Growth, decay and limiting values | CAIE 9709 Pure Mathematics 3 Notes

Pure Mathematics 3

Mathematics9709

Rate proportional to the current amount

\frac{dQ}{dt}=kQ

Q=Q_0e^{kt}

\frac{dQ}{dt}=-kQ

Q=Q_0e^{-kt}

Here $Q_0$ means the initial value $Q(0)$ and $k>0$ . The plus sign gives exponential growth; the minus sign gives exponential decay. The rate changes because it is proportional to the current amount.

Definition

What is meant by equilibrium solution?

Model answer: A constant solution for which the derivative is zero. If the quantity starts at an equilibrium, the model keeps it there.

For both equations above, $Q=0$ is an equilibrium. It would be lost if the equation were divided by Q without first checking $Q=0$ .

A difference from a limit sets the direction

\frac{dQ}{dt}=k(M-Q),\qquad k>0

$M$ is a fixed limiting value. Below M the derivative is positive; above M it is negative; at M it is zero. Change the initial value and k below to see all three cases.

Initial value Q(0)2

Positive constant k0.5

Read at time t3.0

Q(3) = 6.661

dQ/dt = 0.669

increasing toward 8

Exact solution: Q = 8 + (2 − 8)e^(−0.5t)

For dQ/dt = k(8 − Q) with k > 0, the sign of 8 − Q sets the direction and every solution approaches 8.

Q=M+(Q_0-M)e^{-kt}

The exponential factor tends to zero as $t\to\infty$ , so every solution in this model approaches M. It reaches M at a finite time only when it starts at the equilibrium $Q_0=M$ .

A capacity model has two equilibrium values

In $dP/dt=kP(M-P)$ , growth needs both a present population P and remaining capacity $M-P$ . The two equilibrium solutions are $P=0$ and $P=M$ .

Worked example

Solve a bounded-growth model

Solve $dP/dt=\frac1{50}P(10-P)$ , given that $P=2$ when $t=0$ .

\frac{1}{P(10-P)}\,dP=\frac1{50}\,dt

\frac{1}{P(10-P)}

=\frac1{10}\left(\frac1P+\frac1{10-P}\right)

\ln\left(\frac{P}{10-P}\right)=\frac{t}{5}+C

The condition gives $e^C=1/4$ . Solving for P gives:

Answer

P=\dfrac{10}{1+4e^{-t/5}}

This particular solution starts between the equilibria and tends to 10. The complete general solution also includes the constant branches $P=0$ and $P=10$ .

Your turnIndependent bounded-growth check [8 marks]

Solve $dP/dt=\frac1{20}P(6-P)$ , given that $P=2$ when $t=0$ . Give P in terms of t and state its long-term value.

Show worked answer

\frac{1}{P(6-P)}

=\frac16\left(\frac1P+\frac1{6-P}\right)

\ln\left(\frac{P}{6-P}\right)=\frac{3t}{10}+C

At $t=0$ , $P/(6-P)=1/2$ .

Answer

P=\dfrac{6}{1+2e^{-3t/10}}

P\to6\text{ as }t\to\infty

Rate proportional to the current amount

\frac{dQ}{dt}=kQ

Q=Q_0e^{kt}

\frac{dQ}{dt}=-kQ

Q=Q_0e^{-kt}

Definition

What is meant by equilibrium solution?

Model answer: A constant solution for which the derivative is zero. If the quantity starts at an equilibrium, the model keeps it there.

For both equations above, $Q=0$ is an equilibrium. It would be lost if the equation were divided by Q without first checking $Q=0$ .

A difference from a limit sets the direction

\frac{dQ}{dt}=k(M-Q),\qquad k>0

$M$ is a fixed limiting value. Below M the derivative is positive; above M it is negative; at M it is zero. Change the initial value and k below to see all three cases.

Initial value Q(0)2

Positive constant k0.5

Read at time t3.0

Q(3) = 6.661

dQ/dt = 0.669

increasing toward 8

Exact solution: Q = 8 + (2 − 8)e^(−0.5t)

For dQ/dt = k(8 − Q) with k > 0, the sign of 8 − Q sets the direction and every solution approaches 8.

Q=M+(Q_0-M)e^{-kt}

The exponential factor tends to zero as $t\to\infty$ , so every solution in this model approaches M. It reaches M at a finite time only when it starts at the equilibrium $Q_0=M$ .

A capacity model has two equilibrium values

In $dP/dt=kP(M-P)$ , growth needs both a present population P and remaining capacity $M-P$ . The two equilibrium solutions are $P=0$ and $P=M$ .

Worked example

Solve a bounded-growth model

Solve $dP/dt=\frac1{50}P(10-P)$ , given that $P=2$ when $t=0$ .

\frac{1}{P(10-P)}\,dP=\frac1{50}\,dt

\frac{1}{P(10-P)}

=\frac1{10}\left(\frac1P+\frac1{10-P}\right)

\ln\left(\frac{P}{10-P}\right)=\frac{t}{5}+C

The condition gives $e^C=1/4$ . Solving for P gives:

Answer

P=\dfrac{10}{1+4e^{-t/5}}

This particular solution starts between the equilibria and tends to 10. The complete general solution also includes the constant branches $P=0$ and $P=10$ .

Your turnIndependent bounded-growth check [8 marks]

Solve $dP/dt=\frac1{20}P(6-P)$ , given that $P=2$ when $t=0$ . Give P in terms of t and state its long-term value.

Show worked answer

\frac{1}{P(6-P)}

=\frac16\left(\frac1P+\frac1{6-P}\right)

\ln\left(\frac{P}{6-P}\right)=\frac{3t}{10}+C

At $t=0$ , $P/(6-P)=1/2$ .

Answer

P=\dfrac{6}{1+2e^{-3t/10}}

P\to6\text{ as }t\to\infty