10.5 Differential Equationsap Calculus

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Calculus 221, section 10.5 Graphs of Differential Equations notes prepared by Tim Pilachowski Sometimes, we’ll have a differential equation (DE) which can be set up and described, but which is not easily solved—at least with the techniques we’ve got so far. Even though “insoluble”, these DEs can still be useful in. (5 points) Using separation of variables, find a solution to the differential equation 3xy ln(2) dy dx Get more help from Chegg Get 1:1 help now from expert Calculus tutors Solve it with our calculus problem solver and calculator.

CHAPTER 10. Differential Equations

EXPONENTIAL GROWTH AND DECAY

You have probably alluded to exponential growth in everyday conversation without even realizing it. Perhaps you’ve said things like, “Ever since I started carrying raw meat in my pockets, the number of times I’ve been attacked by wild dogs has increased exponentially.' Exponential growth is sudden, quick, and relentless. Mathematically, exponential growth or decay has one defining characteristic (and this is key)', the rate of y’s growth is directly proportional toy itself. In other words, the bigger y is, the faster it grows; the smaller y is, the slower it decays.

Mathematically, something exhibiting exponential growth or decay satisfies the differential equation

where k is called the constant of proportionality. A model ship might be built to a 1:35 scale, which means that any real ship part is 35 times as large as the model. The constant of proportionality in that case is 35. However, k in exponential growth and decay is never so neat and tidy, and it is rarely (if ever) evident from reading a problem. Luckily, it is quite easy to find.

In the first problem set of this chapter (problem 3), you proved that the general solution to is I find the formula easier to remember, however, if you call the constant N instead of C (although that doesn’t amount to a hill of beans mathematically). Why is it easier to remember? It sounds like Roseanne pronouncing “naked”—“nekkit.”

In this formula, N stands for the original amount of material, k is the proportionality constant, t is time, and y is the amount of N that remains after time t has passed. When approaching exponential growth and decay problems, your first goals should be to find N and k; then, answer whatever question is being posed. Don’t be intimidated by these problems—they are very easy.

Example 3: The new theme restaurant in town (Rowdy Rita’s Eat and Hurl) is being tested by the health department for cleanliness. Health inspectors find the men’s room floor to be a fertile ground for growing bacteria. They have determined that the rate of bacterial growth is proportional to the number of colonies. So, they plant 10 colonies and come back in 15 minutes; when they return, the number of colonies has risen to 35. How many colonies will there be one full hour after they planted the original 10?

Solution: The key phrase in the problem is “the rate of bacterial growth is proportional to the number of colonies,” because that means that you can apply exponential growth and decay. They started with 10 colonies, so N = 10 (starting amount). Do not try to figure out what k is in your head—it defies simple calculation. Instead, we know that there will be 35 colonies after t = 15 minutes, so you can set up the equation

Solve this equation for k. Divide by 10 to begin the process.

Now you have a formula to determine the amount of bacteria for any time t minutes after the original planting:

We want the amount of bacteria growth after 1 hour; since we calculated k using minutes, we’ll have to express 1 hour as t = 60 minutes. Now, find the number of colonies.

So, almost 1,501 colonies are partying along the surface of the bathroom floor. In one day, the number will grow to 1.7 X 10⁵³ colonies. You may be safer going to the bathroom in the alley behind the restaurant.

NOTE. All half-life problems automatically satisfy the property by their very nature.

Example 4: The Easter Bunny has begun to express his more malevolent side. This year, instead of hiding real eggs, he’s hiding eggs made of a radioactive substance Nb-95, which has a half-life of 35 days. If the dangerous eggs have a mass of 2 kilograms, and you don’t find the one hiding under your bed, how long will it take that egg to decay to a “harmless” 50 grams?

Solution: The egg starts at a mass of 2,000 g. A half-life of 35 days means that in 35 days, exactly half of the mass will remain. After 70 days, one fourth of the mass will remain, etc. Therefore, after 35 days, the mass will be 1,000. This information will allow us to find k.

TIP. In an exponential decay problem such as this, the k will be negative.

Now that we know N and k, we want to find t when only 50 grams are left. In this case, t will be in days (since days was the unit of time we used when determining k).

You should be safe by Thanksgiving. (Nothing wrong with a little premature hair loss and a healthy greenish glow, is there?)

EXERCISE 4

Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.

YOU MAY USE A GRAPHING CALCULATOR FOR ALL OF THESE PROBLEMS.

1. If Pu-230 (a particularly stinky radioactive substance) has a half-life of 24,360 years, find an equation that represents the amount of Pu-230 left after time t, if you began with N grams.

2. Most men in the world (except, of course, for me, if my wife is reading this) think that Julia Roberts is pretty attractive. If left unchecked (and the practice were legal), we can assume the number of her husbands would increase exponentially. As of right now, she has one husband, but if legal restrictions were lifted she might have 4 husbands 2 years from now. How many years would it take her to marry 100 men if the number of husbands is proportional to the rate of increase?

3. Assume that the world population’s interest in the new boy band, “Hunks o’ Love,” is growing at a rate proportional to the number of its fans. If the Hunks had 2,000 fans one year after they released their first album and 50,000 fans five years after their first album, how many fans did they have the moment the first album was released?

4. Vinny the Talking Dog was an impressive animal for many reasons during his short-lived career. First of all, he was a talking dog, for goodness sakes! However, one of the unfortunate side-effects of this gift was that he increased his size by 1/3 every two weeks. If he weighed 5 pounds at birth, how many days did it take him to reach an enormous 600 pounds (at which point his poor, pitiable, poochie heart puttered out)?

ANSWERS AND EXPLANATIONS

1. Because the rate of decrease is proportional to the amount of substance (the amount decreases by half), we can use exponential growth and decay. In other words, let’s get Nekt. In 24,360 years, N will decrease by half, so we can write

Divide both sides by N, and you get

Therefore, the equation will give you the amount of Pu-230 left after time t if you began with N grams.

2. This problem clearly states the proportionality relationship required to use exponential growth and decay. Here, N = 1, and y = 4 when t = 2 years, so you can set up the equation:

Now that we have k, we need to find t when y = 100.

3. Our job in this problem will be to find N, the original number of fans. We have the following equations based on the given information:

Solve the first equation for N, and you get

Plug this value into the other equation, and you can find k.

Finally, we can find the value of N by plugging k into

NOTE. It should be no surprise that the left-hand side of the equation is 4/3 in the second step, as Vinny’s weight is 4/3 of his original weight every 14 days. In the half-life problems, you may have noticed that this number always turns out to be ½.

4. Oh, cruel fate. If Vinny weighed 5 pounds at birth, he weighed or 6.667 pounds 14 days later. Notice that we will use days rather than weeks as our unit of time, since the final question in the problem asks for days.

We want to find t when y = 600.

The poor guy lived almost 8 months. The real tragedy is that even though he could talk, all he wanted to talk about were his misgivings concerning contemporary U.S. foreign policy. His handlers were relieved at his passing. “It was like having to talk to a furry John Kerry all the time,” they explained.

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Part II. AP CALCULUS AB & BC REVIEW

CHAPTER 10. Differential Equations

HANDS-ON ACTIVITY 10.2: SLOPE FIELDS

Slope fields sound like dangerous places to play soccer but are, instead, handy ways to visualize differential equations. Remember back in the section on linear approximations when we discussed the fact that a derivative has values very close to its original function at the point of tangency? When that fact manifests itself all over the coordinate plane, it’s truly something to behold. So, you’d better be holding on to something when you undertake this activity.

1. Let’s return to the first differential equation from activity 10.1: . What exactly does this equation tell you about the general solution?

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2. If the solution to contained the point (0,1), what could you determine about the graph of the solution at that point?

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3. What would the tangent line to the solution graph look like at the point (—2,0)?

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4. Calculate the slopes at all of the points indicated on the axes below, and draw a small line segment with the correct slope. The slope at point (0,1) has already been drawn as an example.

5. The general solution to (according to our work in the last chapter) was x² + y² = C. How does the solution relate to the drawing you made in problem 4?

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6. What is the purpose of a slope field?

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7. Draw the particular solution to that passes through the point (2,0) on the slope field above.

8. Draw the slope field for on the axes below.

9. Use the slope field to draw an approximate solution graph that contains the point (-1/2,1/2).

10. Find the particular solution to that contains (-1/2,1/2) using separation of variables.

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11. Use your calculator to draw the graph of the particular solution you found in problem 10. It should look a lot like the graph you drew in problem 9.

SELECTED SOLUTIONS TO HANDS-ON ACTIVITY 10.2

Note: All slope fields in the solution are drawn with a computer and hence with more precision than in the exercises. Your slope fields should look similar, though not as detailed. See the technology section at the end of this chapter for a program to help you draw slope fields with your TI-83.

1. It tells you the slope, dy/dx, of the tangent lines to the solution graph at all points (x,y).

2. You would know that the slope of the tangent line to the graph at (0,1) would be The graph would have a horizontal tangent line there.

3. The slope of the tangent line (also called the slope of the solution curve itself) is which is undefined. Therefore, the tangent line there is vertical (since a vertical line has an undefined slope).

4. All you have to do is plug each (x,y) point into dy/dx to get the slope. Draw a small line segment with approximately that slope. It doesn’t have to be exact, but a slope of ½ should be much shallower than a slope of 2.

5. The tangent segments in the slope field trace out the circular shape of the solution graph. Remember that a tangent line has values very close to its original graph near the point of tangency. So, if we draw nothing but little tangent lines so small that all the points on the segment are close to the point of tangency, the result looks like the solution graph.

6. A slope field gives you a basic idea of the shape of the solution graph.

7. The particular solution will be a circle centered at the origin with radius 2.

NOTE. The slope field actually looks like all the circles centered at the origin; that’s because a slope field doesn’t know the value of C, so it draws all possible circles.

10. Divide by y and multiply by dx to separate variables.

To find C, plug in the point (-1/2,1/2).

The final (unsimplified) solution is

EXERCISE 2

Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.

DO NOT USE A CALCULATOR FOR ANY OF THESE PROBLEMS.

1. Sketch the slope field of

2. (a) Draw the slope field for

(b) Find the solution of the differential equation that passes through the point (1,2e).

3. (a) Use the slope field of f'(x) = x/y to draw an approximate graph of f(x) if f(—2) = 0.

(b) Find/(x) specified in 3(a).

4. Explain how the slope field of describes its general solution.

5. Which of the following differential equations has the slope field below?

ANSWERS AND EXPLANATIONS

1. All kinds of weird stuff happens close to the origin. Plot enough points so that you can see what’s going on. Remember, all you have to do is plug in any points (x,y) into the differential equation dy/dx; the result is the slope of the line segment you should draw at that point.

2. a)

(b) Solve the differential equation using separation of variables. You’ll have to factor a y out of the numerator to do so.

Now, plug in the point (1,2e).

2e = C(1)e

C = 2

Therefore, the solution is y = 2xe^x. If you graph it using your calculator (although the problem doesn’t ask you to do so), you’ll see that it fits the slope field perfectly.

3. (a) This slope field sort of looks like the big-bang theory—everything is exploding out of the origin. The solution just screams, “I am a hyperbola! Love me! Accept me! Tell me that I am handsome!”

(b) It’s the revenge of separation of variables. In fact, this problem is very similar to the equation all of us are growing tired of: Begin by cross-multiplying.

Now, find the C (Caspian).

0 = 4 + C

C = -4

All that remains is to plug in everything. Remember that the standard form for a hyberbola always equals 1.

This is the equation of the hyberbola in part 3(a).

4. Elementary integration tells you that if , then y = x³ + C. The slope field for outlines the family of curves y = x³ + C.

5. If you drew all four slope fields, you wasted valuable time; if this were the AP test, you’d have wasted five precious minutes on a very simple question in disguise. Look at the defining characteristic of the slope field: it has horizontal tangents all along thex- and y-axes. In other words, if either the y-coordinate of the point is 0 or the y-coordinate of the point is 0, then dy/dx equals 0. That is only true for one of the four choices: (C). For example, choice (D) is undefined at points that have y = 0; choices (A) and (B) won’t have horizontal tangents anywhere except for the origin.

NOTE. Euler is pronounced “oiler,” not “youler.”

Master AP Calculus AB & BC

Master AP Calculus AB & BC

Differential Equations Solver

10.5 Differential Equationsap Calculus Equation

10.5 Differential Equations Ap Calculus Multiple Choice