Question

A water tank in the shape of an inverted right circular cone that has a height of 12 ft and a base radius 6ft. If water is being pumped into the tank at a rate of 10 ft cubed per min , find the rate at which the water level is rising when the level of the water is 3 ft deep

Answer 1

Answer:

$\dfrac{dh}{dt}=1.41\ ft/min$

Explanation:

Given that

h= 12 ft

r= 6 ft

h= 2 r

r=h/2

We know that volume of cone given as

$V=\dfrac{1}{3}\pi r^2h$

Now by putting the values

r=h/2

$V=\dfrac{1}{3}\pi\left(\dfrac{h}{2}\right)^2h$

$V=\dfrac{1}{12}\pi h^3$

Given that when h= 3 ft

$\dfrac{dV}{dt}=10\ ft^3/min$

$V=\dfrac{1}{12}\pi h^3$

$\dfrac{dV}{dt}=\dfrac{\pi}{12}(3h^2)\times \dfrac{dh}{dt}$

By putting the values

$10=\dfrac{\pi}{12}(3\times 3^2)\times \dfrac{dh}{dt}$

$\dfrac{dh}{dt}=1.41\ ft/min$

So the rate at which the water level is rising  is 1.41 ft/min