# Thread: another rate of change problem.

1. ## another rate of change problem.

A hallow rigght circular cone has height 18cm and base radius 12cm. It is held vertex downwards beneath a tap leaking at the rate of $2 cm^{3} s^{-1}$

Find the rate of rise of water level when the depth is 6cm.
I have no idea how to start this, I know that $V = \frac{1}{3} \pi r^{2} h$

any help appreciated,

thanks

2. Well, you are looking for an expression for dh/dt...once you have this you can substitute 6 for h for your answer.

You have V in terms of r and h. First use the fact that the ratio of r to h stays constant to find a formula for V in terms of h only. Then you can calculate dV/dh.

You are given dV/dt, so no you can use the chain rule to find dh/dt.

3. Originally Posted by mrbohn1
Well, you are looking for an expression for dh/dt...once you have this you can substitute 6 for h for your answer.

You have V in terms of h, so you can calculate dV/dh. And you are given dV/dt.

Now use the chain rule to find dh/dt.

thanks,

The thing I am getting confused with is that we are not given dv/dt. the $2cm^{3} s^{-1}$ does not represent the depth of the cone, that's just the rate that the water is dripping from the tap.

so how can you usethe value for dv/dt?

thanks

4. is a measure of volume per second, which is the rate of change of volume over time, ie dV/dt=2.

5. Originally Posted by Tweety
I have no idea how to start this, I know that $V = \frac{1}{3} \pi r^{2} h$

any help appreciated,

thanks
This exact question has been posted before check it at: http://www.mathhelpforum.com/math-he...es-change.html