Question

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

• A. π
• B. 4 π
• C. 40 π
• D. 4 π/3

Answer 1

The correct option is B

4 π

We have,
$V=\frac{4}{3}\pi r^3\Rightarrow \frac{dv}{dt}=4\pi r^2\frac{dr}{dt}\Rightarrow \frac{dv}{dt}=4\pi \times 100\times 0.01[\because r=10and\frac{dr}{dt}=0.01]\Rightarrow \frac{dv}{dt}=r\pi$
Hence, V is increasing at the rate of $4\pi$ cubic units per unit of time.

Hence (b) is the correct answer.