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JAMB 2002 · UME · Q47

A circle with a radius of 5 cm has its radius increasing at the rate of 0.2 cm s−10.2\text{ cm s}^{-1}. What will be the corresponding rate of increase in the area?

Worked solution (try it first)
  1. The area is A=πr2A = \pi r^2, so dAdr=2πr\frac{dA}{dr} = 2\pi r.
  2. Chain rule: dAdt=dAdr×drdt\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}
    =2π×5×0.2= 2\pi \times 5 \times 0.2.
  3. That is 2π cm2s−12\pi\text{ cm}^2\text{s}^{-1}, option C.

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