Connected Rates of Change: Expanding Sphere

Connected rates is the calculus topic students most often name as “the one I panic on”, yet it is one of the most mechanical once you write the chain out. The whole method is: identify the derivatives, link them with the chain rule, substitute last.

The volume of a sphere is increasing at a constant rate of $50\ \text{cm}^3\,\text{s}^{-1}$. Find the rate at which the radius is increasing at the instant when the radius is $5\ \text{cm}$. [The volume of a sphere is $V = \frac{4}{3}\pi r^3$.] [5]

The working

Step 1, write down what you have and what you want. The word “rate” with “per second” means with respect to time: $\text{Given: } \frac{dV}{dt} = 50 \qquad \text{Want: } \frac{dr}{dt} \text{ when } r = 5$

Step 2, differentiate the volume formula to get the link between $V$ and $r$: $\frac{dV}{dr} = 4\pi r^2 \quad \text{(M1, A1)}$

Step 3, build the chain. The rate you want equals the rate you have, adjusted by the link: $\frac{dr}{dt} = \frac{dr}{dV} \times \frac{dV}{dt} = \frac{1}{\,4\pi r^2\,} \times 50 \quad \text{(M1)}$

Step 4, substitute $r = 5$ last: $\frac{dr}{dt} = \frac{50}{4\pi (5)^2} = \frac{50}{100\pi} = \frac{1}{2\pi} \approx 0.159\ \text{cm s}^{-1} \quad \text{(A1)}$

Where the marks are won and lost

• Substituting $r = 5$ before forming the chain is the classic error. Keep $r$ as a symbol until the final line, or you will differentiate a constant and get zero.
• Units and a decimal both matter on Paper 2. $0.159\ \text{cm s}^{-1}$ (3 s.f.) is the expected form. An exact $\frac{1}{2\pi}$ is also accepted, but a bare decimal with no units risks the final mark.
• The chain must be dimensionally sensible: you want $\frac{dr}{dt}$, so the derivatives on the right have to “cancel” to leave $\frac{dr}{dt}$. If they don’t, you have inverted a term.

Common mistakes

• Writing $\frac{dr}{dt} = \frac{dV}{dr} \times \frac{dV}{dt}$ (using $\frac{dV}{dr}$ instead of its reciprocal). It should be $\frac{1}{4\pi r^2}$, not $4\pi r^2$.
• Differentiating $V = \frac43 \pi r^3$ incorrectly, the $\frac43$ and the $3$ cancel to give exactly $4\pi r^2$.
• Rounding $\pi$ too early and losing the accuracy mark.

Method and more examples: Rates of Change notes. Whole topic: Calculus pillar.