math 1210-3 notes of 1/22/18 6.5 exponential growth …pa/1220/c8.pdfbacterial growth...

Math 1210-3

Notes of 1/22/18

6.5 Exponential Growth and Decay

• Base a exponential:

f(x) = Cax where a > 0 and a ̸= 1

= C!

eln a"x

= Cekx where k = ln a

• C = f(0) is the initial value of f .

• k is the rate constant.

• If a > 1 and k > 0 then we have exponentialgrowth.

• If a < 1 and k < 0 then we have exponentialdecay.

• The defining property of exponential growthand decay is that the growth rate is propor-tional to the derivative:

d

dxf(x) =

d

dxCax =

d

dxCekx = kCekx = kf(x).

Math 1210-3 Notes of 1/22/18 page 1

• Another way of putting this is that over atime interval of a fixed length T (a year, aday, a century) the function grows by a fixedmultiplicative factor:

f(t + T ) = Cak(t+T ) = akT Cakt = akT f(t).

Note that akT is constant. It does not de-pend on t. It does, of course, depend on theconstants a and T .

• Yet another way of putting this is that thepercentage growth over a fixed time intervalis constant.

• The doubling time of an exponential func-tion is the (constant) time it takes for thefunction to double.

• In the case of exponential decay that timewould be negative. In the case we speak aboutthe half life of the function, the time it takesfor the function value to get reduced by a fac-tor 1/2.


Examples


Compound Interest

Suppose you invest money at p percent ef-fective annual interest. (Assume some under-lying percentage rate is compounded continu-ously.) What’s the doubling time of your in-vestment?

• Doubling time rounded to nearest integer:

p : 1 2 3 4 5 6 7 8 9 10DT : 70 35 23 18 14 12 10 9 8 7


Bacterial Growth

A population of bacteria starts with a sin-gle bacterium with a mass of 10−10g. (1 poundequals 453g, and one kg equals 103g.) The bac-teria split in two, and the population doubles,every 30 minutes (assuming unlimited resources,of course). How long will it take the populationto reach a mass of

1 pound ?a person (130lbs) ?

the Earth (5.972 × 1024 kg) ?the Sun (2 × 1030 kg) ?

the milky way (2 × 1042 kg) ?the Universe (1053kg) ?

• Think about the numbers before going on!

• Letting t denote time measured in hours, andletting f(t) be the mass of the bacteria at timet, we get

f(t) = 10−10 × 22t = 10−104t.

• We can easily compute the time at which f(t) =m:


t =ln

!

1010m"

ln 4

Substituting the appropriate numbers we get

1 pound 21 hoursa person (130lbs) 24.5 hours

the Earth (5.972 × 1024 kg) 62.7 hoursthe Sun (2 × 1030 kg) 72 hours

the milky way (2 × 1042 kg) 87 hoursthe Universe (1053kg) 110 hours


Logistic Growth

• Of course, exponential growth cannot go onforever (or even for very long)!

• Let y(t) be the size of a population at time t,and let y0 = y(0).

• Suppose the environment has a carrying ca-pacity L that cannot be exceeded.

• The logistic growth equation is

y′ = ky(L − y), y(0) = y0 (1)

• The growth rate is always positive, but as thepopulation approaches the carrying capacitythe growth rate decreases, and the populationnever exceeds the carrying capacity.

• We expect the solution to look like:


• Exercise 34, page 354: Show that the solutionof (1) is

y(t) =Ly0

y0 + (L − y0)e−kLt. (2)

• It’s a great exercise to substitute the solution(2) in (1) and see that the equation is indeedsatisfied.

• But let’s solve the differential equation di-rectly.


6.6 First order linear differential equations

• Start with a piece of black magic, and thenmake sense of it later.

• Suppose we want to solve the differential equa-tion problem (taken from the textbook):

y′ = 2x − 3y, y(0) = 1.


math 1210-3 notes of 1/22/18 6.5 exponential growth …pa/1220/c8.pdfbacterial growth...

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