Question

an inverted conical water tank with a height of 6 ft and a radius of 3 ft is drained through a hole in the vertex at a rate of 6 ft³/s (see figure). what is the rate of change of the water depth when the water depth is 2 ft? (hint: use similar triangles.) when the water depth is 2 ft, the rate of change of the water depth is about (round to the nearest hundredth as needed.)

Explanation:

Step1: Relate radius and height of water

For similar - triangles in the cone, the ratio of radius to height of the whole cone is $\frac{r}{h}=\frac{3}{6}=\frac{1}{2}$, so $r = \frac{h}{2}$.

Step2: Find volume formula of water in cone

The volume of a cone is $V=\frac{1}{3}\pi r^{2}h$. Substitute $r=\frac{h}{2}$ into the volume formula, we get $V=\frac{1}{3}\pi(\frac{h}{2})^{2}h=\frac{1}{12}\pi h^{3}$.

Step3: Differentiate volume with respect to time

Differentiate $V=\frac{1}{12}\pi h^{3}$ with respect to time $t$ using the chain - rule. $\frac{dV}{dt}=\frac{1}{12}\pi\times3h^{2}\frac{dh}{dt}=\frac{\pi}{4}h^{2}\frac{dh}{dt}$.

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

We know that $\frac{dV}{dt}=- 6$ (negative because the volume is decreasing). When $h = 2$, substitute $\frac{dV}{dt}$ and $h$ into the equation $\frac{dV}{dt}=\frac{\pi}{4}h^{2}\frac{dh}{dt}$.
$-6=\frac{\pi}{4}(2)^{2}\frac{dh}{dt}$.
$-6=\pi\frac{dh}{dt}$.
$\frac{dh}{dt}=-\frac{6}{\pi}\approx - 1.91$ ft/s.

Answer:

$-1.91$ ft/s