Linked rates and net-rate models | CAIE 9709 Pure Mathematics 3 Notes

Chapters

Pure Mathematics 3

Mathematics9709

Build the net rate and connect the variables

Net volume rate controls radius rate

40π

→

radius r

out

0.8πr

dV/dt = 40π − 0.8πr

dV/dt = (dV/dr)(dr/dt)

For linked quantities, write the net rate first and then use the chain rule.

\frac{dV}{dt}=\text{rate in}-\text{rate out}

\frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}

Use the sign of each flow before simplifying.
Differentiate the geometric relation with respect to its variable.
Only then isolate the derivative requested by the question.

2024 source anchor: show, separate, integrate

Worked example

A spherical balloon with inflow and leakage

Air enters at rate $40\pi$ and leaves at rate $0.8\pi r$ . Also $V=\frac43\pi r^3$ .

\frac{dV}{dt}=40\pi-0.8\pi r

\frac{dV}{dr}=4\pi r^2

4\pi r^2\frac{dr}{dt}=40\pi-0.8\pi r

Answer

\dfrac{dr}{dt}=\dfrac{50-r}{5r^2}

To solve for t, invert and separate:

\frac{dt}{dr}=\frac{5r^2}{50-r}

\frac{5r^2}{50-r}

=-5r-250+\frac{12500}{50-r}

t=-\frac52r^2-250r

-12500\ln(50-r)+C

The condition $t=0$ when $r=0$ gives $C=12500\ln 50$ .

Answer

t=12500\ln\dfrac{50}{50-r}

\phantom{t={}}-\dfrac52r^2-250r

At $r=12$ , this gives $t=70.4605\ldots$ , so the required value is $70.5$ to 3 significant figures.

(9709/32/O/N/24 Q10(a,c))

Use the equation to predict the motion

For $0<r<50$ , $dr/dt>0$ , so the balloon grows.
At $r=50$ , inflow equals outflow. The radius is an equilibrium value.
The logarithm grows without bound as $r\to50^-$ , so the model approaches 50 but does not reach it in finite time.
The displayed equation for $dr/dt$ is singular at $r=0$ . The starting condition is used at the endpoint of the integrated relation, while the rate equation describes $r>0$ .

Your turnIndependent linked-rate check [3 marks]

A spherical bubble has $V=\frac43\pi r^3$ . Its volume increases at rate $dV/dt=12\pi r$ . Find $dr/dt$ in terms of r.

Show worked answer

4\pi r^2\frac{dr}{dt}=12\pi r

Answer

\dfrac{dr}{dt}=\dfrac3r\quad(r>0)

Build the net rate and connect the variables

Net volume rate controls radius rate

40π

→

radius r

out

0.8πr

dV/dt = 40π − 0.8πr

dV/dt = (dV/dr)(dr/dt)

For linked quantities, write the net rate first and then use the chain rule.

\frac{dV}{dt}=\text{rate in}-\text{rate out}

\frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}

Use the sign of each flow before simplifying.
Differentiate the geometric relation with respect to its variable.
Only then isolate the derivative requested by the question.

2024 source anchor: show, separate, integrate

Worked example

A spherical balloon with inflow and leakage

Air enters at rate $40\pi$ and leaves at rate $0.8\pi r$ . Also $V=\frac43\pi r^3$ .

\frac{dV}{dt}=40\pi-0.8\pi r

\frac{dV}{dr}=4\pi r^2

4\pi r^2\frac{dr}{dt}=40\pi-0.8\pi r

Answer

\dfrac{dr}{dt}=\dfrac{50-r}{5r^2}

To solve for t, invert and separate:

\frac{dt}{dr}=\frac{5r^2}{50-r}

\frac{5r^2}{50-r}

=-5r-250+\frac{12500}{50-r}

t=-\frac52r^2-250r

-12500\ln(50-r)+C

The condition $t=0$ when $r=0$ gives $C=12500\ln 50$ .

Answer

t=12500\ln\dfrac{50}{50-r}

\phantom{t={}}-\dfrac52r^2-250r

At $r=12$ , this gives $t=70.4605\ldots$ , so the required value is $70.5$ to 3 significant figures.

(9709/32/O/N/24 Q10(a,c))

Use the equation to predict the motion

For $0<r<50$ , $dr/dt>0$ , so the balloon grows.
At $r=50$ , inflow equals outflow. The radius is an equilibrium value.
The logarithm grows without bound as $r\to50^-$ , so the model approaches 50 but does not reach it in finite time.
The displayed equation for $dr/dt$ is singular at $r=0$ . The starting condition is used at the endpoint of the integrated relation, while the rate equation describes $r>0$ .

Your turnIndependent linked-rate check [3 marks]

A spherical bubble has $V=\frac43\pi r^3$ . Its volume increases at rate $dV/dt=12\pi r$ . Find $dr/dt$ in terms of r.

Show worked answer

4\pi r^2\frac{dr}{dt}=12\pi r

Answer

\dfrac{dr}{dt}=\dfrac3r\quad(r>0)