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    IBHardCalculus10–12

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    A sphere is inflated so that its radius increases at 22 cm/s. Find the rate of change of volume when r=5r = 5 cm.

    V=43πr3V = \frac{4}{3}\pi r^{3}. Use implicit differentiation: dVdt=dVdr×drdt\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}.

    Solution

    Step 1 of 3

    1. 1.dVdr=4πr2\frac{dV}{dr} = 4\pi r^{2}