Exponential Astonishment

Copyright © 2015, 2011, 2008 Pearson Education, Inc. Chapter 8, Unit B, Slide 1

Exponential Astonishment8


Unit 8B

Doubling Time and Half-Life


Doubling and Halving Times

The time required for each doubling in exponential growth is called doubling time.

The time required for each halving in exponential decay is called halving time.


After a time t, an exponentially growing quantity with a doubling time of Tdouble increases in size by a factor of . The new value of the growing quantity is related to its initial value (at t = 0) by

New value = initial value × 2t/Tdouble

double2t T

Doubling Time


Example

Compound interest (Unit 4B) produces exponential growth because an interest-bearing account grows by the same percentage each year. Suppose your bank account has a doubling time of 13 years. By what factor does your balance increase in 50 years?

Solution

The doubling time is Tdouble = 13 years, so after t = 50 years your balance increases by a factor of

2t/Tdouble = 250 yr/13 yr = 23.8462 ≈ 14.382

For example, if you start with a balance of $1000, in 50 years it will grow to $1000 × 14.382 = $14,382.


Approximate Double Time Formula(The Rule of 70)

For a quantity growing exponentially at a rate of P% per time period, the doubling time is approximately

This approximation works best for small growth rates and breaks down for growth rates over about 15%.

double

70T

P


Example

World population doubled in the 40 years from 1960 to 2000. What was the average percentage growth rate during this period? Contrast this growth rate with the 2012 growth rate of 1.1% per year.

Solution

We answer the question by solving the approximate doubling time formula for P. Multiplying both sides of the formula by P and dividing both sides by Tdouble, we have

double

70PT


Example (cont)

Substituting Tdouble = 40 years, we find

The average population growth rate between 1960 and 2000 was about P% = 1.75% per year. This is significantly higher (by 0.65 percentage point) than the 2012 growth rate of 1.1% per year.

double

70 701.75/ yr

40 yrP

T


After a time t, an exponentially decaying quantity with a half-life time of Thalf decreases in size by a

factor of . The new value of the decaying quantity is related to its initial value (at t = 0) by

New value = initial value x (1/2)t/Thalf

( ) half1 2t T

Exponential Decay and Half-Life


Example

Radioactive carbon-14 has a half-life of about 5700 years. It collects in organisms only while they are alive. Once they are dead, it only decays. What fraction of the carbon-14 in an animal bone still remains 1000 years after the animal has died?


Example (cont)Solution

The half-life is Thalf = 5700 years, so the fraction of the initial amount remaining after t = 1000 years is

For example, if the bone originally contained 1 kilogram of carbon-14, the amount remaining after 1000 years is approximately 0.885 kilogram.

half/ 1000 yr/5700 yr1 1

0.8852 2

t T


ExampleSuppose that 100 pounds of Pu-239 is deposited at a nuclear waste site. How much of it will still be present in 100,000 years?

Solution

The half-life of Pu-239 is Thalf = 24,000 years. Given an initial amount of 100 pounds, the amount remaining after t = 100,000 years is

About 5.6 pounds of the original 100 pounds of Pu-239 will still be present in 100,000 years.

half/ 100,000 yr/24,000 yr1 1

new value = initial value 100 lb 5.6 lb2 2

t T


For a quantity decaying exponentially at a rate of P% per time period, the half-life is approximately

This approximation works best for small decay rates and breaks down for decay rates over about 15%.

half

70T

P

The Approximate Half-Life Formula


ExampleSuppose that inflation causes the value of the Russian ruble to fall at a rate of 12% per year (relative to the dollar). At this rate, approximately how long does it take for the ruble to lose half its value?

Solution

We can use the approximate half-life formula because the decay rate is less than 15%. The 12% decay rate means we set P = 12/yr.

The ruble loses half its value (against the dollar) in 6 yr.

70 705 8half . yr

P 12/ yrT


Exact Doubling Time and Half-Life Formulas

10

10double

log 2 log (1 + )rT

For more precise work, use the exact formulas. These use the fractional growth rate, r = P/100.

For an exponentially growing quantity with a fractional grow rate r, the doubling time is

For a exponentially decaying quantity, in which the fractional decay rate r is negative (r < 0), the half-life is

10

10half

log 2 log (1 + )rT


ExampleA population of rats is growing at a rate of 80% per month. Find the exact doubling time for this growth rate and compare it to the doubling time found with the approximate doubling time formula.

Solution

The growth rate of 80% per month means P = 80/mo or r = 0.8/mo. The doubling time is

or about 1.2 mo.

10

double10 10

log 2 0.301030 0.3010301.18 mo

log 1 0.8 log 1.8 0.255273T

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