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Calculus I: Lesson 20: Exponential Growth and Decay

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Calculus I

Lesson 20: Exponential Growth and Decay

Suppose we model the growth or decline of a population with the following differential equation. That is, the rate of growth is proportional to the amount present. Let's solve this equation for y.

 dy/dt = k y .

Then,

1*dy/dt/y = d*ln(y)/dt = d/dt*k*t => ln(y) = k*t+C .

Hence,

y = e^ln(y) = e^(k*t+C) = e^(k*t)*e^C

and setting A = e^C we have

y(t) = A*e^kt .

Notice that y(0) = A .

Conclusion: The solution to the differential equation is y(t) = y(0)*e^kt

Example 1
A bacteria culture starts with 500 bacteria and after 3 hours there are 8000

bacteria (assume exponential grwoth model).

a) Find an expression for the number of bacteria after t hours.

b) Find the number of bacteria after 4 hours.

c) When will the population reach 30,000?

d) Plot the expression from part (a)

e) from the Plot in (d) what happens to the population as times increases?

We have that 8000 = 500*e^(3*k) and need to find k. We have:

80/5 = e^(3*k) and hence ln(80/5) = 3*k . Thus,

k = ln(80/5)/3 .

> k:= ln(80/5)/3;

k := 1/3*ln(16)

> evalf(%);

.9241962406

> f1:= t -> 500 * exp(.9241962406 *t);

f1 := proc (t) options operator, arrow; 500*exp(.92...

> plot(f1(t), t = 0..5, color = red);

[Maple Plot]

> f1(4);

20158.73678

(b) Population after 4 hours is: 20158.73678

> fsolve(500 * exp(.9241962406 *t)= 30000,t);

4.430167947

(c) From Maple we have a numerical estimate that at t about 4.430167947 ,

the population is 30,000. More precisely:

30,000 = 500*e^kt => e^kt = 60 => ln(60)/k = t .

Thus at time ln(60)/k = t we have the population being 30,000.

Remember that k is about .9241962406 . Using Maple we have:

> evalf(ln(60)/ .9241962406);

4.430167948

(e) As times increases the population size goes to infinity.

Example 2
Polonium-214 has a half life of 1.4 x 10^(-4) seconds.

a) If a sample has a mass of 50 mg, find a formula for the mass that remains

after t seconds.

b) How long would it take for the mass to decay to 40 mg?

c) Will Maple plot the formula from (a); if not, why?

We have:

25 = 50*e^(.14e-3*k) => .5 = e^(.14e-3*k) => ln(.5) = .00014k.

Thus,

k = ln(.5)/.14e-3 .

> k:= ln(.5)/(.00014);

k := -4951.051290

a) Formula is: y = 50*e^( -4951.051290 *t)

b) We want: 40 = 50*e^( -4951.051290 *t) Thus,

t = ln(4/5) / ( -4951.051290 ).

> evalf(ln(4/5) / (-4951.051290));

.4506993328e-4

Hence it takes about .4506993328e-4 seconds for 40 mg to remain.

> f2:= t -> 50 * e^(-4951.051290 * t);

f2 := proc (t) options operator, arrow; 50*e^(-4951...

> plot(f2(t), t = 0..5);

[Maple Plot]

> evalf(e^(-4951.051290));

1/(e^4951.051290)

The graph of f2 very quickly approaches the x-axis. Hence we do not

get a good plot from Maple (for the x values chosen).

Example 3
 Scientists can determine the age of ancient objects by a method called

radiocarbon dating. The bombardment of the upper atmosphere by cosmic

rays converts bitrogen to a radioactive isotope of carbon, C-14, with a half

life of about 5730 years. Vegetation absorbs carbon dioxide through the

atmosphere and animal life assimilates C-14 through food chains.

When a plant or animal dies it stops replacing its carbon and the amount

of C-14 begins to decrease through radioactive decay. Therefore, the level

of radioactivity must also exponentially decay. A parchment fragment was

discovered that had about 74% as much C-14 radioactivity as does plant material

on the earth today. Estimate the age of the parchment.

Let y(t) denote the level of radioactivity. Then y(t) = y(0)*e^kt .

Using the half-life information we have:

1/2 = e^(5730*k) => ln(1/2)/5730 = k .

Thus we have:

.74 = e^kt => t = ln(.74)/k .

Thus the age of the parchment is: ln(.74)/k .

Using Maple for a numerical estiamte of the age we obtain:

>

> evalf( ln(.74) / ( ln(.5)/5730 ) );

2489.128182

Hence the parchment is about 2489 years old.

Example 4

Newtons Law of Cooling states that the rate of cooling of an object is

proportional to the temperature difference between the object and its surroundings.

A roast turkey is taken from the oven when tis temperature has reached 185 F

and is placed on a table in a room where the temperature is 75 F.

a) If the temperature of the turkey is 150 F after half an hour,

what is the temperature after 45 minutes?

b) When will the turkey have cooled to 100F?

c) Find a formula for the temperature of the turkey at time t.

d) Plot the formula from (c).

Let y(t) be the temperature of the turkey after t minutes. Then

Newtons Law of Cooling states that

dy/dt = k*(y(t)-75) .

Then

d/dt*(y(t)-75) = k*(y(t)-75) and hence y(t)-75 = (y(0)-75)*e^kt .

Thus,

y(t) = 75+(185-75)*e^kt

We have that

y(30) = 150 = 75+110*e^(k*30) => l ln(75/110)/30 = k .

Thus, the temperature of the turkey after t minutes is: y(t) = 75+e^kt , where

k = ln(75/110)/30 .

> evalf(ln(75/110)/30);

-.1276640841e-1

(a) We want y(45).

> y:= t -> 75 + 110 * exp(-.01276640841 * t);

y := proc (t) options operator, arrow; 75+110*exp(-...

> y(45);

136.9292118

Thus the turkey is about 136.9292118 F after 45 minutes.

(b) We want to know when y(t) = 100.

> fsolve( 75 + 110 * exp(-.01276640841 * t) = 100, t);

116.0549227

Numerically we have that it takes 1 hour and 56 minutes for the turkey to cool to 100F.

Without Maple we can solve by:

100 = 75+110*e^kt => 25/110 = e^kt => ln(25/110)/k = t .

Thus it takes ln(25/110)/k minutes for the turkey to cool to 100 F, recall

k = ln(75/110)/30 .

> evalf( ln(25/110) / ( ln(75/110) / 30));

116.0549227

Note that our answers agree.

> plot(y(t),t = 0..120, color = red);

[Maple Plot]