# Thread: help with related rates (draining tank)

1. ## help with related rates (draining tank)

whater drains from the conical tank at the rate of 5 ft^3/min.
the radius of the conical tank is 4' and the hight is 10'
the radius of the water in the tank is r and the hight is h...
a) what is the relationship between the variables h and r in the figure?
the answ for this one is r=(2/5)h .. but why?
B) how fast is the water level dropping when h=6ft??

2. Originally Posted by natty210
whater drains from the conical tank at the rate of 5 ft^3/min.
the radius of the conical tank is 4' and the hight is 10'
the radius of the water in the tank is r and the hight is h...
a) what is the relationship between the variables h and r in the figure?
the answ for this one is r=(2/5)h .. but why?
B) how fast is the water level dropping when h=6ft??
Ok, first things first, here is a drawing to show what I am talking about.

Ok, that is big. The better to see the ratios with my dear... anyway, on to the math. Do you see how the height and radius of the tank form a triangle similar to any corresponding radius and height?

$\frac{r}{h} = \frac{4}{10} = \frac{2}{5}$

You can express the radius in terms of the height.

$r = \frac{2h}{5}$

So now we can set up the related rate problem to determine the rate at a given height.

Standard volume equation:

$V = \frac{1}{3}\pi r^{2}h$

Expressed in terms of height alone:

$V = \frac{1}{3}\pi \left(\frac{2h}{5}\right)^{2}h = \frac{4}{75}\pi h^{3}$

Since we want to know the rate, we need to differentiate with respect to time:

$\frac{dV}{dt} = \frac{4}{25}\pi h^{2}\frac{dh}{dt}$

Now you are ready to begin substituting in known quantites -- the rate and the height. I will let you take over from here. Don't forget that since the water is falling the rate is negative.

### water is draining from a conical tank

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