Question

A filter filled with liquid is in the shape of a vertex-down cone with a height of 9 inches and a diameter of 6 inches at its open (upper) end. If the liquid drips out the bottom of the filter at the constant rate of 4 cubic inches per second, how fast is the level of the liquid dropping when the liquid is 2 inches deep?

Answer 1

Answer: Level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.

Step-by-step explanation:

Since we have given that

Height = 9 inches

Diameter = 6 inches

Radius = 3 inches

So, $\dfrac{r}{h}=\dfrac{3}{9}=\dfrac{1}{3}\\\\r=\dfrac{1}{3}h$

Volume of cone is given by

$V=\dfrac{1}{3}\pi r^2h\\\\V=\dfrac{1}{3}\pi \dfrac{1}{9}h^2\times h\\\\V=\dfrac{1}{27}\pi h^3$

By differentiating with respect to time t, we get that

$\dfrac{dv}{dt}=\dfrac{1}{27}\pi \times 3\times h^2\dfrac{dh}{dt}=\dfrac{1}{9}\pi h^2\dfrac{dh}{dt}$

Now,  the liquid drips out the bottom of the filter at the constant rate of 4 cubic inches per second, ie $\dfrac{dv}{dt}=-4\ in^3$

and h = 2 inches deep.

$-4=\dfrac{1}{9}\times \pi\times (2)^2\dfrac{dh}{dt}\\\\-9\pi =\dfrac{dh}{dt}\\\\-28.28=\dfrac{dh}{dt}$

Hence, level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.