Chapter 8 Solutions

These are explanations that I had emailed to various students while we were working on Chapter 8.  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 8 Review #2 wkst:

See my work on the second and third pages here:

Chp 8 Review (pg. 427) Rs and Ts:

R5b) Ok, here's the "big idea" process for rotation of area problems:

STEP 1: Draw a picture of the situation on your paper.  Seriously, it will really help on these! :)

When you draw the cross-sections on this diagram, they should be horizontal "washers", where the center of both circles is on the y-axis.  The edge of the outer circle touches y = x, and the edge of the inner circle touches y = x^0.25.

STEP 2: Decide whether your problem should be in terms of x or in terms of y.

This shape is being rotated around the y-axis, so our equation should be in terms of y (we need to solve for x).  Our boundaries will also be y-axis boundaries.

Change y = x^0.25 to:

y^4 = x

The other equation doesn't really need to be changed, because it's y = x.

STEP 3: Find the area of one cross-section (usually a circle or a "washer" where it's a big circle with a small circle subtracted from the middle).

See how the radius of each circle is horizontal?  In other, the radius of each circle is an x-length out to the curve?  So, we need to use whatever "x" is equal to for the "radius" in our circle area equations.

Area of the "big circle":

pi * radius^2

pi * y^2

Area of the "little circle":

pi * radius^2

pi * (y^4)^2

pi * y^8

Area of one "washer" cross-section is "big circle" minus "little circle":

pi * y^2 - pi * y^8

STEP 4: Find the boundaries (where do the cross-sections start, and where do they end?)

Draw (or imagine) a lot more cross-sections drawn on your picture.  They would stack up around the y-axis, where they'd start at the bottom y-value (y = 0) and end up at the top y-value (y = 1).  I found these numbers by setting the two equations equal to each other and solving for y:

y = y^4

0 = y^4 - y

0 = y(y^3 - 1)

y = 0 and y = 1

So, 0 and 1 will be the integral boundaries, because we are trying to add up all the cross sections in between those two values.  (Also, these boundaries are in terms of the y-axis).

STEP 5: Integrate the equation for the area of one cross-section from the beginning boundary to the ending boundary.

(integral sign from 0 to 1) (pi*y^2 - pi*y^8) dy

Use your calculator or integrate by hand.  You should get the same answer either way. Hope this helps!

T6) Because we’re rotating around the y-axis, we need all equations in terms of y (in other words, solve for x). We started with y = x^3, which becomes x = y^(1/3).

We need to write an equation for the area of one cross section. Here, because of the rotation, our cross sections will be the shape of circles, so we’ll be finding the area of a circle: pi*r^2.

The radius of this circular cross-section is the distance from the y-axis to the curve. This is why we needed to change our equation to x = y^(1/3). This distance from the y-axis to the curve is a horizontal length, or in other words is the distance of the x-value of the function.

This makes the radius of our circular cross section: y^(1/3)

This makes the area of one circular cross section: pi(y^(1/3))^2

We’re going to integrate this area equation. Our equation is in terms of y, so we’ll use “dy”, and our bounds will be y-values:

integral (from 0 to 8) pi(y^(1/3))^2 dy

You can move the pi to the outside of the integral sign, because it’s a constant, and you can simplify the exponents:

pi * integral (from 0 to 8) y^(2/3) dy

Then integrate! Make the exponent one larger, then divide by the new exponent:

pi * 3/5 * y^(5/3)

Then plug in 8, then plug in 0, then subtract. The answer should be 19.2pi or 60.319.

"Chp 8 + Misc AP Review Problems" #1 worksheet: 

2) A point of inflection only occurs when the concavity CHANGES. So, set f’’(x) = 0, find each possible point of inflection, then check the concavity on either side of it by plugging values into f’’(x) that are chosen from either side of the x-values you just found. In this problem, f’’(x) = 0 when x = 0, 3, 6. So, check the values on either side of 0, 3, and 6. For example, you could find f’’(-1), f’’(1), f’’(4), and f’’(7)

You can pick any numbers to plug into f’’(x), as long as they’re on either side of your values. It’s only an actual point of inflection if the concavity is positive on one side then negative on the other side of the value (or vice versa). 

5) This is asking you to find the derivative of an integral, which would just leave you with the equation inside the integral (called the “integrand”). Here, F’(x) = sqrt(x^3 + 1). You only need to plug in 2 to this equation to find the final answer. It’s an example of the second form of the Fundamental Theorem of Calculus.

If you did this the long way (finding the integral, then finding the derivative, then plugging x = 2 into the derivative), it would look like this:

(anti-derivative with x in it) - (anti-derivative with 0 in it)

Then you’d find the derivative of this to get F’(x). The first part would return to your original equation, but with x as the variable instead of t. The second part would go away, because it’d be the derivative of a constant. So, all you’re left with is the original equation, which you’d then plug x = 2 into. 

6) Find the derivative f’(x). Don’t forget the chain rule.

f’(x) = 2*sec^2 (2x)

Then plug in pi/6.

f’(pi/6) = 2*sec^2 (2*pi/6)

2*sec^2 (pi/3)

2 / cos^2 (pi/3)

2 / (1/2)^2 ………… this step shows that cos (pi/3) = 1/2

2 / (1/4)

2 * 4

8

7) To find the equation of a tangent line, you need the slope and you need a point. They already gave you the slope, which is 1. We can use the information they gave us to solve for the point.

If they told us that f’(x) = 1, we should find f’(x), then set it equal to 1. We can then solve for x.

f’(x) = 4x^3 + 4x

1 = 4x^3 + 4x

To solve for x, you could graph both sides of this equation, then find their intersection points. Or, you could make the equation equal zero, then use your calculator’s Solver function. You should find that x = 0.2367. Plug this x-value back into f(x) to find the y-value of the point we need.

f(0.2367) = .2367^4 + 2(.2367)^2 = 0.1152

Now, you have slope = 1 and you have the point (0.2367, 0.1152) for your tangent line. You could write your tangent line equation in point-slope, then change it to slope-intercept.

y - 0.1152 = 1(x - 0.2367)

y - 0.1152 = x - 0.2367

y = x - 0.122

8) The wording of this problem is the same as saying that F(x) is equal to the integral of the given equation. So, integrate the given equation to find F(x):

You could use u-substitution, where u = ln x and du = 1/x * dx

This makes it: u^3 * du

The integral of this would be: 1/4 u^4 + c

Which is really: 1/4(ln x)^4 + c

They gave you F(1) = 0, which means that when you plug x = 1 into your integrated equation, it should be equal to 0. This will let us solve for the +c.

0 = 1/4(ln 1)^4 + c

0 = 1/4(0)^4 + c

0 = c

So, our particular solution for F(x) = 1/4(ln x)^4

Now find F(9) by plugging in x = 9:

F(9) = 1/4(ln 9)^4 = 5.827

Sec 8.5 Volume of Solids and Slabs GENERAL ADVICE:

We have to think about WHERE the shape is rotating.

If it rotates around the x-axis, we want to integrate the area of each cross section in terms of x.

If it rotates around the y-axis, we want to integrate the area of each cross section in terms of y.

Sec 8.5 (pg. 390):

#1a) It says this shape is rotating about the y-axis.  So, we need our equation in terms of y (in other words, solve for x).

y = 9-x^2

y-9 = -x^2

9-y = x^2

sqrt(9-y) = x

This is the radius of each circular cross section.  Look at the picture…the radius is the length of the x-value (horizontal). So, the area of each circular cross section will be:

pi * radius^2 = pi * sqrt(9-y)^2) = pi * (9-y)

Now, we want to add up all those circular cross sections from y = 0 to y = 9 (notice, our boundaries are referring to the Y-AXIS).  So, we want to integrate the equation for the area of one cross section from 0 to 9:

(integral sign) pi * (9-y)dy

pi (integral sign) (9-y)dy

Now use the FTC, and integrate with respect to y:

pi * (9y - 1/2y^2)

Now plug in 9, then plug in 0, then subtract:

pi * (9*9 - 1/2*9^2) - pi * (9*0 - 1/2*0^2)

pi* (81 - 81/2) - 0

pi * 40.5

127.235

#1b)  Integrating numerically means USE YOUR CALCULATOR and check your work above.

#1c)  This shape is kind of like a cone (but not really) and kind of like a cylinder (but not really).  It should be bigger than a cone with the same dimensions, but smaller than a cylinder with the same dimensions.  So, find the area of a cone, and the area of a cylinder with the same dimensions, and confirm that your rotated shape falls right in between those two answers.  This is one way to confirm geometrically.

#1c continued) You'll need to solve for the intersection of y = 9 - x^2 and the x-axis to find the radius of the "cone" and "cylinder".  Then, the intersection of y = 9 - x^2 and the y-axis will provide the height of the cone and cylinder.

The height of all shapes on the y-axis is 9.

The curve intersects the x-axis at x = 3 and x = -3, so the radius of the shapes is 3.

Volume of cylinder: pi * 3^2 * 9 = pi * 81

Volume of cone: 1/3 * pi * 3^2 * 9 = pi * 27

#2b) Same idea for this problem as 1c.  The intersection of the x-axis and the line y = 10 - 2x will give you the radius of the "cone" and "cylinder".

y = 10 - 2x = 0

10 = 2x

5 = x

So, the radius of both shapes is 5.

#5) It says the shape is in Quadrant 1, and is bounded by y = 1 and y = ln x.  This makes it so the area doesn't "go on forever".  The first step in ALL of these problems is to DRAW THE GRAPH ON YOUR PAPER.  Trust me, it'll help! :)

Now that you have the area for the graph drawn, try drawing the rotated cross-section.  Because it's rotating around the y-axis, it should be a horizontal circle where the radius of the circle goes from the y-axis to the curve y = ln x.  See how the radius is horizontal?  The length of the radius is the x-value of the y = ln x curve.  So, if we want to find the area of this circular cross section, we need an equation for the radius first:

y = ln x 

Solve this for x, to get it in terms of y:

e^y = x

Now you have an equation for the x-length from the y-axis out to the curve, because your equation is now "x=".  This new equation is your radius!  So, the area of one circular cross section is:

pi * radius^2

pi * (e^y)^2

pi * e^(2y)

Great!  Now we have the area of one cross section.  To find the whole volume, we need to "add up" all those circular cross sections from y = 0 up to y = 1.  Let's integrate the equation for the area of one cross-section:

(integral sign from 0 to 1) pi * e^(2y)

Answer = 10.036

BIG IDEA: When you're finding a shape rotated around the X-AXIS, your equation needs to be in terms of x, with horizontal x-axis boundaries.

When you're finding a shape rotated around the Y-AXIS, your equation needs to be in terms of y, with vertical y-axis boundaries.

#6) Because this shape is created by rotating around the Y-AXIS, you need everything in terms of y.  Start by solving your equation for x, to put it in terms of y:

y = x^(3/4)

y^(4/3) = x

Hopefully, you have already drawn a picture for this situation.  It WILL help, even more than just graphing it on your calculator, because you can sketch an idea of what the cross-section looks like.  Since your shape is rotating around the y-axis, you should have drawn a circular horizontal cross-section, where the center is on the y-axis and the outer edge of the circle touches your line x = y^(4/3).

We need to find the area of one circular cross-section, which means we need the radius of that circle.  See how the radius is horizontal, starts at the y-axis, and ends at the line?  Because it's horizontal, the radius length is the x-value of the line.  That's why we solved the equation for x, so that we could figure out what x was equal to!

So, the radius = y^(4/3)

So, the area of one circular cross-section is:

pi * (y^(4/3))^2

= pi * y^(8/3)

To find the volume of the whole shape, we need to use y-axis boundaries, because that shows where it starts and stops, and also everything else in our problem has been in terms of y.  The problem stated that we are bounded by y = 1 and y = 8.  We will integrate the equation for the area of one circular cross-section:

(integral sign from 1 to 8) pi * y^(8/3)

Answer: 1753.865

#7) This one is tough!  First off, DRAW A PICTURE.  It'll help, I promise!  Once you have the picture on your paper, try drawing the circular cross-section as it rotates around the y-axis.  It should be horizontal, with a horizontal radius.  Because it's a horizontal shape (because it rotates around the y-axis), we need all equations and boundaries to be in terms of y.

Solve for x in both of your equations, to put them in terms of y:

y = 8x   becomes…    x = 1/8y

y = x^4   becomes…   x = y^(1/4)

Your horizontal cross-section should look like a "washer".  The outer circle has a center on the y-axis, and touches the x = y^(1/4) curve.  The inner circle has a center on the y-axis, and touches the x = 1/8y line.  We need the area of both circles, because we'll want to subtract the smaller circle from the larger circle.

Area of large circle:

pi * radius^2

pi * (y^(1/4))^2

pi * y^(1/2)

Area of small circle:

pi * radius^2

pi * (1/8y)^2

Area of "washer" is large circle - small circle:

pi * y^(1/2) - pi * (1/8y)^2

Great, we have the area of one cross section!  To find the entire volume, we need to integrate the equation for one cross-section, from the first y-axis boundary to the last y-axis boundary.  We're using y-axis boundaries because we're rotating around the y-axis.  To find the boundaries, find the two y-values where the curves cross:

Find y-boundaries:

1/8y = y^(1/4)

y = 8y^(1/4)

y^(3/4) = 8

y = 8^(4/3)

y = 16  (Also y = 0, because there was one point in this calculation where we divided by y, which took away one of the solutions.  You can also graph on your calculator to find the points where they cross.)

Area of rotated shape:

(integral sign from 0 to 16) pi * y^(1/2) - pi * (1/8y)^2

I'll leave the FTC calculations up to you.  When you're done, check your answer with fnInt on your calculator.

Answer: 67.021 in^3

#13) It will be REALLY helpful to draw a picture of this shape before trying to figure out the integral.  Even though it's shown in the book, it will help to draw it yourself, label the pieces you need, etc.

In this problem, the shape is rotating around the vertical line x = 3.  Because it's a rotation around a vertical line, it's kind of like rotating around the y-axis, so your equations and boundaries used need to be in terms of y.

Look at the picture you've drawn.  One cross section should be a "washer", where the outer circle has a center at x = 3 and goes out to the y-axis on the left side and x = 6 on the right side.  The inner circle has a center at x = 3, and goes out to y = 4-x^2 on the left side.

Before we start anything else, we need to convert our y = 4-x^2 into terms of y instead of terms of x:

y = 4 - x^2

y - 4 = -x^2

-y + 4 = x^2

4 - y = x^2

sqrt(4 - y) = x

To find the area of one cross section, we need "Area of large circle - Area of small circle".

Area of large circle: pi * radius^2, where the radius is from x = 3 to the y-axis (a distance of 3).

= pi * 3^2

= pi * 9

Area of small circle: pi * radius^2, where the radius is from x = 3 to the curve x = sqrt(4 - y)

= pi * (3 - sqrt(4 - y))^2

Area of one cross section:

= pi * 9 - pi * (3 - sqrt(4 - y))^2

Now, for integrating to find the volume, we need to integrate the equation for the area of one cross section from the "beginning" of the shape to the "end" of the shape.  We will use the y-axis boundaries, because this is a rotation around a vertical line.  The shape starts at y = 0 and ends at y = 4.

Integrating to find the volume:

(integral sign from 0 to 4) pi * 9 - pi * (3 - sqrt(4 - y))^2 dy

Use fnInt on your graphing calculator.  Answer = 75.398

Chapter 7 Solutions

These are explanations that I had emailed to various students while we were working on Chapter 7. 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.

General Differential Equation Advice:

"Do you have any tips on how to write differential equations?"

Hmmm, there are a couple things you can do to make it a little easier. Can you look at the wording in Problem #5 on pg. 323? There are a couple key words to look for when you’re writing a differential equation:

• Whenever it says “rate”, it’s telling you how to set up the dy/dt variable in the equation. 
• This problem mostly talks about volume. So, when it says, “The rate at which [the water] flows…”, keep in mind that flowing water is measured by its volume. Here, if you’re talking about the rate of volume, then it’s the same thing as saying the derivative of volume. Hence, dV/dt.

• Other examples you could possibly see: 
• “Rate at which temperature is changing” would be dT/dt
• “Rate at which a plant is growing” would be dh/dt
• “Spending rate of money” would be dM/dt

• Problems will usually follow the word “rate” with the word “is”. “Is” can usually indicate the place your equal sign should go in the equation.

• Anytime you see the words “proportional to”, you need to include a “times k” in your equation. Something being directly “proportional” to something else means that you can multiply one thing by a constant to get the other thing.

• Lastly, the sentence will list whatever object or item that is proportional to the given rate. This will usually just be a plain variable, but if you see any other mathematical words (like “difference”, “square root”, etc.), include them in the equation here.

Here are a couple examples. I’ll try to break them down:

Example #1

“The rate at which [the water] flows…”

dV/dt

“…is…”

=

“…proportional to…”

k*

“…the square root of the depth, y feet.”

sqrt(y)

Full equation: dV/dt = k*sqrt(y)

Example #2 (see pg. 345 #T1)

“The rate of change of pressure…”

dP/dt

“…is…”

=

“…directly proportional to…”

k*

“…the pressure of the remaining oxygen.”

P

Full equation: dP/dt = k*P

Example #3 (see #3 on our in-class Sec 7.3 worksheet)

“The rate of change in the temperature of an object…”

dT/dt

“…is directly proportional to…”

= k*

“…the difference between its temperature and the temperature of the surrounding medium.”

T - 72    (because the surrounding medium was the room temperature of 72 degrees)
Full equation: dT/dt = k*(T - 72)

Chapter 7 Review Worksheet:

Answers are posted in my Google Drive folder! Please let me know if you can't access them for some reason, and I can email them to you instead.

#R3a) When you separate the variables, you don't have to divide by 6.  You can, but you don't have to.  Start like this:

dy/dx = 6y^(1/2)

Then divide by y^(1/2) and multiply by dx:

dy/y^(1/2) = 6dx

Now roll the y^(1/2) up to the numerator and make the exponent negative.

y^(-1/2) dy = 6 dx

Now integrate:

(integral sign) y^(-1/2) dy = (integral sign) 6 dx

2y^(1/2) = 6x + c

Now solve for y.  Start by dividing by 2 on both sides.

y^(1/2) = 3x + c

Now square both sides to get rid of the exponent.

y = (3x + c)^2  This is the general solution.

#R3b) Plug in x = 3 and y = 25, then solve for c.

25 = (3*3 + c)^2

25 = (9 + c)^2

5 = 9 + c

-4 = c

y = (3x - 4)^2  This is the particular solution.

3a) dy/dt = k(y - 42)

3b) Separate the variables:

dy / (y - 42) = k dt

Integrate each side:

ln | y - 42 | = kt + c

Use "e" as a base on each side:

| y - 42 | = e^(kt + c)

Split the right side into two multiplied terms using exponent properties:

| y - 42 | = e^kt * e^c

Rewrite e^c as C, which allows you to drop the absolute value sign from the left side, because C can be either positive or negative:

y - 42 = Ce^kt

Add 42 to the other side to solve for y:

y = Ce^kt + 42 

3c) Let's start from the general solution you found in 1b.

y = Ce^kt + 42

This problem tells you two important pieces of information. First, the soda is 70 degrees at t = 0. Second, the soda is 61 degrees at t = 1/2. Let’s start by plugging in the initial condition to solve for one of our unknown variables:

70 = Ce^k(0) + 42

70 = Ce^0 + 42

70 = C + 42

28 = C

Now plug C = 28 back into your general equation:

y = 28e^kt + 42

Next, plug in the other given data point in order to find k:

61 = 28e^k(1/2) + 42

19 = 28e^0.5k

19/28 = e^0.5k

ln (19/28) = 0.5k

2 ln (19/28) = k

-0.776 = k

So, your final particular solution equation is:

y = 28e^-0.776t + 42

5) There are a couple ways to do this, one of which is to plug in points to each equation, and see which one seems to match the trend. A way to use this strategy in a more effective way is to look for slopes that are all the same. For example, in this slope field, all the x = 1 points have a slope of 0. All the y = 0 points also have a slope of 0. So, look for the equation that would always give you 0 when plugging in x = 1 or when plugging in y = 0.

The other way would be to try to separate the variables and integrate every single answer choice, then see which of those graphs seems to fit the slope field. This way takes a long time, and you can’t always integrate these equations by hand!

6) In this problem, they’re giving you an equation with just x’s and y’s in it, and they’re asking you for dy/dx. In other words, they’re asking you to find the derivative! But, since this isn’t a regular “y =“ equation, we’ll have to use an implicit derivative.

sin (xy) = x

When you find the derivative of the left side, remember to use chain rule. The most outside function is sin, so we’ll find the derivative of that (while leaving the inside the same)…

cos (xy) = x

…then we need to multiply by the derivative of the inside function. To find the derivative of (xy), we’ll need to use product rule, because it’s two functions multiplied together. BIG THING TO REMEMBER: when you’re doing an implicit derivative, and you get to the derivative of a term with “y” in it, you MUST multiply by dy/dx.

Product rule for (xy)…derivative of x is 1, and derivative of y is dy/dx. Then use u’v + uv’.

So, here’s what it looks like when we multiply our problem by the derivative of (xy):

cos (xy) * (1*y + x * dy/dx) = x

We also need to find the derivative of the right side, which will be 1. Here’s our final derivative equation:

cos (xy) * (y + x * dy/dx) = 1

Unfortunately, this isn’t one of the answer choices, and it’s because we haven’t yet solved for dy/dx. Let’s start by dividing both sides by cos (xy):

cos (xy) * (y + x * dy/dx) = 1

y + x * dy/dx = 1 / cos (xy)

Next, we can subtract y from each side:

x * dy/dx = 1 / cos (xy) - y

Before we go too far though, it looks like all the answer choices are single fractions, not two things subtracted. So, let’s combine 1 / cos (xy) and - y by getting a common denominator:

x * dy/dx = 1 / cos (xy) - y * cos (xy) / cos (xy)

Then combine the fractions on the right side:

x * dy/dx = (1 - y * cos (xy) ) / cos (xy)

Lastly, we need to get rid of the “x” on the left side. Multiply both sides by 1 / x:

dy / dx = (1 - y * cos (xy) ) / cos (xy) * 1/x

dy / dx = (1 - y * cos (xy) ) / x cos (xy)

This equation is answer choice D.

10) dy / dt = ky

Separate the variables:

dy / y = k dt

Integrate each side:

ln | y | = kt + c

Use “e” as a base on each side to solve for “y”:

e^ln | y | = e^(kt + c)

| y | = e^(kt + c)

| y | = e^kt * e^c

y = Ce^kt

The only correct answer choice is “B”: 2e^kt. They just picked C = 2.

Sec 7.4 Slope Fields packet

17) Here's an explanation I found elsewhere on the internet:

The slope of the tangent line is simply given by dy/dx, which you are told is equal to: 

dy/dx = x*y/2 

At the point (1,1), 

dy/dx = 1*1/2 = 1/2 

The tangent line must also pass through the point (1,1), so using the point-slope form of the equation of a straight line, we obtain the following equation for the tangent at (1,1): 

y - 1 = (1/2)*(x - 1) 

y = x/2 + 1/2 

Using this equation to estimate the value of y at x = 1.2 gives: 

y(1.2) ~ = 1.2/2 + 0.5 = 1.1 
--------------- 
The original differential equation is separable: 

dy/dx = x*y/2 

dy/y = x/2 dx 

ln(y/c) = (x^2)/4 

y = c*exp((x^2)/4)) 

where c is a constant of integration 

Using the initial condition to solve for the constant c, we have: 

1 = c*exp(1/4) 

c = exp(-1/4) 

So the particular solution is: 

y(x) = exp(-1/4)*exp((x^2)/4) 

y(x) = exp((x^2 - 1)/4) 

The actual value of y at x = 1.2 is: 

y(1.2) = exp((1.2^2 - 1)/4) = 1.116

You can do the comparison and determine if the linear extrapolation overestimated or underestimated the value at x = 1.2 (is 1.1 larger or smaller than 1.116?)

18a) Draw a tiny line that illustrates the slope at each point. For example, at the point (2, 2), when you plug in x = 2 and y = 2 to the given dy/dx equation, you get dy/dx = 2/2 = 1. So, draw a tiny line at (2, 2) that has a slope of 1. Do this at each point.

18b) Start by drawing a point at (0, 1). Then, slowly, SLOWLY draw a curve starting from that point. Every time you pass one of the “tiny slopes” you drew in the slope field, your new curve should be parallel to that tiny slope. It is NOT like connecting the dots as you draw the curve. You don’t need to connect the tiny slope lines. Just be sure to draw your curve parallel to them.

18c) To solve for y, we need to separate the variables, then integrate.

dy / dx = x / y

y * dy = x * dx

Then integrate each side. Make the new exponents “one bigger” and divide by the new exponent:

1/2y^2 = 1/2x^2 + c

y^2 = x^2 + c

y = +-sqrt(x^2 + c)

To find the “c” value, you could plug in the given initial value of (0, 1) at any point in your work. Here’s what it looks like to plug it in to the last step:

1 = +-sqrt(0^2 + c)
1 = +-sqrt(c)
1 = c

y = sqrt(x^2 + 1)
We only need the “positive” square root in the answer because we can see the solution curve is in the positive y-values on the graph.

18d) Follow the same steps as Part B in this problem

18e) Follow the same steps as Part C in this problem. The difference is when you’re solving for “c” and you plug in (0, -1):

-1 = +-sqrt(0^2 + c)
-1 = +-sqrt(c)
1 = c

y = -sqrt(x^2 + 1)
We only need the “negative square root in the answer because we can see the solution curve is in the negative y-values on the graph.

Sec 7.3 (pg. 320):

#1c)  Let's look at each step as we integrate:

(integral sign) dm / (100-km) = (integral sign) dt

Here, you may want to use u-substitution…

u = 100 - km

du = -k dm

-1/k du = dm

So, here it is with u-substitution:

-1/k * (integral sign) 1 / u * du = (integral sign) dt

Then integrate:

-1/k * ln | u | = t + c

Then substitute back in for u:

-1/k * ln | 100 - km | = t + c

Now solve for m.  Start by multiplying both sides by -k.  The "+c" can still be "+c", because it's just a constant multiplied by another constant k.

ln | 100 - km | = -kt + c

| 100 - km | = e^(-kt + c)

| 100 - km | = e^-kt * e^c

100 - km = Ce^-kt

-km = Ce^-kt - 100

Now divide both sides by -k.  The first term on the right (with C) will be a constant divided by a constant k, so we can just leave it as C.

m = Ce^-kt + 100/k

Lastly, it gave you the initial value M = 0 when t = 0.  Plug this in to the equation:

0 = Ce^-k(0) + 100/k

0 = Ce^0 + 100/k

0 = C + 100/k

-100/k = C

Now substitute this value for C back into the equation you had found.

m = -100/k * e^-kt + 100/k

You could leave your answer like this, or factor out 100/k like the textbook did:

m = 100/k (-e^-kt + 1)

#7a) Your starting equation is dy/dx = ky^n

It wants you to plug in n = 1 first:

dy/dx = ky

Then, to solve for y, you'll need to separate the variables and integrate:

Multiply by dx, and divide by y:

dy/y = k*dx

Now integrate:

(integral sign) dy/y = (integral sign) k*dx

ln | y | = kx + c

It also says to make k = 1 and let the constant of integration C = -3.  This is referring to the +c you just got from integrating (hence, "constant of integration).

ln | y | = k - 3

| y | = e^(x - 3)

| y | = e^(x - 3)

| y | = e^x * e^-3

| y | = 0.04978e^x

y = +-0.04978e^x        [This is supposed to be a + or - in front of the coefficient.]