# Homework Help: Rate of change - cones

1. Mar 14, 2012

### jsmith613

2. Mar 14, 2012

### Staff: Mentor

Please post the problem and your work here. IMO, it's a pain in the butt to have to open a web page to see the problem and the work, plus I can't insert a comment at the appropriate place where there's an error.

3. Mar 14, 2012

### jsmith613

dv/dt = 4 cm3 min-1

tan(60) = r/h

r = √3* h

V = (1/3) π r2h

dv/dh = (1/3) π r2

dh/dt = (dh/dv) * (dv/dt)

= 1/(1/3) π r2 * 4

= 12/(pi r2)

for h = 4

dh/dt = 0.079577

The answer given is 0.0265 cm/min. why?

I don't know how do use latex - see post for a clearer solution if you get lost!

4. Mar 14, 2012

### jsmith613

question was:

A hollow cone with a semi-vertical angle of 60 degrees is held vertex down with its axis vertical.
Water drips into the cone at 4 cm3/min
Find the rate at which the depth of water is increasing when the water is 4 cm deep

5. Mar 14, 2012

### Staff: Mentor

You're ignoring the relationship between r and h.

Substitute for r in your volume equation. Then you'll have V purely as a function of h.
What you have is not correct, because V is a function of r and h.