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