Finding Unknown Rates | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Rearrange the chain to find the missing rate

Suppose volume change is known but radius change is wanted. Start with the same chain, then divide by $dV/dr$ :

\begin{aligned}\frac{dV}{dt}&=\frac{dV}{dr}\frac{dr}{dt}\\\frac{dr}{dt}&=\frac{dV/dt}{dV/dr}\end{aligned}

This division requires $dV/dr\ne0$ . Equivalently, the inverse derivative is $dr/dV=1/(dV/dr)$ wherever the original derivative is non-zero and different nearby radii give different volumes. Do not multiply by $dV/dr$ when you need its reciprocal.

Worked example

A spherical balloon has dV/dt = 12π cm³/s. Find dr/dt when V = 36π cm³.

MF19 gives the sphere’s volume as $V=\tfrac43\pi r^3$ . First use the given volume to locate the instant:

\tfrac43\pi r^3=36\pi\quad\Rightarrow\quad r=3

Differentiate the general volume formula: $dV/dr=4\pi r^2$ . At this instant it equals $36\pi$ , so

\frac{dr}{dt}=\frac{12\pi}{36\pi}=\frac13\text{ cm/s}

Check: multiplying $36\pi\text{ cm}^2$ by $1/3\text{ cm/s}$ recovers the given volume rate.

Radius, diameter and area are different quantities

A sphere’s diameter is $D=2r$ , so $dD/dt=2\,dr/dt$ . A given diameter rate must be halved before using it as a radius rate.

MF19 also gives its surface area: $A=4\pi r^2$ , not the circle formula $\pi r^2$ . To connect volume rate to surface-area rate, first find $dr/dt$ , then use $dA/dt=8\pi r\,dr/dt$ . For the balloon above, this gives $8\pi\text{ cm}^2/\text{s}$ .

Your turnPractice

A spherical object loses volume at $6\pi\text{ cm}^3/\text{s}$ . When its radius is 3 cm, find the signed rates of change of radius, diameter and surface area.

Show worked answer

Loss means $dV/dt=-6\pi$ . Thus $dr/dt=-6\pi/(36\pi)=-1/6$ cm/s. The diameter rate is $-1/3$ cm/s. Finally $dA/dt=8\pi(3)(-1/6)=-4\pi$ cm²/s.

Common mistake

“Decreases at 5 units/s” gives a signed derivative of −5 units/s. Report a positive magnitude only when explicitly describing the rate of decrease.