1. ## Rate of change

An inverted cone of base radius 4cm and height 8cm is initially filled with water. Water drips out from the vertex at a constant rate of 2pi cm^3/s. Find the rate of decrease in the depth of the water in the cone 16seconds after dripping starts.

From what i understand:

dv/dt = -2pi cm^3/s
dh/dr = ?

Vol.of cone = (1/3)(pi)(r^2)(h)
dv/dh = 16pi/3

dv/dt = dh/dt * dv/dh
....
dh/dt = -2pi * (3/16pi) = -0.375cm/s
16seconds after dripping starts = 16 * -0.375

My ans was wrong, any idea what went wrong? Thanks in advance!

2. Where did you get dv/dh? I get $\displaystyle 4\pi\sqrt[3]{4}$ for 16 seconds in.

Thank you for showing some of your work, but you didn't show this important piece.

Did you notice that r = h/2 is a constant relationship?

3. ## Re: Rate of change

Using r = h/2,
Vol.of cone = (1/3)pi(r^2)h = (1/3)pi[(h/2)^2]h = (pi/12)h^3
Therefore, dv/dh = (pi/12)(3h^2) = (pi/4)h^2
How do I bring in the '16seconds after dripping starts' from here on?

4. ## Re: Rate of change

Start with the volume formula. How much have you lost after 16 seconds? Solve for 'h' at that moment.

5. ## Re: Rate of change

Rate of increase is constant therefore rate of decrease is constant so shouldn't they be the same?

6. ## Re: Rate of change

What rate of increase? Shouldn't what be the same as what?