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We will be discussing the exponential rates while working on integrals. We will also see the topics: radioactive decay, half-life, and Newton's Law of cooling.

1.
6.4 Exponential Growth

Glacier National Park, Montana and Decay

Photo by Vickie Kelly, 2004 Greg Kelly, Hanford High School, Richland, Washington

Glacier National Park, Montana and Decay

Photo by Vickie Kelly, 2004 Greg Kelly, Hanford High School, Richland, Washington

2.
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:

dy

ky

dt

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:

dy

ky

dt

3.
dy Rate of change is proportional

ky

dt to the amount present.

dy k dt Divide both sides by y.

1

y dy k dt Integrate both sides.

ln y kt C

ky

dt to the amount present.

dy k dt Divide both sides by y.

1

y dy k dt Integrate both sides.

ln y kt C

4.
1

y dy k dt Integrate both sides.

ln y kt C

ln y kt C Exponentiate both sides.

e e

C kt When multiplying like bases, add

y e e exponents. So added exponents

can be written as multiplication.

y dy k dt Integrate both sides.

ln y kt C

ln y kt C Exponentiate both sides.

e e

C kt When multiplying like bases, add

y e e exponents. So added exponents

can be written as multiplication.

5.
ln y kt C

Exponentiate both sides.

e e

C kt When multiplying like bases, add

y e e exponents. So added exponents

can be written as multiplication.

C kt

y e e

kt

y Ae Since eC is a constant, let e

C

A .

Exponentiate both sides.

e e

C kt When multiplying like bases, add

y e e exponents. So added exponents

can be written as multiplication.

C kt

y e e

kt

y Ae Since eC is a constant, let e

C

A .

6.
C kt

y e e

kt

y Ae Since eC is a constant, let e

C

A .

1

k 0

y0 Ae At t 0 , y y0 .

y0 A

kt This is the solution to our original initial

y y0 e value problem.

y e e

kt

y Ae Since eC is a constant, let e

C

A .

1

k 0

y0 Ae At t 0 , y y0 .

y0 A

kt This is the solution to our original initial

y y0 e value problem.

7.
kt

Exponential Change: y y0 e

If the constant k is positive then the equation

represents growth. If k is negative then the equation

represents decay.

Note: This lecture will talk about exponential change

formulas and where they come from. The problems in

this section of the book mostly involve using those

formulas. There are good examples in the book, which I

will not repeat here.

Exponential Change: y y0 e

If the constant k is positive then the equation

represents growth. If k is negative then the equation

represents decay.

Note: This lecture will talk about exponential change

formulas and where they come from. The problems in

this section of the book mostly involve using those

formulas. There are good examples in the book, which I

will not repeat here.

8.
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:

kt

r

A t A0 1

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.

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:

kt

r

A t A0 1

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.

9.
Of course, the bank does not employ some clerk to

continuously calculate your interest with an adding machine.

kt

r

We could calculate: lim A0 1

k

k

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

(The TI-89 can do it now if you would like to try it.)

Since the interest is proportional to the amount present,

the equation becomes:

You may also use:

Continuously Compounded

rt

Interest:

rt

A Pe

A A0 e

which is the same thing.

continuously calculate your interest with an adding machine.

kt

r

We could calculate: lim A0 1

k

k

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

(The TI-89 can do it now if you would like to try it.)

Since the interest is proportional to the amount present,

the equation becomes:

You may also use:

Continuously Compounded

rt

Interest:

rt

A Pe

A A0 e

which is the same thing.

10.
Radioactive Decay

The equation for the amount of

a radioactive element left after

time t is:

kt

y y0e

This allows the decay constant, k,

to be positive.

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

The equation for the amount of

a radioactive element left after

time t is:

kt

y y0e

This allows the decay constant, k,

to be positive.

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

11.
Half-life

1 kt

y0 y0 e

2

1

ln ln e kt

2

0

ln1 ln 2 kt

Half-life:

ln 2 kt

ln 2

ln 2 half-life

t k

k

1 kt

y0 y0 e

2

1

ln ln e kt

2

0

ln1 ln 2 kt

Half-life:

ln 2 kt

ln 2

ln 2 half-life

t k

k

12.
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.)

dT

If we solve the differential equation: k T Ts

dt

we get:

Newton’s Law of Cooling

T Ts T0 Ts e kt

where Ts is the temperature

of the surrounding medium,

which is a constant.

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.)

dT

If we solve the differential equation: k T Ts

dt

we get:

Newton’s Law of Cooling

T Ts T0 Ts e kt

where Ts is the temperature

of the surrounding medium,

which is a constant.