Question

PLEASE SHOW WORK

Answer 1

(C)

Step-by-step explanation:

The volume of the conical pile is given by

$V = \dfrac{\pi}{3}r^2h$

Taking the derivative of V with respect to time, we get

$\dfrac{dV}{dt} = \dfrac{\pi}{3}\dfrac{d}{dt}(r^2h)$

$\:\:\:\:\:\:\:= \dfrac{\pi}{3}\left(2rh\dfrac{dr}{dt} + r^2\dfrac{dh}{dt}\right)$

Since r is always equal to h, we can set

$\dfrac{dr}{dt} = \dfrac{dh}{dt}$

so that our expression for dV/dt becomes

$\dfrac{dV}{dt} = \dfrac{\pi}{3}\left(3r^2\dfrac{dh}{dt}\right)$

$\:\:\:\:\:\:\:= \pi r^2\dfrac{dh}{dt}$

Solving for dh/dt, we get

$\dfrac{dh}{dt} = \dfrac{1}{\pi r^2}\dfrac{dV}{dt}$

$\:\:\:\:\:\:\:= \dfrac{1}{9\pi\:\text{m}^2}(36\:\text{m}^3\text{/s})$

$\:\:\:\:\:\:\:= \dfrac{4}{\pi}\:\text{m/s}$