Question:
If $V=\frac{4}{3} \pi r^{3}$, at what rate in cubic units is $V$ increasing when $r=10$ and $\frac{d r}{d t}=0.01 ?$
(a) $\Pi$
(b) $4 \pi$
(c) $40 \pi$
(d) $4 \pi / 3$
Solution:
(b) $4 \pi$
Given: $V=\frac{4}{3} \pi r^{3}, r=10$ and $\frac{d r}{d t}=0.01$
$\Rightarrow \frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t}$
$\Rightarrow \frac{d V}{d t}=4 \pi(10)^{2} \times 0.01$
$\Rightarrow \frac{d V}{d t}=4 \pi$