Connected Rates of Change: Expanding Circle

Connected rates use the chain rule to link two changing quantities through a shared variable: $\frac{dA}{dt} = \frac{dA}{dr}\times\frac{dr}{dt}.$ Differentiate the area formula for $\frac{dA}{dr}$, insert the given $\frac{dr}{dt}$, and evaluate.

The radius of a circle is increasing at a constant rate of $0.2$ cm/s. Find the rate at which its area is increasing at the instant when the radius is $5$ cm. [5]

The working

Step 1, area formula and its derivative: $A = \pi r^2 \implies \frac{dA}{dr} = 2\pi r \quad \text{(M1)}$

Step 2, note the given rate: $\frac{dr}{dt} = 0.2\ \text{cm/s} \quad \text{(B1)}$

Step 3, apply the chain rule: $\frac{dA}{dt} = \frac{dA}{dr}\times\frac{dr}{dt} = 2\pi r \times 0.2 \quad \text{(M1)}$

Step 4, evaluate at $r = 5$: $\frac{dA}{dt} = 2\pi(5)(0.2) = 2\pi \approx 6.28\ \text{cm}^2/\text{s} \quad \text{(A1, A1)}$

Where the marks are won and lost

• Set up the chain rule with the right derivatives: you have $\frac{dr}{dt}$ and want $\frac{dA}{dt}$, so the bridge is $\frac{dA}{dr}$.
• Evaluate $\frac{dA}{dr} = 2\pi r$ at the given radius ($r = 5$), not in general.
• Keep the units: the answer is a rate of area, cm$^2$/s.

Common mistakes

• Multiplying by the radius instead of by $\frac{dr}{dt}$.
• Forgetting to substitute $r = 5$ and leaving the answer in terms of $r$.
• Using circumference $2\pi r$ as the area, or $\pi r^2$ as its own derivative.

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