Exponential growth and decay

Warm-up

Three quick questions from earlier lessons.

Discover

The blue curve is $y = a\,e^{kt}$ , a solution of $y' = ky$ with $y(0) = a$ . Move $k$ through 0.

Solutions y = a e^(kt) of y' = ky. For k > 0 they grow faster and faster; for k < 0 they decay towards 0; k = 0 gives a constant.

For $k < 0$ , what happens to $y$ ?

It decays towards 0
It falls in a straight line and crosses 0

Key idea

Rate proportional to amount means exponential.

\dfrac{dy}{dt} = ky

with

y(0) = y_0

has the solution

y = y_0 e^{kt}

.

k > 0

: growth, doubling time

\tfrac{\ln 2}{k}

.

k < 0

: decay, half-life

\tfrac{\ln 2}{|k|}

.

Worked example

Bacteria double every 3 hours, starting from 500. How many after 12 hours?

500 \cdot 2^4

12 hours = 4 doubling times.

= 8000

Compute.

y = 500e^{(\ln 2/3)t}

Equivalently k = ln 2 / 3.

Your turn, guided

Your turn: half-life 5 years, find $k$ .

k = -\dfrac{\ln 2}{5}

e^{5k} = ½.

Practice ★

A population grows with $\dfrac{dP}{dt} = kP$ . It starts at 104 and doubles every 9 years. What is it after 27 years?
A population grows with $\dfrac{dP}{dt} = kP$ . It starts at 469 and doubles every 4 years. What is it after 12 years?

Solve $\dfrac{dy}{dt} = -2y$ with $y(0) = 5$ .
Solve $\dfrac{dy}{dt} = 4y$ with $y(0) = 8$ .

Watch out!

Practice ★★

A substance decays with $\dfrac{dm}{dt} = km$ and has a half-life of 27 years. Find $k$ (exact).
A substance decays with $\dfrac{dm}{dt} = km$ and has a half-life of 25 years. Find $k$ (exact).

Exit ticket

Three questions to check you have got it.

Solve $\dfrac{dy}{dt} = -2y$ with $y(0) = 5$ .
Solve $\dfrac{dy}{dt} = 4y$ with $y(0) = 8$ .
Solve $\dfrac{dy}{dt} = 3y$ with $y(0) = 4$ .

Summary card

Exponential models

$y' = ky$ , $y(0) = y_0 \Rightarrow y = y_0e^{kt}$ .
Decay: $k < 0$ ; $k = -\ln 2 / (\text{half-life})$ .

Watch out: equal times multiply by equal factors.

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