Answer to Question #148146 in Calculus for Sean

Question

A water tank in the form of an inverted right-circular cone is being emptied at the rate
of 6m^3 / min. The altitude of the cone is 24m, and the base radius is 12m. Find how fast the water level is lowering when the water is 10m deep?

Answer 1

Let V be the volume of the water.

dV/dt=6m³/min

Let h be the height of the water

we want dh/dt

Need to find a relationship between dV/dt and dh/dt

V=1/3 πr²h

If V is only the function of h, insted of h and r, we could take derivatives.

By similar triangles r/h= 12/24

r=h/2

$V=\frac{1}{3}π(\frac{h}{2})^2h=\frac{h^3π}{12}$

$\frac{dV}{dt}=\frac{3πh^2}{12}\frac{dh}{dt}=\frac{πh^2}{4}\frac{dh}{dt}$

Where looking for dh/dt when h=10

$6=\frac{π10^2}{4}\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{24}{100π}=\frac{6}{25π}$

Answer:$\frac{6}{25π}$ m/min