Question

question: an inverted cone is partially filled with water. the radius of the inverted cone is 16 mm. the height of the inverted cone is 18 mm. if the volume of the water is increasing at a rate of 548 cubic mm per hour, what is the rate, in mm per hour, at which the height of the water is changing when the height of the water is 2 mm? remember that the volume of a cone is $v=\frac{1}{3}pi r^{2}h$. provide your answer below.

Explanation:

Step1: Relate radius and height

For similar - cones, $\frac{r}{h}=\frac{16}{18}=\frac{8}{9}$, so $r = \frac{8}{9}h$.

Step2: Substitute into volume formula

$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi(\frac{8}{9}h)^{2}h=\frac{64\pi}{243}h^{3}$.

Step3: Differentiate with respect to time

$\frac{dV}{dt}=\frac{64\pi}{81}h^{2}\frac{dh}{dt}$.

Step4: Solve for $\frac{dh}{dt}$

Given $\frac{dV}{dt}=548$ and $h = 2$, then $548=\frac{64\pi}{81}(2)^{2}\frac{dh}{dt}$.
$\frac{dh}{dt}=\frac{548\times81}{64\pi\times4}\approx54$.

Answer:

54