6.4 exponential growth and decay. the number of bighorn sheep in a population increases at a rate...

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6.4 Exponential Growth and Decay

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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:

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Page 1: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

6.4 Exponential Growth and Decay

Page 2: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

eb =ac =lnd

x =e3 −2x =ln5

x =ln2ln1.5 y =e2x−3 −1

Page 3: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

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 kydt

=

Page 4: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

dy kydt

=

1 dy k dty

=

1 dy k dty

=∫ ∫

ln y kt C= +

Rate of change is proportional to the amount present.

Divide both sides by y.

Integrate both sides.

Page 5: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

1 dy k dty

=∫ ∫

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= ⋅

Page 6: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

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± =

Page 7: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

C kty e e=±

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

At , .0t = 0y y=00

ky Ae ⋅=

0y A=

1

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

value problem.

Page 8: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

0kty y e=Exponential Change:

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.

Page 9: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

Modeling GrowthAt the beginning of summer, the population of a hive of hornets is growing at a rate proportional to the population. From a population of 10 on May 1, the number of hornets grows to 50 in thirty days. If the growth continues to follow the same model, how many days after May 1 will the population reach 100?

Page 10: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

y = y0 ekt and y0 =10, y(30)= 50

y = 10 ekt

50 = 10 e 30k

ln 530

= k

100 = 10 eln 530

t

y = 10 eln 530

t

y = 10 eln 530

t

10 = eln 530

t

t =

30 ln 10ln 5

= 42.920 days

Page 11: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

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 1ktrA 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.

Page 12: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

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

We could calculate: 0lim 1kt

k

rAk→∞

⎛ ⎞+⎜ ⎟⎝ ⎠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.

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

Page 13: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

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.

Page 14: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

Half-life

0 012

kty y e−=

( )1ln ln2

kte−⎛ ⎞=⎜ ⎟⎝ ⎠

ln1 ln 2 kt− =−0

ln 2 kt=

ln 2 tk

=

Half-life:

ln 2half-lifek

=

Page 15: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

Using Carbon-14 Dating

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

Page 16: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

A = A0

12

⎛⎝⎜

⎞⎠⎟

t5700

.9 A0 = A0

12

⎛⎝⎜

⎞⎠⎟

t5700

ln (.9) =

t5700

ln 12

t = 5700

ln .9ln .5

= 866 yrs old

Page 17: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

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

Page 18: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

Newton’s Law of Cooling

dTdt

= - k T - Ts( )

dT

T - Ts( ) = - k∫ ∫ dt

ln T - Ts = - k⋅t + C

T - Ts = e−k⋅t + C

T - Ts = C e−k⋅t at t = 0, C = T0 - Ts

T - Ts = T0 - Ts( ) e−k⋅t

T = T0 - Ts( ) e−k⋅t + Ts

Page 19: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

Newton’s Law of Cooling

Newton's Law of Cooling: For T0 = initial temperature of object

Ts = temperature of surrounding environment

k = constant of variation, found by using givenThen, T = T0 - Ts( )⋅e−kt + Ts

Page 20: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

Using Newton’s Law of Cooling

A hard boiled egg at 98ºC is put in a pan under running 18ºC water to cool. After 5 minutes, the egg’s temperature is found to be 38ºC. How much longer will it take the egg to reach 20ºC?

T = T0 - Ts( )⋅e−kt + Ts

Page 21: 6.4 Exponential Growth and Decay. The number of bighorn sheep in a population increases at a rate that…

Ts = 18 and T0 = 98, then T - 18 = (98 - 18) e- k t

T = 18 + 80 e- k t Find k by using T = 38 when t = 5

38 = 18 + 80 e- 5k

e- 5k =

14

−5 k = ln

14 = - ln 4

k =

15

ln 4

T = 18 + 80 e(-.2 ln 4) t

Solve 20 = 18 + 80 e(-.2 ln 4) t

t = 13.3 min

13.3 - 5 = 8.3 secs longer