Question

9 a conical tank has height 10 ft and radius 5 ft at the top. water flows into the tank at a rate of 2 ft³/min. how fast is the water level rising when it is 4 ft?

Explanation:

Step1: Relate radius and height of water in cone

The ratio of radius to height for the whole cone is $\frac{r}{h}=\frac{5}{10}=\frac{1}{2}$, so $r = \frac{h}{2}$. The volume of a cone of water $V=\frac{1}{3}\pi r^{2}h$. Substituting $r=\frac{h}{2}$ into the volume formula, we get $V=\frac{1}{3}\pi(\frac{h}{2})^{2}h=\frac{1}{12}\pi h^{3}$.

Step2: Differentiate volume with respect to time

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

Step3: Solve for $\frac{dh}{dt}$

We know that $\frac{dV}{dt} = 2$ $ft^{3}/min$ and $h = 4$ ft. Substitute these values into the equation $\frac{dV}{dt}=\frac{\pi}{4}h^{2}\frac{dh}{dt}$. So $2=\frac{\pi}{4}(4)^{2}\frac{dh}{dt}$. First, simplify the right - hand side: $\frac{\pi}{4}(4)^{2}\frac{dh}{dt}=4\pi\frac{dh}{dt}$. Then solve for $\frac{dh}{dt}$: $\frac{dh}{dt}=\frac{2}{4\pi}=\frac{1}{2\pi}$ ft/min.

Answer:

$\frac{1}{2\pi}$ ft/min