Thread: 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!

Where did you get dv/dh? I get 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?

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?

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

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

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