6.4

Exponential Growth and Decay

Quick Review

1. Rewrite the equation in exponential form or logarithmic

form.

a. ln

b.

2. Solve for .

a. ln( 2) 3

b. 2 10

c. 1.5 2

3. Solve for given ln( 1) 2 -3.

c

x

x

a b

e d

x

x

e

y y x

aeb cd ln

23 ex

5lnx

5.1ln

2lnx

132 xey

What you’ll learn about Separable Differential Equations Law of Exponential Change Continuously Compounded Interest Modeling Growth with Other Bases Newton’s Law of Cooling

Essential QuestionHow do we gain new insight into exponentialgrowth and decay by understanding the differentialequation dy/dx = ky?

Separable Differential Equation

A differential equation of the form

is called separable. We separate the variables by writing it in the form

xgyfdx

dy

dxxgdyyf

1

The solution is found by antidifferentiating each side with respect to its thusly isolated variable.

Example Solving by Separation of Variables

.0 when 3 and if for Solve 1. 22 xyyxdx

dyy

22 yxdx

dy

dxxdyy 22

dxxdyy 22

1 y3

3x C

3

1 C0

3

1C

3

1

3

31 x

y

3

131 x

y

31

3

xy

Only in the domain when x < 1.

Pg. 357, 6.4 #1-10

xxdxx 5cos15sin , 5sin 1. 223

Use the given trig identity to set up a u-substitution and then evaluate the indefinite integral.

xxdx

dyln 4 2. 5

Solve the differential equation.

If y changes at a rate proportional to the amount present

(that is, if ), and if y = yo when t = 0, thenkydx

dy

The Law of Exponential Change

tkoeyy

The constant k is the growth constant if k > 0 or the decay constant if k < 0.

Continuously Compounded Interest

If the interest is added continuously at a rate proportional to the amount in the account, you can model the growth of the account with the initial value problem:

Differential equation: Ardt

dA

Initial condition: oAA 0

The amount of money in the account after t years at an annual interest rate r:

troeAtA

Example Compounding Interest Continuously

2. Suppose you deposit $500 in an account that pays 5.3% annual interest. How much will you have 4 years later if the interest is (a) compounded continuously? (b) compounded monthly?

.053.0 and 500Let rAo

4 a. A 500 e 053.0 4 07.618

4 b. A 500

1 053.0

12

12 4

79.617

Example Finding Half-Life

The half-life of a radioactive substance with rate constant k ( k > 0 ) is:

k

2lnlife-half

3. Find the half-life of a radioactive substance with decay equationtk

oeyy The half-life is the solution to the equation o

tko yey

2

1

2

1 tke

2

1ln tk2

1ln

1

kt

k

2ln

Newton’s Law of Cooling

The rate at which an object’s temperature is changing at any given time is roughly proportional to the difference between its temperature and the temperature of the surrounding medium.

sTTkdt

dT

Since dT = d ( T – TS ), rewrite as:

ss TTkTTdt

d

Its solution, by the law of exponential change, is:

tksos eTTTT

Where To is the temperature at time t = 0.

Example Using Newton’s Law of Cooling

4. A temperature probe is removed from a cup of coffee and placed in water that has a temperature of TS = 4.5o C. Temperature readings T, as recorded in the table below are taken after 2 sec, 5 sec, and every other 5 sec thereafter.Estimate.

a. The coffee’s temperature at the time the temperature probe was removed.

b. The time when the temperature probe reading will be 8o C.

tko eTT 5.45.4

tT 9277.066.615.4 tT 9277.066.615.4 ,0at t 16.66T t9277.066.615.48

t9277.066.615.48

t9277.066.615.3

t9277.00568.0 0568.0log 9277.0t

22.38t

Pg. 357, 6.4 #11-43 odd