative of this equation. Use implicit differentiation:
d/dx ( tan y = 4x^2 )
sec^2 y * dy/dx = 8x ... then get dy/dx by itself:
dy/dx = 8x / sec^2 y
dy/dx = 8x * cos^2 y
We can use that right triangle to figure out what cos y is equal to, because "y" was an angle in our triangle. We know cos = adj/hyp. In our triangle the ADJACENT was 1 and the HYPOTENUSE was sqrt(16x^4 + 1).
cos y = 1 / sqrt(16x^4 + 1)
Then... dy/dx = 8x * cos^2 y ...becomes...
dy/dx = 8x * (1 / sqrt(16x^4 + 1))^2
dy/dx = 8x * (1 / (16x^4 + 1))
dy/dx = 8x / (16x^4 + 1)
26) This is probably the toughest problem on the entire assignment. I strongly encourage you to look at the work that’s written on my answer key in the Google Drive folder. But, to help you decode all those steps, here are the big steps you take along the way:
Step 1) You’re given an equation with x and y on the same side, so you need to use implicit differentiation. Remember to multiply by dy/dx every time you find a derivative involving y.
After you find the derivative, get all terms with dy/dx on one side, and all other terms on the other side. Then, factor out the dy/dx. Then, get dy/dx by itself by dividing.
Step 2) Next, you need to use the points “where x = -2”. But, they only told you that x = -2, and they didn’t tell you what y is equal to at that point. Use your ORIGINAL EQUATION to find the necessary y-values. Plug in x = -2 everywhere x shows up, then solve for y (you’ll need to use quadratic formula to solve for y at some point in this work).
Step 3) To actually find the slope at x = -2, now you’ll use the dy/dx equation (because the derivative equation will tell you the slope at any point), and you’ll plug in the (x, y) values you found earlier.
These links are Educreations videos I made for Implicit Derivatives. Make sure your sound is on!
Sec 4.5:
These links are Educreations videos I made for Derivatives of Inverses. Make sure your sound is on!
13) There are two main ways to do this problem: Derivative of y = sin^-1(4x)
sin = opp / hyp, so draw a triangle with opposite side = 4x and hypotenuse = 1.
Use Pythagorean Theorem to find missing side of triangle:
(4x)^2 + b^2 = 1^2
16x^2 + b^2 = 1
b^2 = 1 - 16x^2
b = sqrt(1 - 16x^2)
When you're solving with SOH CAH TOA, we know that cos = adj / hyp. Because sec is the reciprocal of cos, then sec must be equal to the reciprocal of the cos proportion.
sec = hyp / adj
sec(y) = 1 / sqrt(1 - 16x^2)
Substitute this back into the derivative equation you found, dy/dx:
dy/dx = 4 * sec(y)
dy/dx = 4 * 1 / sqrt(1 - 16x^2)
13) For this problem, the most "outside function" is ^(5/4).
y' = 40 (sec 4x)^(5/4) * tan 4x
21) y = sec x csc x
24) The equation can simplify before you even find the derivative. cot x is the same as cos x / sin x, so the cos x part will cancel out with the denominator, leaving you with 1 / sin x. This is the same as csc x. See below:
Remember, my Google Drive folder contains copies of worksheets and ANSWERS (on the last page of each file). CLICK HERE TO ACCESS.
Extra topics we've covered in class:
Educreations Video:
YouTube Video:
Chapter 4 Review Packet:
These links are Educreations videos I made. Make sure your sound is on! (I might refer to the book problem number instead of the packet problem number in the video, since I made the videos last year).
13) You're given y = tan^-1 (4x^2). Start by moving "tan" to the other side:
tan y = 4x^2
Next, set up a right triangle where "y" is an angle. From SOH CAH TOA, we know tan = opp/adj. Here, we have tan y = 4x^2 / 1. So, your triangle should have a side OPPOSITE of y that says 4x^2, and ADJACENT to y that says 1.
To find the missing hypotenuse side, use the Pythagorean Theorem:
(4x^2)^2 + 1^2 = c^2
16x^4 + 1 = c^2
sqrt(16x^4 + 1) = c
Back to the starting equation: tan y = 4x^2
d/dx ( tan y = 4x^2 )
sec^2 y * dy/dx = 8x ... then get dy/dx by itself:
dy/dx = 8x / sec^2 y
dy/dx = 8x * cos^2 y
We can use that right triangle to figure out what cos y is equal to, because "y" was an angle in our triangle. We know cos = adj/hyp. In our triangle the ADJACENT was 1 and the HYPOTENUSE was sqrt(16x^4 + 1).
cos y = 1 / sqrt(16x^4 + 1)
Then... dy/dx = 8x * cos^2 y ...becomes...
dy/dx = 8x * (1 / sqrt(16x^4 + 1))^2
dy/dx = 8x * (1 / (16x^4 + 1))
dy/dx = 8x / (16x^4 + 1)
24f) Because you have both x and y on the same side, you need to do an implicit derivative. Remember, every time you find a derivative with y in it, you need to multiply by dy/dx (or y’ for short).
We’re starting with:
5x^3*y^7 = y^1.6
Let’s find the derivative on the right of the = sign first. Just do the power rule, but multiply by y’:
= 1.6y^0.6 * y’
To find the derivative on the left side, we need to use product rule, because we have two functions multiplied. It will help to write them out as “u” and “v”, then find u’ and v’.
u = 5x^3
u’ = 15x^2
v = y^7
v’ = 7y^6 * y’
Now use product rule, where the derivative will equal u’v + uv’.
15x^2 * y^7 + 5x^3 * 7y^6 * y’ = 1.6y^0.6 * y’
Now get all terms with y’ on one side, and all terms without y’ on the other side. I’ve bolded the terms with y’ and underlined the terms without y’.
15x^2 * y^7 + 5x^3 * 7y^6 * y’ = 1.6y^0.6 * y’
You can subtract to move the 5x^3 * 7y^6 * y’ to the other side.
15x^2 * y^7 = 1.6y^0.6 * y’ - 5x^3 * 7y^6 * y’
Now factor out the y’, because it’s common to both terms on the right side.
15x^2 * y^7 = y’(1.6y^0.6 - 5x^3 * 7y^6)
Now divide to get y’ by itself. This is your final answer.
(15x^2 * y^7) / (1.6y^0.6 - 5x^3 * 7y^6) = y’
We’re starting with:
5x^3*y^7 = y^1.6
Let’s find the derivative on the right of the = sign first. Just do the power rule, but multiply by y’:
= 1.6y^0.6 * y’
To find the derivative on the left side, we need to use product rule, because we have two functions multiplied. It will help to write them out as “u” and “v”, then find u’ and v’.
u = 5x^3
u’ = 15x^2
v = y^7
v’ = 7y^6 * y’
Now use product rule, where the derivative will equal u’v + uv’.
15x^2 * y^7 + 5x^3 * 7y^6 * y’ = 1.6y^0.6 * y’
Now get all terms with y’ on one side, and all terms without y’ on the other side. I’ve bolded the terms with y’ and underlined the terms without y’.
15x^2 * y^7 + 5x^3 * 7y^6 * y’ = 1.6y^0.6 * y’
You can subtract to move the 5x^3 * 7y^6 * y’ to the other side.
15x^2 * y^7 = 1.6y^0.6 * y’ - 5x^3 * 7y^6 * y’
Now factor out the y’, because it’s common to both terms on the right side.
15x^2 * y^7 = y’(1.6y^0.6 - 5x^3 * 7y^6)
Now divide to get y’ by itself. This is your final answer.
(15x^2 * y^7) / (1.6y^0.6 - 5x^3 * 7y^6) = y’
25) My answer key is actually a little shorter than it should be for this problem. Because you’re finding the derivative of an inverse trig function, this is one of those problems where you need to set up a SOH CAH TOA triangle and do an implicit derivative. Here are the steps:
First, change the equation so it’s not an inverse trig function any more:
y = 4 sin^-1 (5x^3)
y/4 = sin^-1 (5x^3)
sin (y/4) = 5x^3
Now, do two things with this equation. One, find the implicit derivative. Two, set up a SOH CAH TOA triangle with it.
Implicit derivative (remember to multiply by dy/dx any time you find a derivative involving y):
cos (y/4) * 1/4 * dy/dx = 15x^2
Then solve for dy/dx:
cos (y/4) * dy/dx = 60x^2
dy/dx = 60x^2 / cos (y/4)
SOH CAH TOA triangle:
Use your equation from before taking the implicit derivative: sin (y/4) = 5x^3. It’s always sin of the angle, so the angle must be y/4. Then, sin is equal to opp / hyp, so the opposite side must be 5x^3 and the hypotenuse must be 1. I’ll do my best to type out a triangle:
----
1 ---- |
---- | 5x^3
y/4__________
Use Pythagorean Theorem to find the missing side (adj side):
A^2 + (5x^3)^2 = 1^2
A^2 + 25x^6 = 1
A^2 = 1 - 25x^6
A = sqrt(1 - 25x^6) ----
1 ---- |
---- | 5x^3
y/4__________
sqrt(1 - 25x^6)
The reason we’re doing this is because our derivative equation left off as: dy/dx = 60x^2 / cos (y/4) but we really don’t want any “y”s to show up in it. So, we’ll use this triangle to figure out what cos (y/4) is equal to. Remember, cos = adj / hyp, so…
cos (y/4) = sqrt(1 - 25x^6) / 1
cos (y/4) = sqrt(1 - 25x^6)
Back to our derivative! Now we can plug this in for cos (y/4):
dy/dx = 60x^2 / cos (y/4)
dy/dx = 60x^2 / sqrt(1 - 25x^6)
First, change the equation so it’s not an inverse trig function any more:
y = 4 sin^-1 (5x^3)
y/4 = sin^-1 (5x^3)
sin (y/4) = 5x^3
Now, do two things with this equation. One, find the implicit derivative. Two, set up a SOH CAH TOA triangle with it.
Implicit derivative (remember to multiply by dy/dx any time you find a derivative involving y):
cos (y/4) * 1/4 * dy/dx = 15x^2
Then solve for dy/dx:
cos (y/4) * dy/dx = 60x^2
dy/dx = 60x^2 / cos (y/4)
SOH CAH TOA triangle:
Use your equation from before taking the implicit derivative: sin (y/4) = 5x^3. It’s always sin of the angle, so the angle must be y/4. Then, sin is equal to opp / hyp, so the opposite side must be 5x^3 and the hypotenuse must be 1. I’ll do my best to type out a triangle:
----
1 ---- |
---- | 5x^3
y/4__________
Use Pythagorean Theorem to find the missing side (adj side):
A^2 + (5x^3)^2 = 1^2
A^2 + 25x^6 = 1
A^2 = 1 - 25x^6
A = sqrt(1 - 25x^6) ----
1 ---- |
---- | 5x^3
y/4__________
sqrt(1 - 25x^6)
The reason we’re doing this is because our derivative equation left off as: dy/dx = 60x^2 / cos (y/4) but we really don’t want any “y”s to show up in it. So, we’ll use this triangle to figure out what cos (y/4) is equal to. Remember, cos = adj / hyp, so…
cos (y/4) = sqrt(1 - 25x^6) / 1
cos (y/4) = sqrt(1 - 25x^6)
Back to our derivative! Now we can plug this in for cos (y/4):
dy/dx = 60x^2 / cos (y/4)
dy/dx = 60x^2 / sqrt(1 - 25x^6)
26) This is probably the toughest problem on the entire assignment. I strongly encourage you to look at the work that’s written on my answer key in the Google Drive folder. But, to help you decode all those steps, here are the big steps you take along the way:
Step 1) You’re given an equation with x and y on the same side, so you need to use implicit differentiation. Remember to multiply by dy/dx every time you find a derivative involving y.
After you find the derivative, get all terms with dy/dx on one side, and all other terms on the other side. Then, factor out the dy/dx. Then, get dy/dx by itself by dividing.
Step 2) Next, you need to use the points “where x = -2”. But, they only told you that x = -2, and they didn’t tell you what y is equal to at that point. Use your ORIGINAL EQUATION to find the necessary y-values. Plug in x = -2 everywhere x shows up, then solve for y (you’ll need to use quadratic formula to solve for y at some point in this work).
Step 3) To actually find the slope at x = -2, now you’ll use the dy/dx equation (because the derivative equation will tell you the slope at any point), and you’ll plug in the (x, y) values you found earlier.
Sec 4.8:
These links are Educreations videos I made for Implicit Derivatives. Make sure your sound is on!
Sec 4.5:
These links are Educreations videos I made for Derivatives of Inverses. Make sure your sound is on!
13) There are two main ways to do this problem: Derivative of y = sin^-1(4x)
Short way, but requires memorization:
According to the table on pg. 150, the derivative of y = sin^-1(x) is
1 / sqrt(1 - x^2)
Apply the chain rule to this problem! The derivative of the "outside", which is sin^-1, will be the formula given above. Keep the "inside", which is 4x, the same. Then multiply by the derivative of the inside.
1 / sqrt(1 - (4x)^2) * 4
= 4 / sqrt(1 - 16x^2)
Long way, but you don't have to memorize anything:
Start by finding "sin" of each side of the equation:
y = sin^-1(4x)
sin(y) = sin(sin^-1(4x))
sin(y) = 4x
Next, find the implicit derivative of each side of the equation:
cos(y) * dy/dx = 4
dy/dx = 4 / cos(y)
dy/dx = 4 * sec(y)
We can't leave our answer in terms of "y", so we will draw a right triangle using the initial information to figure out what "sec(y)" is equal to.
Initial information: sin(y) = 4x
Use Pythagorean Theorem to find missing side of triangle:
(4x)^2 + b^2 = 1^2
16x^2 + b^2 = 1
b^2 = 1 - 16x^2
b = sqrt(1 - 16x^2)
When you're solving with SOH CAH TOA, we know that cos = adj / hyp. Because sec is the reciprocal of cos, then sec must be equal to the reciprocal of the cos proportion.
sec = hyp / adj
sec(y) = 1 / sqrt(1 - 16x^2)
Substitute this back into the derivative equation you found, dy/dx:
dy/dx = 4 * sec(y)
dy/dx = 4 * 1 / sqrt(1 - 16x^2)
dy/dx = 4 / sqrt(1 - 16x^2)
23) You need to do product rule with the first part of the equation, then chain rule with the second part. Here's what each piece would look like when you find the derivative, then simplify:
x * sin^-1 x
When you find the derivative:
1 * sin^-1 x + x * 1/sqrt(1-x^2)
The derivative of sin^-1 x is given on page 150 in the textbook. We can simplify this equation to:
sin^-1 x + x/sqrt(1-x^2)
For the next part, use the chain rule: (1-x^2)^1/2
1/2(1-x^2)^-1/2 * -2x
Then simplify. The 1/2 * 2 will cancel out, leaving just -x. We can move the parentheses to the denominator to make the exponent positive. Then, once it's a positive 1/2 exponent, we can rewrite it as a square root. Simplified version:
-x/sqrt(1-x^2)
So, here's what we have for the derivative of v:
v' = sin^-1 x + x/sqrt(1-x^2) - x/sqrt(1-x^2)
Notice that the last two terms are the same, except one is positive x and one is negative x, so they will cancel. We are left with:
v' = sin^-1 x
25a) tan (theta) = opp/adj, so tan (theta) = x/100. You can then solve for (theta) by taking the inverse tan of both sides to get:
(theta) = tan^-1 (x/100)
25b) To find d(theta)/dx, you need to find the derivative of the equation from Part "a" of this problem.
(theta) = tan^-1 (x/100)
Find the derivative with respect to x on each side. This problem will be easier to just use the FORMULA for derivative of tan^-1. Be sure to apply chain rule, because you have an outside function (tan^-1) and an inside function (x/100):
d(theta)/dx = 1 / (1 + (x/100)^2) ... then also multiplied by 1/100, which is the derivative of x/100.
d(theta)/dx = 1 / (1 + (x/100)^2) * 1/100
We can algebraically simplify this equation a lot! Start by squaring the x/100:
d(theta)/dx = 1 / (1 + x^2/10000) * 1/100
Then, because you have two fractions multiplied together, we can multiply the top and then the bottom:
d(theta)/dx = 1 / (100 + 100x^2/10000)
d(theta)/dx = 1 / (100 + x^2/100)
It's awkward to have a fraction within the denominator of another fraction, so we should find a way to combine the two terms in the denominator. For just these two terms, find a common denominator, then add them together:
d(theta)/dx = 1 / (10000/100 + x^2/100)
d(theta)/dx = 1 / ((10000+ x^2)/100)
Then, instead of having this fraction in the denominator, you can multiply by its reciprocal:
d(theta)/dx = 1 * 100/(10000+ x^2)
d(theta)/dx = 100/(10000+ x^2)
WHEW! Finally, to find d(theta)/dt, let's think about our alternate definition of the chain rule, and how we could get from d(theta)/dx to d(theta)/dt:
d(theta)/dt = d(theta)/?? * ??/dt
The missing pieces..."??"...can be "dx":
d(theta)/dt = d(theta)/dx * dx/dt
So, we can use the d(theta)/dx equation we already found, and simply multiply it by dx/dt to force it to be the d(theta)/dt equation we're looking for:
d(theta)/dt = 100/(10000+ x^2) * dx/dt
25c) Plug in x = 500 and d(theta)/dt = -0/.04:
-0.04 = 100/(10000+ 500^2) * dx/dt
When this question asks you, "How fast is the truck going?", they're really just asking you what dx/dt is equal to. This is because the "x" units are feet, and the "t" units are seconds, so "dx/dt" is ft/sec.
When you solve this equation algebraically for dx/dt, you get 104, so your answer is "104 ft/sec". You can convert this to miles per hour by using pre-calc unit conversion strategies?
104 ft * 60 sec * 60 min * 1 mile = 71 mil/hr
sec 1 min 1 hr 5280 ft
25a) tan (theta) = opp/adj, so tan (theta) = x/100. You can then solve for (theta) by taking the inverse tan of both sides to get:
(theta) = tan^-1 (x/100)
25b) To find d(theta)/dx, you need to find the derivative of the equation from Part "a" of this problem.
(theta) = tan^-1 (x/100)
Find the derivative with respect to x on each side. This problem will be easier to just use the FORMULA for derivative of tan^-1. Be sure to apply chain rule, because you have an outside function (tan^-1) and an inside function (x/100):
d(theta)/dx = 1 / (1 + (x/100)^2) ... then also multiplied by 1/100, which is the derivative of x/100.
d(theta)/dx = 1 / (1 + (x/100)^2) * 1/100
We can algebraically simplify this equation a lot! Start by squaring the x/100:
d(theta)/dx = 1 / (1 + x^2/10000) * 1/100
Then, because you have two fractions multiplied together, we can multiply the top and then the bottom:
d(theta)/dx = 1 / (100 + 100x^2/10000)
d(theta)/dx = 1 / (100 + x^2/100)
It's awkward to have a fraction within the denominator of another fraction, so we should find a way to combine the two terms in the denominator. For just these two terms, find a common denominator, then add them together:
d(theta)/dx = 1 / (10000/100 + x^2/100)
d(theta)/dx = 1 / ((10000+ x^2)/100)
Then, instead of having this fraction in the denominator, you can multiply by its reciprocal:
d(theta)/dx = 1 * 100/(10000+ x^2)
d(theta)/dx = 100/(10000+ x^2)
WHEW! Finally, to find d(theta)/dt, let's think about our alternate definition of the chain rule, and how we could get from d(theta)/dx to d(theta)/dt:
d(theta)/dt = d(theta)/?? * ??/dt
The missing pieces..."??"...can be "dx":
d(theta)/dt = d(theta)/dx * dx/dt
So, we can use the d(theta)/dx equation we already found, and simply multiply it by dx/dt to force it to be the d(theta)/dt equation we're looking for:
d(theta)/dt = 100/(10000+ x^2) * dx/dt
25c) Plug in x = 500 and d(theta)/dt = -0/.04:
-0.04 = 100/(10000+ 500^2) * dx/dt
When this question asks you, "How fast is the truck going?", they're really just asking you what dx/dt is equal to. This is because the "x" units are feet, and the "t" units are seconds, so "dx/dt" is ft/sec.
When you solve this equation algebraically for dx/dt, you get 104, so your answer is "104 ft/sec". You can convert this to miles per hour by using pre-calc unit conversion strategies?
104 ft * 60 sec * 60 min * 1 mile = 71 mil/hr
sec 1 min 1 hr 5280 ft
Sec 4.4:
13) For this problem, the most "outside function" is ^(5/4).
The next inside function is sec.
The most inside function is 4x.
So, take the derivative of the ^(5/4) first:
y = 8(sec 4x)^(5/4)
y' = 8 * 5/4(sec 4x)^(1/4)
Then multiply by the derivative of sec:
y' = 8 * 5/4(sec 4x)^(1/4) * sec 4x * tan 4x
Then multiply by the derivative of 4x:
y' = 8 * 5/4(sec 4x)^(1/4) * sec 4x * tan 4x * 4
This is your entire derivative! The last step is to simplify by combining the constant numbers that are multiplied together.
8 * 5/4 * 4 = 40
Also, you have (sec 4x)^(1/4) and (sec 4x). These can be combined, because they are both sec and they both have 4x on the inside. Because the first one has an exponent of 1/4 and the second one has an exponent of 1, they combine to:
(sec 4x)^(5/4)
So, the equation you got when you found the derivative…
y' = 8 * 5/4(sec 4x)^(1/4) * sec 4x * tan 4x * 4
…now simplifies to this:
21) y = sec x csc x
This is two functions multiplied, so we need to use the product rule to find the derivative.
u = sec x
u' = sec x tan x
v = csc x
v' = -csc x cot x
So, the product rule gives us:
y' = sec x tan x * csc x + sec x * - csc x cot x
Next, try changing everything to terms of sin and cos to see if anything will cancel out. After you cancel things out, you can change everything back to sec, csc, tan, cot. See my work below:
24) The equation can simplify before you even find the derivative. cot x is the same as cos x / sin x, so the cos x part will cancel out with the denominator, leaving you with 1 / sin x. This is the same as csc x. See below:
Then you can just find the derivative of y = csc x:
y' = - csc x cot x
Sec 4.2 (pg. 134):
General Help) This link is an Educreations video I made to teach you how to use the Product Rule. There is NO sound on this one:
Video: Product Rule Example Problems
(There is a mistake in this video where I magically change a 20 to a 45. Sorry! I bet you can still figure out how to do the problem.)
29) ODD FUNCTION PROOF:
An odd function starts with the property f(-x) = -f(x). We will find the derivative of this equation, and will show that the end result is the same as an even function.
f(-x) = -f(x)
Find the derivative of each side, using the chain rule where appropriate:
f'(-x) * -1 = -f'(x)
Rearranged to show:
-f'(-x) = -f'(x)
Divide both sides by -1:
f'(-x) = f'(x)
This is the same as the property of an even function, so we've shown that the derivative of an odd function will give you an even function.
EVEN FUNCTION PROOF:
An even function starts with the property f(-x) = f(x). We will find the derivative of this equation, and will show that the end result is the same as an odd function.
f(-x) = f(x)
Find the derivative of each side, using chain rule where appropriate:
f'(-x) * -1 = f'(x)
Rearranged to show:
-f'(-x) = f'(x)
Divide both sides by -1:
f'(-x) = -f'(x)
This is the same as the property of an odd function, so we've shown that the derivative of an even function will give you an odd function.
35a) Start with A = L*W. When they're asking for dA/dt, they are asking you to find the derivative of your A equation with respect to t (this just means the derivative of A where t is the variable). This sounds confusing, but let's use the product rule like normal, and it should work out:
A = L*W
Product rule says: y' = u'v + uv'
The derivative of L with respect to time t will be: dL/dt
The derivative of W with respect to time t will be: dW/dt
dA/dt = dL/dt * W + L * dW/dt
35b) They're basically giving you the equations for L and W, and want you to plug them into the equation you found in 35a. Let's start by finding the derivative of L and of W:
W = 2 + 2 cos t
dW/dt = -2 sin t
L = 3 + 2 sin 2t
dL/dt = 4 cos 2t
We actually just automatically found the derivative of W and L with respect to time t because the variable in both of these equations is t. Now let's plug all four equations back into dA/dt:
dA/dt = dL/dt * W + L * dW/dt
dA/dt = (4 cos t)*(2 + 2 cos t) + (3 + 2 sin 2t)*(-2 sin t)
Look! Without even trying, we now have the derivative of A with respect to t, because the variables in our dA equation are t.
To actually answer this book problem, we need to use dA/dt. When they ask, "At what rate is the area changing when t = 4 sec?", they are asking about the rate of change of A, or in other words the derivative of A when t = 4. So, plug t = 4 into dA/dt. Use your calculator!
dA/dt = (4 cos 4)*(2 + 2 cos 4) + (3 + 2 sin 2*4)*(-2 sin 4)
dA/dt = 7.133 ft^2 / sec
This is a positive rate of change, which means the area is INCREASING at t = 4. To figure out the units, think about the units of A (ft^2) divided by the units of t (sec). This gives you ft^2 / sec. Do the same thing for t = 5, but I'll let you figure that out on your own.
Sec 4.1 (pg. 131):
6) "Just for fun, see if you can randomly guess these answers somehow..."
The real way to do this problem is to use the Product Rule and the Quotient Rule. It's OK if you weren't able to get these answers while doing Sec 4.1 for the first time; I just wanted you guys to try it out!
p'(2) = f'(2) * g(2) + f(2) * g'(2)
q'(2) = [f'(2) * g(2) - f(2) * g'(2)] / [g(2)]^2
General Help) This link is an Educreations video I made to teach you how to use the Product Rule. There is NO sound on this one:
Video: Product Rule Example Problems
(There is a mistake in this video where I magically change a 20 to a 45. Sorry! I bet you can still figure out how to do the problem.)
29) ODD FUNCTION PROOF:
An odd function starts with the property f(-x) = -f(x). We will find the derivative of this equation, and will show that the end result is the same as an even function.
f(-x) = -f(x)
Find the derivative of each side, using the chain rule where appropriate:
f'(-x) * -1 = -f'(x)
Rearranged to show:
-f'(-x) = -f'(x)
Divide both sides by -1:
f'(-x) = f'(x)
This is the same as the property of an even function, so we've shown that the derivative of an odd function will give you an even function.
EVEN FUNCTION PROOF:
An even function starts with the property f(-x) = f(x). We will find the derivative of this equation, and will show that the end result is the same as an odd function.
f(-x) = f(x)
Find the derivative of each side, using chain rule where appropriate:
f'(-x) * -1 = f'(x)
Rearranged to show:
-f'(-x) = f'(x)
Divide both sides by -1:
f'(-x) = -f'(x)
This is the same as the property of an odd function, so we've shown that the derivative of an even function will give you an odd function.
35a) Start with A = L*W. When they're asking for dA/dt, they are asking you to find the derivative of your A equation with respect to t (this just means the derivative of A where t is the variable). This sounds confusing, but let's use the product rule like normal, and it should work out:
A = L*W
Product rule says: y' = u'v + uv'
The derivative of L with respect to time t will be: dL/dt
The derivative of W with respect to time t will be: dW/dt
dA/dt = dL/dt * W + L * dW/dt
35b) They're basically giving you the equations for L and W, and want you to plug them into the equation you found in 35a. Let's start by finding the derivative of L and of W:
W = 2 + 2 cos t
dW/dt = -2 sin t
L = 3 + 2 sin 2t
dL/dt = 4 cos 2t
We actually just automatically found the derivative of W and L with respect to time t because the variable in both of these equations is t. Now let's plug all four equations back into dA/dt:
dA/dt = dL/dt * W + L * dW/dt
dA/dt = (4 cos t)*(2 + 2 cos t) + (3 + 2 sin 2t)*(-2 sin t)
Look! Without even trying, we now have the derivative of A with respect to t, because the variables in our dA equation are t.
To actually answer this book problem, we need to use dA/dt. When they ask, "At what rate is the area changing when t = 4 sec?", they are asking about the rate of change of A, or in other words the derivative of A when t = 4. So, plug t = 4 into dA/dt. Use your calculator!
dA/dt = (4 cos 4)*(2 + 2 cos 4) + (3 + 2 sin 2*4)*(-2 sin 4)
dA/dt = 7.133 ft^2 / sec
This is a positive rate of change, which means the area is INCREASING at t = 4. To figure out the units, think about the units of A (ft^2) divided by the units of t (sec). This gives you ft^2 / sec. Do the same thing for t = 5, but I'll let you figure that out on your own.
Sec 4.1 (pg. 131):
6) "Just for fun, see if you can randomly guess these answers somehow..."
The real way to do this problem is to use the Product Rule and the Quotient Rule. It's OK if you weren't able to get these answers while doing Sec 4.1 for the first time; I just wanted you guys to try it out!
p'(2) = f'(2) * g(2) + f(2) * g'(2)
q'(2) = [f'(2) * g(2) - f(2) * g'(2)] / [g(2)]^2
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