exponential growth and decay. the sequence of numbers: 1 2 4 8 16 32 is an example of exponential...

EXPONENTIAL GROWTH AND DECAY

The sequence of numbers: 1 2 4 8 16 32

is an example of exponential growth.

i.e. the numbers are doubling.

An equation which would generate this sequence is: y = 2x

20

= 1, 21

= 2, 22

= 4, 23

= 8, etc.

Exponential growth arises from a situation in which the rate ofgrowth of a quantity is proportional to the size of the quantity.

i.e. from an equation of the form: = k ydydx

Example 1: The rate of increase of a population is proportional to the size of the population, p, at any time. At the start of the year 2000, the population was 12 million and at the start of 2005 the population was 12.7 million. Find an expression for the population, t years after 2000. Hence find an estimate of the population in 2020. Find also the year in which the population will reach 18 million.

α pdpdt

= k p

dtk dp p1

kt + cAt t = 0

p = 12ln p – ln 12 = kt

ln p12 = kt

p12 = ekt

= kdpdt

1p

dpdt

ln p = kt + ln 12

Integrating both sides of the equation w.r.t. t :

e ktp = 12

Given:

This is a crucial step:. . . (1)

. . . (2)

ln p =c = ln 12}

When t = 5p = 12.7

ln 12.7 12

= 5k

= 15.05

Now we have: ln p12 = kt . . . (1)

e ktp = 12 . . . (2)and

} sub in (1)

k = 0.01134

In 2020, t = 20 sub in (2) p = 12 e0.01134 × 20

So, in 2020, the population will be approximately 15.1 million

Also, when p = 18, sub in (1) = 0.01134 t

Hence the population will reach 18 million in the year 2035.

ln 1812

t ≈ 35.8

Example 2: A radioactive substance decays at a rate, proportional to the mass, m of the substance present at any time. The initialmass of the substance is 6kg. After 50 years the mass of the substanceis half of its initial mass. Find the time taken for the substance to be reduced to one third of its initial mass. Find also the mass of thesubstance after 100 years.

α mdm d t

= – kt + cAt t = 0 m = 6

ln m – ln 6 = – kt

ln m 6 = – kt = e – kt m

6 m = 6 e – kt

ln m

= – kmdm d t

dm d t

1m

= – k

Integrating both sides of the equation w.r.t. t : 1m dtk –dm

ln m = – kt + ln 6

Given:

c = ln 6}

Now we have:

and

ln m 6

= – kt … (1)

m = 6 e – kt . . . (2)

When t = 50

m = 3} sub in (1) ln 1

2= – 50k

( – ln 2 = – 50 k )

k = ln 2 50

After t years: sub in (1): ln 1 3

= – ln 2 50

t

Mass is reduced to one third of its initial mass after approx. 79 years.

When t = 100, sub in (2):

≈ 0. 01386

m = 2 ,

t ≈ 79.25

m = 6 e – 100k m = 1.5kg mass,

Example 2: A radioactive substance decays at a rate, proportional to the mass, m of the substance present at any time. The initialmass of the substance is m0. After 50 years the mass of the substanceis half of its initial mass. Find the time taken for the substance to be reduced to one third of its initial mass. Find also the mass of thesubstance after 100 years.

α mdm d t

= – kt + cAt t = 0 m = m0

ln m – ln m0 = – kt

ln m m0

= – kt = e – kt m

m0

m = m0 e – kt

ln m

= – kmdm d t

dm d t

1m

= – k

Integrating both sides of the equation w.r.t. t :

1m dtk –dm

ln m = – kt + ln m0

Given:

c = ln m0}

A harder version of the previous example:

Now we have:

and

ln m m0

= – kt … (1)

m = m0 e – kt . . . (2)

When t = 50

m = 12

m0

} sub in (1) ln 1 2

= – 50k

( – ln 2 = – 50 k )

k = ln 2 50

After t years: sub in (1): ln 1 3

= – ln 2 50

t

Mass is reduced to one third of its initial mass after approx. 79 years.

When t = 100, sub in (2):

≈ 0. 01386

m = 13

m0 ,

t ≈ 79.25

m = m0 e – 100k m = 14

m0 mass,

Summary of key points:

This PowerPoint produced by R.Collins ; Updated Mar. 2013.

Exponential growth arises from a situation in which the rate ofgrowth of a quantity is proportional to the size of the quantity.

i.e. from an equation of the form: = k ydydx

To solve this, we divide by y, and integrate with respect to x.

Note the following results: c ba ln a1 d

ba1 xxx

= ktln ab

leads to: a = bekt

and: