MCQ
If $\text{V}=\frac{4}{3}\pi\text{r}^3,$ at What rate in cubic units is $V$ increasing when $\text{r}=10\frac{\text{dr}}{\text{dt}}=0.01\ ?$
  • A
    $\pi$
  • $4\pi$
  • C
    $40\pi$
  • D
    $4=\frac{\pi}{3}$

Answer

Correct option: B.
$4\pi$
Given$:\ \text{V}=\frac{4}{3}\pi\text{r}^3,$ and $\frac{\text{dr}}{\text{dt}}=0.01$
$\frac{\text{dV}}{\text{dt}}=4\pi\text{r}^2\frac{\text{dr}}{\text{dt}}$
$\frac{\text{dV}}{\text{dt}}=4\pi(10)^2\times0.01$
$\frac{\text{dV}}{\text{dt}}=4\pi$

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