Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 53 ppt

The differential equation dydt = ky has the general solution yt = Cekt, where C is an arbitrary constant.. Modeling a Situation with a Differential Equation EXAMPLE 15.8 For each of th

General and Particular Solutions to Differential Equations

The families of functions we found as solutions to the differential equations in Example

15.6 are not simply more general solutions but in fact are the general solutions to each of the differential equations By this we mean that any particular solution to the differential

equation can be expressed in this form

Terminology:The graph of a solution is called a solution curve.

The table below gathers together examples of some differential equations and their solutions

y

t

Differential Equation Some Particular Solutions The General Solution Some Solution Curves

y = C

y = 3x + C

y = 7

y = –12

= 0

dy

dt

= 3

dy

dx

y = 3x

y = 3x + 2

y = t2+ C

= 2t

dy

dt

y = t2 + 1

y = t2 – 4

y = Ce t

= y

dy

dt

y = e t

y = –5e t

y = 0

y = Ce kt

= ky

dy

dt

y = 5e kt

y = –2e kt

y

x

y

t

y

t

y

t y

t

k > 0 k < 0

In this section we will focus primarily on the differential equation dydt = ky It is ubiquitous, arising in fields as varied as economics, biology, and chemistry

The differential equation dy

dt = ky has the general solution y(t ) = Cekt, where C is an arbitrary constant

In order to pick out a particular solution it is only necessary to know one point on the solution curve.12This is equivalent to being given exactly one data point, that is, one value of the independent variable along with the corresponding value of the dependent variable Such a

piece of information is called an initial condition.

EXERCISE 15.3 Let k be an arbitrary constant Show that for any point P in the plane there is a value of C

such that the curve y = Cektpasses through P

EXAMPLE 15.7 For each of the differential equations below, find the particular solution corresponding to

the initial condition given

(a)dQdt = −0.2Q, where Q(0) = 10 (b) dWdx = 3W , where W (2) = −7

SOLUTION

(a) The general solution is Q(t) = Ce−0.2t To find the particular solution, we use the information that Q = 10 when t = 0

10 = Ce−0.2(0)

10 = C The particular solution is Q(t) = 10e−0.2t

y

t

10

Q (t) = 10e–0.2t

Figure 15.7

(b) This differential equation is basically of the same form as the one in part (a) The role

of t is now played by x; the role of Q is now played by W The general solution is

W = Ce3x

12 This is true for any differential equation involving only first derivatives.

To find the particular solution, we use the information that W = −7 when x = 2.

−7 = Ce3(2)

C = −7e−6≈ −0.01735

So W (x) = −7e−6e3x

Modeling a Situation with a Differential Equation

EXAMPLE 15.8 For each of the scenarios given below, write a differential equation modeling the situation

(a) Economics Money is being kept in a savings account with a nominal interest rate of

5% per year compounded continuously If no deposits and no withdrawals are made,

then the rate of change of the amount of money in the account is proportional to the

amount of money in the account at any given time

(b) Population Biology The rate of growth of a population under ideal circumstances

(unlimited food, space, etc.) is proportional to the size of the population itself at any time

(c) Chemistry A radioactive substance decays at a rate proportional to the amount of the

substance present at any given time

SOLUTION

(a) Let M = M(t) be the amount of money in the account as a function of time ThendMdt is

the rate of change of money in the account The rate of change of the amount of money

is proportional to the amount of moneytranslates to

dM

dt = kM, where k is a constant of proportionality

We know that k = 0.05 because that is the rate at which interest accumulates Therefore

dM

dt = 0.05M

Note that we’ve just shown this in Section 15.1; if money is compounded contin-uously at a rate of 5%, then the amount of money in the account after t years is given

by M(t) = limn→∞M0(1 +0.05n )nt = M0e0.05t If M(t) = M0e0.05t, then dM/dt =

0.05M(t) The proportionality constant is 0.05 not 1.05, because we are looking at the rate of increase of M; the increase is due to interest and the interest rate is 5% of

M(t ), not 105% of M(t)

(b) Let P = P (t) represent the size of the population as a function of time Then dPdt is the rate of growth of the population Our task is really just a matter of translating

prose into mathematical notation The population’s rate of change is proportional to the population itself translates to

dP

dt = kP , where k is the constant of proportionality (k > 0)

(c) Let S(t) represent the number of milligrams of the substance present at time t Then

we can model the situation with the differential equation

dS

dt = kS

Because S is decreasing over time, dSdt should be negative That means that kS must also be negative Because S (the amount of the substance) has to be positive, k must be

a negative constant.13

EXERCISE 15.4 Suppose that the radioactive substance above decays at a continuous rate of 3% Is the value

of k 0.03, or −0.03, or 0.97, or −0.97?

The answer is supplied at the end of the section.

EXERCISE 15.5 Realistically speaking, most populations do not exhibit true exponential growth in the long

run, because resources are not, in practice, unlimited Consider, for example, the population

of fish in a pond In the short run, the number of fish may appear to grow exponentially, but in the long run, we don’t expect this to continue If it did, the pond would eventually

be solidly packed with fish Instead, in the long run, we expect the number of fish to level off at what ecologists refer to as the carrying capacity of the pond Suppose the carrying capacity of this particular pond is 700 fish In this problem we model the long-run behavior

of the population (When we return to differential equations, we will construct a model that

reflects the situation both in the long run and at the beginning.)

I We can model the situation by saying that due to limited resources, the difference between the number of fish and the carrying capacity of the pond decays exponentially (a) Use statement (i) in Section 15.2 characterizing exponential functions to write an equation for D(t), where D(t) is the difference between the number of fish at time

tand the carrying capacity of the pond Make all constants in your answer positive (b) Use the equation from part (a) to solve for F (t), the number of fish at time t Note that D(t) = 700 − F (t), where F (t) is the number of fish in the pond at time t (c) Graph F (t), assuming that F (0) < 700

II Equivalently, we can model the situation by saying that due to limited resources, the difference D(t) changes at a rate proportional to itself

(a) Write a differential equation involving D and reflecting the situation Again, use a positive proportionality constant k

(b) Use the differential equation from part II(a) to write a differential equation involving

F (t ) Write a differential equation reflecting the situation but this time put it in the form dF /dt =

(c) Translate the differential equation you obtained in part II(b) to a sentence about the rate at which the fish population is growing

III Check that the function F (t) obtained in part I(b) is a solution to the differential equation obtained in part II(b)

The answer is supplied at the end of the section.

EXAMPLE 15.9 A population is growing at a rate proportional to itself When the population is 11,500 it is

growing at a rate of 225 per year How fast is it growing when the population has reached 19,000?

13 Alternatively, we could write dS

dt = −kS, where k is a positive constant Often it is nice to keep all constants positive;

dS

= −kS, k > 0 reminds us that the rate of change is negative.

SOLUTION Let P = P (t) represent the size of the population at time t Notice that it is much simpler to

write a differential equation modeling this scenario than it is to write an amount function

dP

dt = kP

We are told thatdPdt = 225 when P = 11,500, so we can find k

225 = k(11,500)

k =11,500225

k = 9 460

SodPdt =4609 P If P = 19,000 thendPdt =4609 · 19,000 ≈ 371.7

When the population is 19,000, it is growing at a rate of roughly 372 individuals per year

Answer to Exercise 15.4

The value of k should be −0.03 k must be negative, because the quantity is decreasing Why does k = −0.03? The rate of decay is 3% of itself, so the change in S is 3% of S, not 97% of S

Answers to Exercise 15.5

I (a) D(t) = Ce−kt

(b) 700 − F (t) = Ce−kt

− F (t) = −700 + Ce−kt

F (t ) = 700 − Ce−kt (c)

700

t F

Figure 15.8

II (a) dDdt = −kD

(b) dtd(700 − F ) = −k(700 − F )

−dFdt = −k(700 − F )

dF

dt = k(700 − F ) (c) The fish population is growing at a rate proportional to the difference between the carrying capacity of the pond and the number of fish in the pond

III We want to show that F (t) = 700 − Ce−kt is a solution to the differential equation

dF

dt = k(700 − F )

dF

dt = k(700 − F ) d

dt

700 − Ce−kt ?

= k[700 − (700 − Ce−kt)]

Cke−kt ?

= kCe−kt kCe−kt= kCe−kt √

Exploratory Problems for Chapter 15

Population Studies

We’ve focused a lot of attention on the differential equation

dy

dt = ky

In the problems that follow you will use substitution to extend your

knowledge about this differential equation to a much larger class of

differential equations

Problem. In a laboratory, a colony of fruit flies are under scrutiny Let

P = P (t) be the number of flies in the colony at time t

Given ample food, the population would grow at a rate

propor-tional to itself according todPdt = kP But flies are continually being

siphoned off to another lab at a constant rate of C flies per day

The rate of change of the fly population is the rate at which it is

increasing due to reproduction minus the rate at which flies are being

siphoned off

We model the situation as follows

(Rate of change) = (rate of increase) − (rate of decrease)

dP

dt = kP − C

We want to solve this differential equation for P (t)

1.Let’s work on the equation

dP

dt = 0.02P − 2

First we’ll rewrite it as

dP

dt = 0.02(P − 100)

We’ll convert this differential equation to the form that we are

familiar with by making the substitution y = P − 100

(a) Express the differential equation dP

dt = 0.02(P − 100) in terms of y

(b) Solve the differential equation you got in part (a)

(c) Knowing that y = P − 100, find P (t) There will be an

arbi-trary constant in your answer because you will have found the

general solution to the differential equation

(d) Suppose that P (0) = 3000 Find the solution corresponding

to this particular initial condition

2.Use similar substitution techniques to solve the following

differ-ential equations

(a)dQdt = 2Q − 6

(b) dM= 0.1M − 200

P R O B L E M S F O R S E C T I O N 1 5 2

1 (a) Suppose a population grows at a rate of 5% per year: P = P0(1.05)t

i Express this in the form P = P0ert

ii Compute dPdt iii Find the proportionality constant k so that dPdt = kP (b) Suppose a population grows according to P = P0e0.05t

i Find the proportionality constant k so that dPdt = kP

ii By what percent does the population grow each year?

Look back over this problem and think about it Do your answers make sense to

you?

2 Forestland in Borneo is being destroyed at a rapid rate This is factual; the figures given

in this problem are not Assume that no new forestland is being created Let F (t) be the number of acres of forestland in Borneo

(a) Suppose that each year 10% of the existing forestland in Borneo is destroyed What

is F (t)? How many years will it take for half of the forestland in the country to be destroyed?

(b) Suppose instead that forestland in Borneo is being destroyed at a continuous rate of 10% per year What is F (t)? How many years will it take for half of the forestland

in the country to be destroyed?

(c) If forestland in Borneo is being destroyed at a continuous rate of 10% per year, what isdFdt? Does the sign of your answer make sense? Write your answer in the formdFdt = kF

3 Solve the following differential equations Give a general solution and then a particular solution corresponding to the initial condition given

(a) dydt = 3y initial condition: y(0) = 5 (b) dydx= −0.01y initial condition: y(2) = 1 (c) dw

ds = w initial condition: w(0) = π

4 Solve the following differential equations For each differential equation, find the general solution and then find a solution passing through the point (0,√

2)

(a) dydt = −2y (b) dydt = −2t (c) dydt = −2

5 (a) Show that P = Ce2t (where C is any constant) is a solution to the differential equation dPdt = 2P That is, show that if you computedPdt, you get 2P

(b) Show that P = e2t+ C is not a solution to the differential equationdPdt = 2P

6 Consider the differential equation dydt = y − 2 Which of the functions below are solutions? There could be more than one answer

(a) y = et

+ 2 (b) y = et+ 3 (c) y = Cet

+ 2 (d) y = C(et

+ 2)

7 (a) Is y = et

+ ln t a solution to the differential equation dydt = y − yt? (b) Is y = teta solution to the differential equationdydt = y −yt?

8 A wet dish towel is put on the back of a kitchen chair to dry It dries at a rate proportional

to the difference in moisture content between the dishtowel and the kitchen air Assume that the moisture content in the air is fixed and is given by M

(a) Set up the differential equation involving W = W (t), the amount of water in the dish towel at time t

(b) Find and sketch the solution

9 When a population has unlimited resources and is free from disease and strife, the rate

at which the population grows is proportional to the population Assume that both the bee and the mosquito populations described below behave according to this model

In both scenarios you are given enough information to find the proportionality constant k In one case, the information allows you to find k solely using the differential equation, without requiring that you solve it In the other scenario, you must actually solve the differential equation in order to find k

(a) Let M = M(t) be the mosquito population at time t, t in weeks At t = 0, there are

1000 mosquitoes Suppose that when there are 5000 mosquitoes, the population

is growing at a rate of 250 mosquitoes per week Write a differential equation reflecting the situation Include a value for k, the proportionality constant (b) Let B = B(t) be the bee population at time t, t in weeks At t = 0, there are 600 bees When t = 10, there are 800 bees Write a differential equation reflecting the situation Include a value for k, the proportionality constant

10 Newton’s law of cooling in its more general form tells us that the rate at which the temperature between an object and its environment changes is proportional to the difference in temperatures In other words, if D(t) is the temperature difference, then

dD

dt = kD

(a) Solve the differential equationdDdt = kD for D(t)

(b) Suppose a hot object is placed in a room whose temperature is kept constant at R degrees Let T (t) be the temperature of the object Newton’s law says that the hot object will cool at a rate proportional to the difference in temperature between the object and its environment Write a differential equation reflecting this statement and involving T Explain why this differential equation is equivalent to the previous one

(c) What is the sign of the constant of proportionality in the equation you wrote in part (b)? Explain

(d) Suppose that instead of a hot object we now consider a cold object Suppose that

we are interested in the temperature of a cold cup of lemonade as it warms up

to room temperature Let L(t) represent the temperature of the lemonade at time

t and assume that it sits in a room that is kept at 65 degrees At time t = 0, the lemonade is at 40 degrees 15 minutes later it has warmed to 50 degrees

i Sketch a graph of L(t) using your intuition and the information given

ii Is L(t) increasing at an increasing rate, or a decreasing rate?

iii Write a differential equation reflecting the situation Indicate the sign of the proportionality constant

iv Find L(t) Your final answer should have no undetermined constants Does the graph of the function you got for L(t) match the graph you drew?

v How long will it take the lemonade to reach a temperature of 55 degrees?

11 Suppose that in a certain country the population grows at a rate proportional to itself with proportionality constant 0.02 Further suppose that due to a drought people are leaving the country at a constant rate of 1000 people per year Let P = P (t) be the population of the country at time t, where t is in years Write a differential equation modeling the situation

12 Money in a certain trust-fund account is earning 5% interest per year compounded continuously Suppose money is being withdrawn from the account at a constant rate

of $2000 per year For the sake of our model, assume that money is being withdrawn continuously The account begins with $30,000 Let M = M(t) be the amount of money

in the account at time t, where t is in years Write a differential equation modeling the situation What is the initial condition?

13 Solve the following differential equations Use substitution to convert them to the form

dy

dt = ky

(a) dydt = 3y − 6 (b) dydt = y + 1 (c) dydt = 4 − 2y

14 Solve dydt = 2y − 6 with the initial condition y(0) = 2000

15 Return to Problem 12 Determine the amount of money in the account at time t by solving the differential equation and using the initial condition

16 A boarding school with 800 students has been hit by a flu epidemic If we assume that every student is either sick or healthy, that sick students will infect healthy ones, and that the disease is quite long in duration (it’s a nasty flu) then we can model the epidemic using the following assumption

The rate at which students are getting sick is proportional to the product of the number of sick students and the number of healthy ones

(a) Let S = S(t) be the number of sick students at time t Translate the statement above into mathematical language

(b) S(t) is an increasing function The rate at which students are getting sick is a quadratic function of S When the rate at which students are getting sick is highest, how many students are sick?

17 In the beginning of a chemical reaction there are 800 moles of substance A and none

of substance B Over the course of the reaction, the 800 moles of substance A are converted to 800 moles of substance B (Each molecule of A is converted to a molecule

of B via the reaction.) Suppose the rate at which A is turning into B is proportional to the product of the number of moles of A and the number of moles of B