If $V=\frac{4}{3} \pi r^3$, at what rate in cubic units is V increasing when r = 10 and $\frac{d r}{d t}=0.01$ ?

$4 \pi$

We have,

$V=\frac{4}{3} \pi r^3$

$\Rightarrow \frac{d V}{d t}=4 \pi r^2 \frac{d r}{d t}$

$\Rightarrow \frac{d V}{d t}=4 \pi \times 100 \times 0.01$        [∵ r = 10 and $\frac{d r}{d t}$ = 0.01]

$\Rightarrow \frac{d V}{d t}=4 \pi$

Hence, V is increasing at the rate of $4 \pi$ cubic units per unit of time.