Unit 4, lesson 42

Exponential Growth and Decay

Why a quantity that changes at a rate proportional to its size must be an exponential function, and how that one fact models populations, radioactive decay, cooling and interest.

  • Theory part 83
  • Practice part 84
  • Modelling and Applications

Theory, part 83

Practice, part 84

A colony of bacteria with twice as many cells produces twice as many new cells per hour. A lump of radioactive material with half as many atoms loses half as many atoms per year. In both cases the rate of change is proportional to the amount present. That single rule pins the quantity down completely, and the answer is always an exponential function.

Key terms

Constant relative growth rate

Definition For a positive differentiable quantity y(t)y(t), the relative growth rate at time tt is

1ydydt=y′(t)y(t).\frac{1}{y}\frac{dy}{dt} = \frac{y'(t)}{y(t)}.

The quantity has constant relative growth rate kk on an interval II if y′(t)=k y(t)y'(t) = k\,y(t) for every tt in II.

In plain words The relative growth rate is growth per unit of what is already there. A rate of 0.030.03 per year means growing by about 33 percent of the current size each year, at every moment.

Example If y(t)=7e0.03ty(t) = 7e^{0.03t}, then y′(t)=0.21e0.03ty'(t) = 0.21e^{0.03t}, so

y′(t)y(t)=0.21e0.03t7e0.03t=0.03\frac{y'(t)}{y(t)} = \frac{0.21e^{0.03t}}{7e^{0.03t}} = 0.03

for every tt. If instead y(t)=t2y(t) = t^2 for t>0t > 0, the relative rate is

2tt2=2t,\frac{2t}{t^2} = \frac{2}{t},

which is not constant.

The law of natural growth and decay

Definition The differential equation

dydt=ky,\frac{dy}{dt} = ky,

with a constant kk, is the law of natural growth when k>0k > 0 and of natural decay when k<0k < 0. On an interval containing 00, the differentiable solutions are exactly the functions

y(t)=y(0)ekt.y(t) = y(0)e^{kt}.

To see that there are no others, note that every solution satisfies

ddt(y(t)e−kt)=(y′(t)−ky(t))e−kt=0,\frac{d}{dt}\bigl(y(t)e^{-kt}\bigr) = \bigl(y'(t) - ky(t)\bigr)e^{-kt} = 0,

so y(t)e−kty(t)e^{-kt} is constant on the interval and equals y(0)y(0).

In plain words If the rate of change is always the same multiple of the amount, then the amount is its starting value times ekte^{kt}. Nothing else fits.

Example If y′=−0.2yy' = -0.2y and y(0)=50y(0) = 50, then y(t)=50e−0.2ty(t) = 50e^{-0.2t}. At t=5t = 5 the amount is 50e−1≈18.450e^{-1} \approx 18.4.

Population growth

Definition A population P(t)P(t) with constant relative growth rate k>0k > 0 satisfies P′=kPP' = kP, so P(t)=P0ektP(t) = P_0 e^{kt} where P0=P(0)P_0 = P(0). Two measurements at different times determine both P0P_0 and kk. The model fits as long as food and space do not limit growth.

In plain words Find kk from the data, then use P0ektP_0 e^{kt} to look forward or back.

Example A culture has 20002000 bacteria at t=0t = 0 and 60006000 at t=2t = 2 hours. Then 2000e2k=60002000e^{2k} = 6000, so e2k=3e^{2k} = 3 and

k=ln⁡32≈0.549k = \frac{\ln 3}{2} \approx 0.549

per hour. After 55 hours,

P(5)=2000e5k=2000⋅35/2≈31 177 bacteria.P(5) = 2000e^{5k} = 2000 \cdot 3^{5/2} \approx 31\,177 \text{ bacteria}.

Radioactive decay and half-life

Definition The mass m(t)m(t) of a radioactive substance satisfies m′=−rmm' = -rm for a constant r>0r > 0, so m(t)=m0e−rtm(t) = m_0 e^{-rt}. The half-life TT is the time needed for any given amount to fall to half of itself. From e−rT=12e^{-rT} = \frac{1}{2},

T=ln⁡2r,m(t)=m0⋅2−t/T.T = \frac{\ln 2}{r}, \qquad m(t) = m_0 \cdot 2^{-t/T}.

The half-life does not depend on the starting amount or the starting time.

In plain words Every half-life, whatever is left is cut in half.

Example Carbon-14 has a half-life of about 57305730 years. Of 1010 grams, after 20002000 years there remain 10⋅2−2000/5730≈7.8510 \cdot 2^{-2000/5730} \approx 7.85 grams. To fall to 22 grams takes tt years with 2−t/5730=152^{-t/5730} = \frac{1}{5}, so t=5730log⁡25≈13 300t = 5730\log_2 5 \approx 13\,300 years.

Newton’s law of cooling

Definition Newton’s law of cooling says that the temperature T(t)T(t) of an object in surroundings held at constant temperature TsT_s changes at a rate proportional to the difference,

dTdt=k(T−Ts),k<0.\frac{dT}{dt} = k(T - T_s), \qquad k < 0.

Applying the law of natural decay to y=T−Tsy = T - T_s gives

T(t)=Ts+(T0−Ts)ekt,T0=T(0).T(t) = T_s + (T_0 - T_s)e^{kt}, \qquad T_0 = T(0).

In plain words It is the difference from the room temperature that decays exponentially, not the temperature itself.

Example Coffee at 9090 degrees Celsius is left in a room at 2020 degrees, and after 1010 minutes it is at 6060 degrees. Then 70e10k=4070e^{10k} = 40, so

k=110ln⁡47≈−0.0560k = \frac{1}{10}\ln\frac{4}{7} \approx -0.0560

per minute. It reaches 4040 degrees when 70ekt=2070e^{kt} = 20, that is

t=ln⁡(2/7)k=10 ln⁡(2/7)ln⁡(4/7)≈22.4 minutes.t = \frac{\ln(2/7)}{k} = 10\,\frac{\ln(2/7)}{\ln(4/7)} \approx 22.4 \text{ minutes}.

Continuously compounded interest

Definition An amount A0A_0 invested at annual interest rate rr compounded nn times a year grows after tt years to

A0(1+rn)nt.A_0\left(1 + \frac{r}{n}\right)^{nt}.

For fixed tt, as n→∞n \to \infty this tends to

A(t)=A0ert,A(t) = A_0 e^{rt},

which is called continuous compounding. It satisfies A′(t)=rA(t)A'(t) = rA(t), the law of natural growth with k=rk = r.

In plain words Compounding more and more often approaches a limit, and that limit is exponential growth at the interest rate.

Example Put 10001000 dollars at 55 percent for 1010 years.

CompoundingAmount after 10 years
yearly1000(1.05)10≈1628.891000(1.05)^{10} \approx 1628.89
monthly1000(1+0.0512)120≈1647.011000\left(1 + \frac{0.05}{12}\right)^{120} \approx 1647.01
continuous1000e0.5≈1648.721000e^{0.5} \approx 1648.72

Checking a model by differentiation

Definition A function yy solves the initial value problem y′=F(t,y)y' = F(t, y), y(t0)=y0y(t_0) = y_0 on an interval II containing t0t_0 if yy is differentiable on II, y′(t)=F(t,y(t))y'(t) = F(t, y(t)) for every tt in II, and y(t0)=y0y(t_0) = y_0. Checking a proposed answer means verifying both conditions directly.

In plain words Differentiate your answer, put it into the equation, and check the starting value. It takes a minute and catches most mistakes.

Example For the coffee, T(t)=20+70ektT(t) = 20 + 70e^{kt} gives

T′(t)=70kekt=k(T(t)−20)T'(t) = 70ke^{kt} = k\bigl(T(t) - 20\bigr)

and T(0)=90T(0) = 90, so it checks. A common slip is T(t)=90ektT(t) = 90e^{kt}. Its derivative is kTkT, not k(T−20)k(T - 20), so it fails the equation and would wrongly predict the coffee cooling toward 00 degrees.

Check yourself

  1. A substance has a half-life of 88 days. What fraction of it is left after 2424 days?
Show the answer

Since 2424 days is 33 half-lives, the fraction left is (1/2)3=1/8(1/2)^3 = 1/8.

  1. A population doubles every 33 hours. What is its relative growth rate?
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From e3k=2e^{3k} = 2 we get

k=ln⁡23≈0.231k = \frac{\ln 2}{3} \approx 0.231

per hour.

  1. Check that y(t)=5e0.3ty(t) = 5e^{0.3t} solves y′=0.3yy' = 0.3y with y(0)=5y(0) = 5.
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The derivative is

y′(t)=1.5e0.3t=0.3⋅5e0.3t=0.3 y(t),y'(t) = 1.5e^{0.3t} = 0.3 \cdot 5e^{0.3t} = 0.3\,y(t),

and y(0)=5e0=5y(0) = 5e^0 = 5. Both conditions hold.

How this lesson connects

It also uses elementary separable equations.