Question

A water tank has the shape of an inverted right circular cone with base radius 3 meters and height 6 meters. Water is being pumped into the tank at the rate of 12 meters3/sec. Find the rate, in meters/sec, at which the water level is rising when the water is 2 meters deep. Give 2 decimal places for your answer. Type your answer in the space below. If your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35).

Answer 1

Answer:

the rate of change is 3.8m/s

Step-by-step explanation:

The volume of the right circular cone is

$V= \frac{\pi*r^2h}{3}.$

The rate of change of volume is

$\frac{dV}{dt} =\frac{d}{dt}( \frac{\pi*r^2h}{3})=\frac{\pi}{3} \frac{d}{dt}(r^2h)=\frac{\pi}{3}(2rh\frac{dr}{dt}+r^2\frac{dh}{dt})$

In order to proceed further we have to define r in terms of h so tht we can compute the derivative above.

The ratio between h and r is

$\frac{h}{r}=\frac{6}{3} =2$

Therefore $r=\frac{h}{2}$.

We plug that into the derivative above and get:

$2rh\frac{dr}{dt}+r^2\frac{dh}{dt}=\frac{h^2}{2}\frac{dh}{dt}+\frac{h^2}{4}\frac{dh}{dt}=\frac{3h^2}{4}\frac{dh}{dt}.$

Thus

$\frac{dV}{dt}=\frac{\pi}{3}\frac{3h^2}{4}\frac{dh}{dt}$

Now for the numerical part.

The rate of change of volume $\frac{dV}{dt}$ is $12\frac{m^3}{s}$, so when the water is 2 meters deep $h=2$, therefore:

$12=\frac{\pi*3*2^2}{4*3} \frac{dh}{dt} \\\\\therefore \boxed{ \frac{dh}{dt}=3.8m/s.}$