Question

Find the rate of change of the volume V=4(pi)(r^3)/3 of a sphere of raius r with respect to its surface area A=4(pi)(r^2) when r=10.

Answer 1

Take the Total Differential of V and A respectively is: \[1) \text{ }\text{Dt}[V]=4 \pi r^2 \text{Dt}[r]\]\[2) \text{ Dt}[A]=8 \pi r \text{Dt}[r] \]Divide the LHS of 1) by the LHS of 2 and the RHS of 1 by the RHS of 2 and simplify.\[\frac{\text{Dt}[V]}{ \text{Dt}[A]}= \frac{1}{2} r\]Replace r in the above expression with 10 and simplify.