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

Step2: Differentiate with respect to time $t$

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

Step3: Substitute known values

We know that $\frac{dV}{dt}=2\mathrm{ft}^{3}/\mathrm{min}$ and $h = 4\mathrm{ft}$. Substitute these values into the equation $\frac{dV}{dt}=\frac{\pi}{4}h^{2}\frac{dh}{dt}$. So, $2=\frac{\pi}{4}(4)^{2}\frac{dh}{dt}$.

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

First, simplify the right - hand side of the equation: $\frac{\pi}{4}(4)^{2}\frac{dh}{dt}=4\pi\frac{dh}{dt}$. Then, solve for $\frac{dh}{dt}$ from $2 = 4\pi\frac{dh}{dt}$. We get $\frac{dh}{dt}=\frac{2}{4\pi}=\frac{1}{2\pi}\mathrm{ft}/\mathrm{min}$.

Answer:

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