Question

If $V=\frac{4}{3}\mathrm{\pi }{r}^3$, at what rate in cubic units is V increasing when r = 10 and $\frac{dr}{dt}=0.01$?
(a) π
(b) 4π
(c) 40π
(d) 4π/3

Answer 1

(b) 4π
$$\begin{gathered} \text{Given:}V=\frac{4}{3}\mathrm{\pi }{r}^3,r=10\mathrm{and}\frac{dr}{dt}=0.01 \\ \Rightarrow \frac{dV}{dt}=4\mathrm{\pi }{r}^2\frac{dr}{dt} \\ \Rightarrow \frac{dV}{dt}=4\mathrm{\pi }{\left(10\right)}^2\times 0.01 \\ \Rightarrow \frac{dV}{dt}=4\mathrm{\pi } \end{gathered}$$