Warmup

1)

2)

6.4: Exponential Growth and Decay

The number of bighorn sheep in a population increases at a rate that is proportional to the number of sheep present (at least for awhile.)

So does any population of living creatures. Other things that increase or decrease at a rate proportional to the amount present include radioactive material and money in an interest-bearing account.

If the rate of change is proportional to the amount present, the change can be modeled by:

dyky

dt

dyky

dt

1 dy k dt

y

1 dy k dt

y

ln y kt C

Rate of change is proportional to the amount present.

Divide both sides by y.

Integrate both sides.

1 dy k dt

y

ln y kt C

Integrate both sides.

Exponentiate both sides.

When multiplying like bases, add exponents. So added exponents can be written as multiplication.

ln y kt Ce e

C kty e e

ln y kt Ce e

C kty e e

Exponentiate both sides.

When multiplying like bases, add exponents. So added exponents can be written as multiplication.

C kty e e

kty Ae Since is a constant, let .Ce Ce A

C kty e e

kty Ae Since is a constant, let .Ce Ce A

At , .0t 0y y00

ky Ae

0y A

1

0kty y e This is the solution to our original initial

value problem.

0kty y eExponential Change:

If the constant k is positive then the equation

represents growth. If k is negative then the equation represents decay.

Continuously Compounded Interest

If money is invested in a fixed-interest account where the interest is added to the account k times per year, the

amount present after t years is:

0 1kt

rA t A

k

If the money is added back more frequently, you will make a little more money.

The best you can do is if the interest is added continuously.

Of course, the bank does not employ some clerk to continuously calculate your interest with an adding machine.

We could calculate: 0lim 1kt

k

rA

k

but we won’t learn how to find this limit until chapter 8.

Since the interest is proportional to the amount present, the equation becomes:

Continuously Compounded Interest:

0rtA A e

You may also use:

rtA Pe

which is the same thing.

Suppose you deposit $600 in an account that pays 2.1% annualinterest. How much will you have 10 years later if the interest is (a) compounded annually (b) compounded quarterly (c) compounded continuously

Bank of America currently offers a whopping 0.05% compounding

Radioactive Decay

The equation for the amount of a radioactive element left after time t is:

0kty y e

This allows the decay constant, k, to be positive.

The half-life is the time required for half the material to decay.

Half-life

0 0

1

2kty y e

1ln ln

2kte

ln1 ln 2 kt 0

ln 2 kt

ln 2t

k

Half-life:

ln 2half-life

k

Scientists who do carbon-14 dating use 5700 years for its half-life. Find the age of a sample in which 10% of theradioactive nuclei originally present have delayed.

ex. 3 p. 332

Half-life:

ln 2half-life

k5700

2ln

2ln5700

k

k

0kty y e

t

eyy

5700

2ln

009.0t

e

5700

2ln

9.0t

e

5700

2ln

ln9.0ln

t5700

2ln9.0ln

8669.0ln2ln

5700 t

Newton’s Law of Cooling

Espresso left in a cup will cool to the temperature of the surrounding air. The rate of cooling is proportional to the difference in temperature between the liquid and the air.

(It is assumed that the air temperature is constant.)

If we solve the differential equation: sdT

k T Tdt

we get:Newton’s Law of Cooling

0kt

s sT T T T e

where is the temperature of the surrounding medium, which is a constant.

sT

ex. 4 p.333

A hard boiled egg at 98oC is put under running 18oC waterto cool. After 5 minutes, the egg’s temperature is found to be38oC. How much longer will it take the egg to reach 20o?

Newton’s Law of Cooling

0kt

s sT T T T e

where is the temperature of the surrounding medium, which is a constant.

sT

98

18

0

T

Ts

ktss eTTTT ][ 0

kteT ]1898[18

)5(]1898[1838 ke )5(]80[20 ke

)5(

4

1 ke

)5(ln4

1ln ke

k54

1ln

k 5

41

ln

k5

4ln1ln

k5

4ln1ln

k

5

4ln0

k

k

4ln2.05

4ln

teT )4ln2.0(]1898[18

teT )4ln2.0(8018 How much longer will it take the egg to reach 20o?

te )4ln2.0(801820 Solve algebraically or graphically min3.13t

minutes more 8.35 -13.3 so

Group Problem:

When an object is removed from a furnace and placedin an environment with a constant temperature of 80oF, itscore temperature is 1500oF. One hour after it is removed,the core temperature is 1120oF. Find the core temperature5 hours after the object is removed from the furnace.

Solution

The end