Help w/ 'A coffee filter has the shape of an inverted cone.'

[Sarah]

I need help with this question please...

A coffee filter has the shape of an inverted cone. Water drains out of the filter at a rate of 10cm^3/min. When the depth of the water in the cone is 8 cm, the depth is decreasing at 2 cm/min. What is the ratio of the height of the cone to it's radius.

I have the solutions from the teachers answer book but it doesn't explain the steps and I'm confused with a few things.

The Solution

v=1/3 pi r^2 h

dv/dt = -10cm^3/min
dh/dt = -2 cm/min
Find h/r when h=8cm

let h/r = k, k is a constant
v=1/3 pi r^2 h and r=h/k
Therefore v = 1/3 pi h^2/k^2 h
v= 1/3k^2 pi h^3
dv/dt = 1/k^2 pi h^2 dh/dt
At a specific time, h=8cm,

-10 = 1/k^2 pi (8)^2(-2)
-10= (-128 pi)/k^2
k^2 = (64 pi) / 5
k = [(8)(squareroot pi)]/5
Therefore h/r = 8 (squareroot pi/5)
=8(squareroot 5 pi)/5

Therefore, the ratio of the height to the radius is 8(squareroot 5 pi) / 5


dv/dt = 1/k^2 pi h^2 dh/dt

I don't understand why in this line (i have bolded it in the solution) the 3 is somehow divided out.

 

[marcmtlca]

You consider v and h as functions of t, and take the derivative:

\(\displaystyle v(t)= \frac{1}{3k^2} \pi h(t)^3\)

then:

\(\displaystyle v'(t)=\frac{1}{3k^2} \pi 3 h(t)^2 h'(t)\)

which is just:

\(\displaystyle \frac{dv}{dt}=\frac{\pi}{k^2}h^2 \frac{dh}{dt}\)