10.5 Differential Equationsap Calculus

Master AP Calculus AB & BC

Now put these into the differential equation to see if the equality holds.1116. Y” that is going to be 4e ⁺x + 6, that is the y” part, and then, + 6x.1129. The question is, does it equal the first derivative which is 4e ⁺x + 6x + 6.1145. Yes, this is a solution of that differential equation.1157.

Part II. AP CALCULUS AB & BC REVIEW

CHAPTER 10. Differential Equations

EXPONENTIAL GROWTH AND DECAY

• 10.5 = 3(1 3) – 2.5(1) + 9(1) + C; 10.5 = 3 – 2.5 + 9 + C; C = 1; The solution to the differential equation is y = 3x 3 – 2.5x 2 + 9x + 1.
• This is a lesson from the tutorial, Differential Calculus and you are encouraged to log in or register, so that you can track your progress.
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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)?

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.

10.5 Differential Equationsap Calculus Solver

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.

Learning Objectives

In this section, we strive to understand the ideas generated by the following important questions:

• How can we use differential equations to describe phenomena in the world around us?
• How can we use differential equations to better understand these phenomena?

In our work to date, we have seen several ways that differential equations arise in the natural world, from the growth of a population to the temperature of a cup of coffee. In this section, we will look more closely at how differential equations give us a natural way to describe various phenomena. As we’ll see, the key is to focus on understanding the different factors that cause a quantity to change.

Preview Activity (PageIndex{1})

Any time that the rate of change of a quantity is related to the amount of a quantity, a differential equation naturally arises. In the following two problems, we see two such scenarios; for each, we want to develop a differential equation whose solution is the quantity of interest.

1. Suppose you have a bank account in which money grows at an annual rate of 3%.
1. If you have \$10,000 in the account, at what rate is your money growing?
2. Suppose that you are also withdrawing money from the account at \$1,000 per year. What is the rate of change in the amount of money in the account? What are the units on this rate of change?
2. Suppose that a water tank holds 100 gallons and that a salty solution, which contains 20 grams of salt in every gallon, enters the tank at 2 gallons per minute.
1. How much salt enters the tank each minute?
2. Suppose that initially there are 300 grams of salt in the tank. How much salt is in each gallon at this point in time?
3. Finally, suppose that evenly mixed solution is pumped out of the tank at the rate of 2 gallons per minute. How much salt leaves the tank each minute?
4. What is the total rate of change in the amount of salt in the tank?

Developing a Differential Equation

Preview activity (PageIndex{1}) demonstrates the kind of thinking we will be doing in this section. In each of the two examples we considered, there is a quantity, such as the amount of money in the bank account or the amount of salt in the tank, that is changing due to several factors. The governing differential equation results from the total rate of change being the difference between the rate of increase and the rate of decrease.

Example (PageIndex{1}): Lake Michigan

In the Great Lakes region, rivers flowing into the lakes carry a great deal of pollution in the form of small pieces of plastic averaging 1 millimeter in diameter. In order to understand how the amount of plastic in Lake Michigan is changing, construct a model for how this type pollution has built up in the lake.

Solution

First, some basic facts about Lake Michigan.

• The volume of the lake is (5 times 10^{12}) cubic meters.
• Water flows into the lake at a rate of (5 times 10^{10}) cubic meters per year. It flows out of the lake at the same rate.
• Each cubic meter flowing into the lake contains roughly (3 times 10^{−8}) cubic meters of plastic pollution.

Let’s denote the amount of pollution in the lake by (P(t)), where (P) is measured in cubic meters of plastic and (t) in years. Our goal is to describe the rate of change of this function; in other words, we want to develop a differential equation describing (P(t)).

First, we will measure how (P(t)) increases due to pollution flowing into the lake. We know that (5 times 10^10) cubic meters of water enters the lake every year and each cubic meter of water contains (3 times 10^{−8}) cubic meters of pollution. Therefore, pollution enters the lake at the rate of

(left(5 cdot 10^{10} dfrac{m^3 text{water}}{text{year}}right)cdotleft(3 cdot 10^{-8} dfrac{m^3 text{plastic}}{m^3 text{water}}right) = 1.5 cdot 10^{3})

Second, we will measure how (P(t)) decreases due to pollution flowing out of the lake. If the total amount of pollution is (P) cubic meters and the volume of Lake Michigan is (5 cdot 10^{12}) cubic meters then the concentration of plastic pollution in Lake Michigan is

(frac{P}{5 cdot 10^{12}}) cubic meters of plastic per cubic meter of water.

Differential Equations Calculator

Since (5 cdot 10^{10}) cubic meters of water flow out each year, then the plastic pollution leaves the lake at the rate of

(left(dfrac{P}{5cdot10^{12}} dfrac{m^3 text{plastic}}{m^3 text{water}}right)cdotleft(5cdot10^{10} dfrac{m^3 text{water}}{text{year}}right)= dfrac{P}{100}) cubic meters of plastic per cubic meter of water.

The total rate of change of (P) is thus the difference between the rate at which pollution enters the lake minus the rate at which pollution leaves the lake; that is,

(begin{align} dfrac{dP}{dt} &= 1.5 cdot 10^3 − dfrac{P}{100} &= dfrac{1}{ 100} (1.5 cdot 10^5 − P). end{align} )

We have now found a differential equation that describes the rate at which the amount of pollution is changing. To better understand the behavior of (P(t)), we now apply some of the techniques we have recently developed.

Since this is an autonomous differential equation, we can sketch (dP/dt) as a function of (P) and then construct a slope field, as shown in Figure (PageIndex{1}).

Figure (PageIndex{1}): Plots of ( dfrac{dP}{dt}) vs. (P) and the slope field for the differential equation (dfrac{dP}{dt} = dfrac{1}{100} (1.5 cdot 10^5 − P)).

These plots both show that (P = 1.5 cdot 10^5) is a stable equilibrium. Therefore, we should expect that the amount of pollution in Lake Michigan will stabilize near (1.5 cdot 10^5) cubic meters of pollution. Next, assuming that there is initially no pollution in the lake, we will solve the initial 6 and we assume that each cubic meter of water that flows out carries with it the plastic pollution it contains value problem

(dfrac{dP}{dt} = dfrac{1}{100} (1.5 cdot 10^5 − P)), (P(0) = 0).

Separating variables, we find that

10.5 Differential Equations Ap Calculus Multiple Choice

( dfrac{1}{1.5 cdot 10^5− P} dfrac{dP}{dt} = dfrac{1}{100} .)

Integrating with respect to (t), we have

(int dfrac{1}{1.5 cdot10^5− P} dfrac{dP}{dt} dt = int dfrac{1}{100} dt )

and thus changing variables on the left and antidifferentiating on both sides, we find that

(int dfrac{dP}{1.5 cdot 10^5 P } = int dfrac{1}{100} dt )

(- ln 1.5 cdot 10^5 - P = dfrac{1}{100}t + C)

Finally, multiplying both sides by −1 and using the definition of the logarithm, we find that

(1.5cdot 10^5 − P = Ce^{−t/100}. tag{7.1}label{7.1} )

This is a good time to determine the constant (C). Since (P = 0) when (t = 0), we have

( 1.5 cdot 10^5 -0 = Ce^0 = C.)

In other words, ( C = 1.5 times 10^5)

Using this value of (C) in Equation ((ref{7.1})) and solving for (P), we arrive at the solution

(P(t) = 1.5 cdot 10^5 (1 - e^{-t/100}))

Superimposing the graph of P on the slope field we saw in Figure (PageIndex{1}), we see, as shown in Figure (PageIndex{2}) We see that, as expected, the amount of plastic pollution stabilizes around (1.5 cdot 10^5) cubic meters. There are many important lessons to learn from Example (PageIndex{1}). Foremost is how we can develop a differential equation by thinking about the “total rate = rate in - rate out” model. In addition, we note how we can bring together all of our available understanding (plotting ( dfrac{dP}{dt}) vs. (P), creating a slope field, solving the differential equation) to see how the differential equation describes the behavior of a changing quantity.

Of course, we can also explore what happens when certain aspects of the problem change. For instance, let’s suppose we are at a time when the plastic pollution entering

Figure (PageIndex{2}):The solution ( P(t)) and the slope field for the differential equation (dfrac{dP}{dt}= dfrac{1}{100} (1.5 times 10^5 − P)).

Lake Michigan has stabilized at (1.5 times 10^5) cubic meters, and that new legislation is passed to prevent this type of pollution entering the lake. So, there is no longer any inflow of plastic pollution to the lake. How does the amount of plastic pollution in Lake Michigan now change? For example, how long does it take for the amount of plastic pollution in the lake to halve?

Restarting the problem at time (t = 0), we now have the modified initial value problem

(dfrac{dP}{dt} = − dfrac{1}{100} P), (P(0) = 1.5 cdot 10^5.)

It is a straightforward and familiar exercise to find that the solution to this equation is (P(t) = 1.5 cdot 10^5 e^{−t/100}.) The time that it takes for half of the pollution to flow out of the lake is given by (T) where (P(T) = 0.75 cdot 10^5). Thus, we must solve the equation

(0.75 times 10^5 = 1.5 times 10^5 e −T/100 , )

or

(dfrac{1}{2} = e^{−T/100} .)

It follows that

(T = −100 ln left(dfrac{1}{2}right) approx 69.3), years.

In the upcoming activities, we explore some other natural settings in which differential equation model changing quantities.

Activity (PageIndex{1}): Accrued Savings

Suppose you have a bank account that grows by 5% every year. Let (A(t)) be the amount of money in the account in year ( t).

1. What is the rate of change of (A) with respect to (t)?
2. Suppose that you are also withdrawing \$10,000 per year. Write a differential equation that expresses the total rate of change of (A).
3. Sketch a slope field for this differential equation, find any equilibrium solutions, and identify them as either stable or unstable. Write a sentence or two that describes the significance of the stability of the equilibrium solution.
4. Suppose that you initially deposit \$100,000 into the account. How long does it take for you to deplete the account?
5. What is the smallest amount of money you would need to have in the account to guarantee that you never deplete the money in the account?
6. If your initial deposit is \$300,000, how much could you withdraw every year without depleting the account?

Activity (PageIndex{2}): Morphine

A dose of morphine is absorbed from the bloodstream of a patient at a rate proportional to the amount in the bloodstream.

1. Write a differential equation for (M(t)), the amount of morphine in the patient’s bloodstream, using (k) as the constant proportionality.
2. Assuming that the initial dose of morphine is (M_0), solve the initial value problem to find (M(t)). Use the fact that the half-life for the absorption of morphine is two hours to find the constant (k).
3. Suppose that a patient is given morphine intravenously at the rate of 3 milligrams per hour. Write a differential equation that combines the intravenous administration of morphine with the body’s natural absorption.
4. Find any equilibrium solutions and determine their stability.
5. Assuming that there is initially no morphine in the patient’s bloodstream, solve the initial value problem to determine (M(t)). What happens to (M(t)) after a very long time?
6. To what rate should a doctor reduce the intravenous rate so that there is eventually 7 milligrams of morphine in the patient’s bloodstream?

Summary In this section, we encountered the following important ideas:

• Differential equations arise in a situation when we understand how various factors cause a quantity to change.
• We may use the tools we have developed so far—slope fields, Euler’s methods, and our method for solving separable equations—to understand a quantity described by a differential equation. 