If V=43πr3, at what rate in cubic units is V increasing when r=10 and drdt=0.01?

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

A
$π$
B
$4 π$
C
$40 π$
D
$\frac{4 π}{3}$

Solution:

$\begin{array}{l} V = \frac{4}{3} π r^{3} \\ \Rightarrow \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} \end{array}$

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

$\Rightarrow \frac{d V}{d t} = 4 π$

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

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