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If v=43πr3v = \frac{4}{3} \pi r^3, at what rate is cubic / unit sec is increasing when r=10r = 10, and drdt=0.01\frac{dr}{dt} = 0.01 ?

Solution

✅ Correct Option: 2

dVdt=4πr2drdt=4π(10)2(0.01)=4π\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} = 4 \pi (10)^2 (0.01) = 4\pi cubic units per second.

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