Harder question on connected rates of change

**Big_Joe** · February 21st 2009, 05:51 AM

This is the question:

A hollow right circular cone has height 18 cm and base radius 12 cm.
It is held vertex downwards beneath a tap leaking at the rate of 2 cm/s.
Find the rate of rise of water level when the depth is 6 cm.

I have no idea where to go with this question. This is what i've tried:

$\frac{dh water}{dt}=(\frac{dh water}{dv})(\frac{dv}{dt})$

Then i can find out the volume of the cone easily in respect to either r or h but i don't understand how i can use this in order to find the change in the height of the depth of the water.

Can someone please explain how i can proceed to complete this question. Thanks in advance.

**skeeter** · February 21st 2009, 06:17 AM

Originally Posted by Big_Joe

This is the question:

A hollow right circular cone has height 18 cm and base radius 12 cm.
It is held vertex downwards beneath a tap leaking at the rate of 2 cm/s.
Find the rate of rise of water level when the depth is 6 cm.

I have no idea where to go with this question. This is what i've tried:

$\frac{dh water}{dt}=(\frac{dh water}{dv})(\frac{dv}{dt})$

Then i can find out the volume of the cone easily in respect to either r or h but i don't understand how i can use this in order to find the change in the height of the depth of the water.

Can someone please explain how i can proceed to complete this question. Thanks in advance.

$V = \frac{\pi}{3} r^2 h$

$\frac{r}{h} = \frac{12}{18} = \frac{2}{3}$

$r = \frac{2h}{3}$

$V = \frac{\pi}{3}\left(\frac{2h}{3}\right)^2 h$

$V = \frac{4\pi}{27}h^3$

find the time derivative, sub in your known values, and determine the value of $\frac{dh}{dt}$

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