Exponential Growth and Decay - Tianran Chentrchen.weebly.com/uploads/7/2/7/4/7274826/ch1-7.pdf ·...

Exponential Growth and Decay

An exponential function P(t) = P0at

can also be written asI P(t) = P0(1 + r)t (with a = (1 + r))I P(t) = P0ekt (with a = ek)

Both r and k represents a relative rate of change:r percentage rate of changek continuous rate of change

An exponential function models. . .exponential growth if r > 0 or k > 0 (i.e. a > 1).exponential decay if r < 0 or k < 0 (i.e. a < 1).

An exponential function P(t) = P0at can also be written asI P(t) = P0(1 + r)t (with a = (1 + r))I P(t) = P0ekt (with a = ek)

How Fast Does an Exponential Function Grow/Decay?

Both the percentage and the continuous rate of change tell ushow fast an exponential function grow/decay.

There are other ways of measuring how fast an exponentialfunction changes: by measuring how “long” does it take for thefunction to double or decrease by a half.

I For an exponential growth function, the “time” it takes forthe function to double is called its doubling time.

I For an exponential decay function, the “time” it takes forthe function to decrease to half of the initial value is calledits half-life.

Half-life and Doubling timeHalf-life is the value T such that P(T) = 1

2 P(0)Doubling time is the value T such that P(T) = 2P(0)

Half-life is usually only used in exponential decay whiledoubling time is usually only used in exponential growth.

Both half-life and doubling time are independent from theinitial value of the exponential function.

Formulae:

half-life =ln(1/2)

ln(1/2)ln(1 + r)

=ln(1/2)

doubling time =ln 2ln a

ln(1 + r)=

Formulae:

half-life =ln(1/2)

ln(1/2)ln(1 + r)

=ln(1/2)

ln(1 + r)=

Formulae:

half-life =ln(1/2)

ln(1/2)ln(1 + r)

=ln(1/2)

ln(1 + r)=

Formulae:

half-life =ln(1/2)

ln(1/2)ln(1 + r)

=ln(1/2)

ln(1 + r)=

ExampleFind the doubling time of the function P(t) = 100 · e0.05t.

Solution:Quick way:

doubling time =ln 2

=ln 20.05

≈ 13.8629

Slow way: we want to findthe value t such that

2 · P(0) = P(t)

2 · 100 = 100 · e0.05t

2 = e0.05t

ln 2 = ln e0.05t = 0.05t

t =ln 20.05

≈ 3.6889

Solution:Quick way:

doubling time =ln 2

=ln 20.05

≈ 13.8629

2 · P(0) = P(t)

2 · 100 = 100 · e0.05t

2 = e0.05t

ln 2 = ln e0.05t = 0.05t

t =ln 20.05

≈ 3.6889

Solution:Quick way:

doubling time =ln 2

=ln 20.05

≈ 13.8629

2 · P(0) = P(t)

2 · 100 = 100 · e0.05t

2 = e0.05t

ln 2 = ln e0.05t = 0.05t

t =ln 20.05

≈ 3.6889

Solution:Quick way:

doubling time =ln 2

=ln 20.05

≈ 13.8629

2 · P(0) = P(t)

2 · 100 = 100 · e0.05t

2 = e0.05t

ln 2 = ln e0.05t = 0.05t

t =ln 20.05

≈ 3.6889

Solution:Quick way:

doubling time =ln 2

=ln 20.05

≈ 13.8629

2 · P(0) = P(t)

2 · 100 = 100 · e0.05t

2 = e0.05t

ln 2 = ln e0.05t = 0.05t

t =ln 20.05

≈ 3.6889

Solution:Quick way:

doubling time =ln 2

=ln 20.05

≈ 13.8629

2 · P(0) = P(t)

2 · 100 = 100 · e0.05t

2 = e0.05t

ln 2 = ln e0.05t = 0.05t

t =ln 20.05

≈ 3.6889

Solution:Quick way:

doubling time =ln 2

=ln 20.05

≈ 13.8629

2 · P(0) = P(t)

2 · 100 = 100 · e0.05t

2 = e0.05t

ln 2 = ln e0.05t = 0.05t

t =ln 20.05

≈ 3.6889

Solution:Quick way:

doubling time =ln 2

=ln 20.05

≈ 13.8629

2 · P(0) = P(t)

2 · 100 = 100 · e0.05t

2 = e0.05t

ln 2 = ln e0.05t = 0.05t

t =ln 20.05

≈ 3.6889

Solution:Quick way:

doubling time =ln 2

=ln 20.05

≈ 13.8629

2 · P(0) = P(t)

2 · 100 = 100 · e0.05t

2 = e0.05t

ln 2 = ln e0.05t = 0.05t

t =ln 20.05

≈ 3.6889

Solution:Quick way:

doubling time =ln 2

=ln 20.05

≈ 13.8629

2 · P(0) = P(t)

2 · 100 = 100 · e0.05t

2 = e0.05t

ln 2 = ln e0.05t = 0.05t

t =ln 20.05

≈ 3.6889

ExampleThe human body eliminates an antidepressant at a rate of 10%per hour. Find the half-life of this antidepressant in the body.

Solution: First, the per hour percentage decay rate is r = −0.1(-10%). So the formula for this function is

P(t) = P0(1 − 0.1)t = P0(0.9)t

Quick way:

half-life =ln(1/2)

ln(1 + r)

=ln(1/2)ln 0.9

≈ 6.5788 (hours)

Slow way:

1/2P0 = P0(0.9)t

1/2 = (0.9)t

ln(1/2) = ln(0.9)t = t ln 0.9

t =ln(1/2)ln 0.9

≈ 6.5788 (hours)

Solution: First, the per hour percentage decay rate is r = −0.1(-10%).

So the formula for this function is

P(t) = P0(1 − 0.1)t = P0(0.9)t

Quick way:

half-life =ln(1/2)

ln(1 + r)

=ln(1/2)ln 0.9

≈ 6.5788 (hours)

Slow way:

1/2P0 = P0(0.9)t

1/2 = (0.9)t

ln(1/2) = ln(0.9)t = t ln 0.9

t =ln(1/2)ln 0.9

≈ 6.5788 (hours)

P(t) = P0(1 − 0.1)t = P0(0.9)t

Quick way:

half-life =ln(1/2)

ln(1 + r)

=ln(1/2)ln 0.9

≈ 6.5788 (hours)

Slow way:

1/2P0 = P0(0.9)t

1/2 = (0.9)t

ln(1/2) = ln(0.9)t = t ln 0.9

t =ln(1/2)ln 0.9

≈ 6.5788 (hours)

P(t) = P0(1 − 0.1)t = P0(0.9)t

Quick way:

half-life =ln(1/2)

ln(1 + r)

=ln(1/2)ln 0.9

≈ 6.5788 (hours)

Slow way:

1/2P0 = P0(0.9)t

1/2 = (0.9)t

ln(1/2) = ln(0.9)t = t ln 0.9

t =ln(1/2)ln 0.9

≈ 6.5788 (hours)

P(t) = P0(1 − 0.1)t = P0(0.9)t

Quick way:

half-life =ln(1/2)

ln(1 + r)

=ln(1/2)ln 0.9

≈ 6.5788 (hours)

Slow way:

1/2P0 = P0(0.9)t

1/2 = (0.9)t

ln(1/2) = ln(0.9)t = t ln 0.9

t =ln(1/2)ln 0.9

≈ 6.5788 (hours)

P(t) = P0(1 − 0.1)t = P0(0.9)t

Quick way:

half-life =ln(1/2)

ln(1 + r)

=ln(1/2)ln 0.9

≈ 6.5788 (hours)

Slow way:

1/2P0 = P0(0.9)t

1/2 = (0.9)t

ln(1/2) = ln(0.9)t = t ln 0.9

t =ln(1/2)ln 0.9

≈ 6.5788 (hours)

P(t) = P0(1 − 0.1)t = P0(0.9)t

Quick way:

half-life =ln(1/2)

ln(1 + r)

=ln(1/2)ln 0.9

≈ 6.5788 (hours)

Slow way:

1/2P0 = P0(0.9)t

1/2 = (0.9)t

ln(1/2) = ln(0.9)t = t ln 0.9

t =ln(1/2)ln 0.9

≈ 6.5788 (hours)

P(t) = P0(1 − 0.1)t = P0(0.9)t

Quick way:

half-life =ln(1/2)

ln(1 + r)

=ln(1/2)ln 0.9

≈ 6.5788 (hours)

Slow way:

1/2P0 = P0(0.9)t

1/2 = (0.9)t

ln(1/2) = ln(0.9)t = t ln 0.9

t =ln(1/2)ln 0.9

≈ 6.5788 (hours)

P(t) = P0(1 − 0.1)t = P0(0.9)t

Quick way:

half-life =ln(1/2)

ln(1 + r)

=ln(1/2)ln 0.9

≈ 6.5788 (hours)

Slow way:

1/2P0 = P0(0.9)t

1/2 = (0.9)t

ln(1/2) = ln(0.9)t = t ln 0.9

t =ln(1/2)ln 0.9

≈ 6.5788 (hours)

P(t) = P0(1 − 0.1)t = P0(0.9)t

Quick way:

half-life =ln(1/2)

ln(1 + r)

=ln(1/2)ln 0.9

≈ 6.5788 (hours)

Slow way:

1/2P0 = P0(0.9)t

1/2 = (0.9)t

ln(1/2) = ln(0.9)t = t ln 0.9

t =ln(1/2)ln 0.9

≈ 6.5788 (hours)

P(t) = P0(1 − 0.1)t = P0(0.9)t

Quick way:

half-life =ln(1/2)

ln(1 + r)

=ln(1/2)ln 0.9

≈ 6.5788 (hours)

Slow way:

1/2P0 = P0(0.9)t

1/2 = (0.9)t

ln(1/2) = ln(0.9)t = t ln 0.9

t =ln(1/2)ln 0.9

≈ 6.5788 (hours)

ExampleThe human body eliminates an antidepressant at a rate of 10%per hour. The initial dose is 500mg. Estimate the length of timeuntil the amount of this antidepressant is reduced to 100mg.

Solution: As in the previous example, r = −0.1. It is also giventhat P0 = 500 (mg). So the amount of the antidepressant left is

P(t) = 500(1 − 0.1)t = 500(0.9)t

And we are trying to find the t value such that P(t) = 100.

100 = 500(0.9)t

1/5 = (0.9)t

ln(1/5) = ln(0.9)t = t ln 0.9

t =ln(1/5)ln 0.9

≈ 15.2755 (hours)

Solution: As in the previous example, r = −0.1.

It is also giventhat P0 = 500 (mg). So the amount of the antidepressant left is

P(t) = 500(1 − 0.1)t = 500(0.9)t

100 = 500(0.9)t

1/5 = (0.9)t

ln(1/5) = ln(0.9)t = t ln 0.9

t =ln(1/5)ln 0.9

≈ 15.2755 (hours)

Solution: As in the previous example, r = −0.1. It is also giventhat P0 = 500 (mg).

So the amount of the antidepressant left is

P(t) = 500(1 − 0.1)t = 500(0.9)t

100 = 500(0.9)t

1/5 = (0.9)t

ln(1/5) = ln(0.9)t = t ln 0.9

t =ln(1/5)ln 0.9

≈ 15.2755 (hours)

P(t) = 500(1 − 0.1)t = 500(0.9)t

100 = 500(0.9)t

1/5 = (0.9)t

ln(1/5) = ln(0.9)t = t ln 0.9

t =ln(1/5)ln 0.9

≈ 15.2755 (hours)

P(t) = 500(1 − 0.1)t = 500(0.9)t

100 = 500(0.9)t

1/5 = (0.9)t

ln(1/5) = ln(0.9)t = t ln 0.9

t =ln(1/5)ln 0.9

≈ 15.2755 (hours)

P(t) = 500(1 − 0.1)t = 500(0.9)t

100 = 500(0.9)t

1/5 = (0.9)t

ln(1/5) = ln(0.9)t = t ln 0.9

t =ln(1/5)ln 0.9

≈ 15.2755 (hours)

P(t) = 500(1 − 0.1)t = 500(0.9)t

100 = 500(0.9)t

1/5 = (0.9)t

ln(1/5) = ln(0.9)t = t ln 0.9

t =ln(1/5)ln 0.9

≈ 15.2755 (hours)

P(t) = 500(1 − 0.1)t = 500(0.9)t

100 = 500(0.9)t

1/5 = (0.9)t

ln(1/5) = ln(0.9)t = t ln 0.9

t =ln(1/5)ln 0.9

≈ 15.2755 (hours)

P(t) = 500(1 − 0.1)t = 500(0.9)t

100 = 500(0.9)t

1/5 = (0.9)t

ln(1/5) = ln(0.9)t = t ln 0.9

t =ln(1/5)ln 0.9

≈ 15.2755 (hours)

ExampleA picture supposedly painted by Vermeer (1632-1675) contains99.5% of its carbon-14 (half-life 5730 years). From thisinformation decide whether the picture is a fake.

Solution: We can construct the model for the amount of carbon-14 asa function of time.Then we can determine how long does it take for carbon-14 todecrease to 99.5% of its original amount. And that will be the age ofthe painting.We can use P(t) = P0ekt where t is the number of years since thepainting was created. Use the half-life information, we get equation

P0 = P0ek·5730

I.e.12= e5730k ⇒ k =

ln(1/2)5730

≈ −0.00012

Solution: We can construct the model for the amount of carbon-14 asa function of time.

Then we can determine how long does it take for carbon-14 todecrease to 99.5% of its original amount. And that will be the age ofthe painting.We can use P(t) = P0ekt where t is the number of years since thepainting was created. Use the half-life information, we get equation

P0 = P0ek·5730

I.e.12= e5730k ⇒ k =

ln(1/2)5730

≈ −0.00012

Solution: We can construct the model for the amount of carbon-14 asa function of time.Then we can determine how long does it take for carbon-14 todecrease to 99.5% of its original amount. And that will be the age ofthe painting.

We can use P(t) = P0ekt where t is the number of years since thepainting was created. Use the half-life information, we get equation

P0 = P0ek·5730

I.e.12= e5730k ⇒ k =

ln(1/2)5730

≈ −0.00012

Solution: We can construct the model for the amount of carbon-14 asa function of time.Then we can determine how long does it take for carbon-14 todecrease to 99.5% of its original amount. And that will be the age ofthe painting.We can use P(t) = P0ekt where t is the number of years since thepainting was created.

Use the half-life information, we get equation

P0 = P0ek·5730

I.e.12= e5730k ⇒ k =

ln(1/2)5730

≈ −0.00012

P0 = P0ek·5730

I.e.12= e5730k ⇒ k =

ln(1/2)5730

≈ −0.00012

P0 = P0ek·5730

I.e.12= e5730k

⇒ k =ln(1/2)

5730≈ −0.00012

P0 = P0ek·5730

I.e.12= e5730k ⇒ k =

ln(1/2)5730

≈ −0.00012

(. . . continue)Then the function P(t) is given by (approximately)

P(t) = P0e−0.00012t

Note the that the initial amount is not given, and it is irreleventin this problem.To find the age of the painting, we shall use the fact that 99.5%of the original amount is still left. I.e.,

0.995P0 = P(t) = P0e−0.00012t

0.995 = e−0.00012t ⇒ t =ln(0.995)−0.00012

≈ 41.8

So the age of the painting is approximately 42 years and henceit must be a fake.

P(t) = P0e−0.00012t

Note the that the initial amount is not given, and it is irreleventin this problem.

To find the age of the painting, we shall use the fact that 99.5%of the original amount is still left. I.e.,

0.995P0 = P(t) = P0e−0.00012t

0.995 = e−0.00012t ⇒ t =ln(0.995)−0.00012

≈ 41.8

P(t) = P0e−0.00012t

0.995P0 = P(t) = P0e−0.00012t

0.995 = e−0.00012t ⇒ t =ln(0.995)−0.00012

≈ 41.8

P(t) = P0e−0.00012t

0.995P0 = P(t) = P0e−0.00012t

0.995 = e−0.00012t

⇒ t =ln(0.995)−0.00012

≈ 41.8

P(t) = P0e−0.00012t

0.995P0 = P(t) = P0e−0.00012t

0.995 = e−0.00012t ⇒ t =ln(0.995)−0.00012

≈ 41.8

P(t) = P0e−0.00012t

0.995P0 = P(t) = P0e−0.00012t

0.995 = e−0.00012t ⇒ t =ln(0.995)−0.00012

≈ 41.8

Financial Application: Compound InterestCompound interest (as opposed to “simple interest”) arisewhen interest is allowed to be added to the principal.

I.e., the interest itself is allowed to generate more interest.Compound interest is standard in finance and it is modeled byexponential functions.

If P0 is the principal (the initial amount in an account) and r isthe interest rate, then the balance in the account after t years isgiven by

P(t) = P0(1 + r)t if interest is compounded annually

P(t) = P0ekt if interest is compounded continuously

P(t) = P0(1 + r/n)nt if interest is compounded n times per year

Financial Application: Compound InterestCompound interest (as opposed to “simple interest”) arisewhen interest is allowed to be added to the principal.I.e., the interest itself is allowed to generate more interest.

Compound interest is standard in finance and it is modeled byexponential functions.

Financial Application: Compound InterestCompound interest (as opposed to “simple interest”) arisewhen interest is allowed to be added to the principal.I.e., the interest itself is allowed to generate more interest.Compound interest is standard in finance and it is modeled byexponential functions.

Example$5000 is deposited in a saving account earning interest at a rateof %3 per year, compounded continuously. Find the balanceafter 20 years.

Solution: Since the interest is compounded continuously, weshould use the formula P(t) = P0ekt with P0 = 5000 andk = 0.03. I.e.,

P(t) = 5000e0.03t

Then the balance after 20 years is given by

P(20) = 5000e0.03·20 ≈ 9110.59 dollars

Solution: Since the interest is compounded continuously, weshould use the formula P(t) = P0ekt

with P0 = 5000 andk = 0.03. I.e.,

P(t) = 5000e0.03t

P(20) = 5000e0.03·20 ≈ 9110.59 dollars

Solution: Since the interest is compounded continuously, weshould use the formula P(t) = P0ekt with P0 = 5000 andk = 0.03.

I.e.,P(t) = 5000e0.03t

P(20) = 5000e0.03·20 ≈ 9110.59 dollars

P(t) = 5000e0.03t

P(20) = 5000e0.03·20 ≈ 9110.59 dollars

P(t) = 5000e0.03t

P(20) = 5000e0.03·20 ≈ 9110.59 dollars

Example$5000 is deposited in a saving account earning interest at a rateof %2 per year, compounded annually. How long will it takefor the balance to reach $8000?

Solution: Since the interest is compounded annually, we should useP(t) = P0(1 + r)t with P0 = 5000 and r = 0.02.

P(t) = 5000(1 + 0.02)t = 5000(1.02)t

To find out how long it will take for the balance to reach 8000, wesimply have to solve the equation

8000 = P(t) = 5000(1.02)t ⇒ 1.02t =80005000

and so

ln(1.02t) = t ln 1.02 = ln 1.6 ⇒ t =ln 1.6

ln 1.02≈ 23.73

Solution: Since the interest is compounded annually, we should useP(t) = P0(1 + r)t

with P0 = 5000 and r = 0.02.

P(t) = 5000(1 + 0.02)t = 5000(1.02)t

8000 = P(t) = 5000(1.02)t ⇒ 1.02t =80005000

and so

ln(1.02t) = t ln 1.02 = ln 1.6 ⇒ t =ln 1.6

ln 1.02≈ 23.73

P(t) = 5000(1 + 0.02)t = 5000(1.02)t

8000 = P(t) = 5000(1.02)t ⇒ 1.02t =80005000

and so

ln(1.02t) = t ln 1.02 = ln 1.6 ⇒ t =ln 1.6

ln 1.02≈ 23.73

P(t) = 5000(1 + 0.02)t = 5000(1.02)t

8000 = P(t) = 5000(1.02)t

⇒ 1.02t =80005000

and so

ln(1.02t) = t ln 1.02 = ln 1.6 ⇒ t =ln 1.6

ln 1.02≈ 23.73

P(t) = 5000(1 + 0.02)t = 5000(1.02)t

8000 = P(t) = 5000(1.02)t ⇒ 1.02t =80005000

and so

ln(1.02t) = t ln 1.02 = ln 1.6

⇒ t =ln 1.6

ln 1.02≈ 23.73

P(t) = 5000(1 + 0.02)t = 5000(1.02)t

8000 = P(t) = 5000(1.02)t ⇒ 1.02t =80005000

and so

ln(1.02t) = t ln 1.02 = ln 1.6 ⇒ t =ln 1.6

ln 1.02≈ 23.73

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