A cone whose height is always equal to its diameter is increasing in volume at the rate of 40 $\displaystyle\frac{cm^3}{sec}$. At the what rate is the radius increasing when its circular base area is 1 $m^2$?

Using the cone volume formula V = (1/3) * pi * r squared * h, and given that the height equals the diameter (h = 2r), the volume simplifies to V = (2/3) * pi * r cubed. Differentiating with respect to time gives dV/dt = 2 * pi * r squared * dr/dt. The circular base area is 1 m squared, which equals 10000 cm squared, meaning pi * r squared = 10000. We are given dV/dt = 40 cm cubed/sec; substituting these values gives 40 = 2 * 10000 * dr/dt. Solving this gives dr/dt = 0.002 cm/sec.

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