# Thread: Related Rates problem...

1. ## Related Rates problem...

A filter filled with liquid is in the shape of a vertex-down cone with a height of 16 inches and a diameter of 24 inches at its open (upper) end. If the liquid drips out the bottom of the filter at the constant rate of 7 cubic inches per second, how fast is the level of the liquid dropping when the liquid is 4 inches deep?

I know that the volume of a cone is $\frac{1}{3} \pi r^2 h$.

From earlier examples I know that I have to somehow rewrite this equation using only one variable, and this is where I always get stuck with these problems. The book gives a hint that the equation can be rewritten
with respect to either the height or radius because the height and radius of the cone is fixed.

The answer to this problem is $\frac{112}{144\pi}$.

Thanks.

2. ## Re: Related Rates problem...

In a cone the ratio between height and radius is constant $\frac{h}{r}=\frac{16}{12}$
That gives you one equation so you can do it like the earlier examples

3. ## Re: Related Rates problem...

Start by drawing a diagram.
You are told the base radius and height of the complete cone, so you should, by the use of similar right-angled triangles, be able to find a relationship between r and h for any depth of water.
That can be used to eliminate one of the variables, (eliminating r is best, and is assumed), from the formula for the volume.
Now use the chain rule,

$\frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt}.$

Substitute what is known, the value of $dV/dt,$ and the value of $dV/dh,$ (calculated after differentiating the formula for the volume wrt h).

The only thing that you don't know, $dh/dt$ can now be calculated.