A cone whose height is always equal to its diameter is increasing in volume at the rate of 40 cm3/sec. At what rate is the radius increasing when its circular base area is 1 m2?

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KollegeApply · Accepted Answer

0.002 cm/sec \text { Let r be the radius,h be the height and V be the volume of the cone at any time t. Then }, V = $\frac{1}{3}$$\pi$ r^2 h \Rightarrow V = $\frac{2}{3}$$\pi$ r^3 \left[ \because h = 2r \right] \Rightarrow dV/dt = 2 $\times$ {10}^4 dr/dt \left[ \because $\pi$ r^2 = 1 m^2 \text { or } {10}^4 {cm}^2 \right] \Rightarrow dr/dt = \frac{1}{2 $\times$ {10}^4}dV/dt \Rightarrow dr/dt = \frac{40}{2 $\times$ {10}^4} \Rightarrow dr/dt = 0 . 002 \ cm/\sec

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