If $V=\frac{4}{3} \pi r^3$, at what rate in cubic units is V increasing when r = 10 and $\frac{d r}{d t}=0.01$ ?
Answer & explanation
Correct answer: option 2
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.