# related ratesA conical paper cup is 20 cm tall with a radius of 10 cm. The cup is being filled with water at a rate of `(16pi)/3` cm^3/sec. How fast is the water level rising when the water level...

related rates

A conical paper cup is 20 cm tall with a radius of 10 cm. The cup is being filled with water at a rate of `(16pi)/3` cm^3/sec. How fast is the water level rising when the water level is 5 cm?

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### 1 Answer

Since cup is being filled at constant rate of `(16pi)/3 (cm^3)/s` it follows that current *speed* at which the water level is rising is *inversely proportional* to *area of section* of our paper cup (section is circle because cup is conical) at the current water level i.e. as cup is being filled the water level will rise slowe and slower because section area is getting bigger and bigger.

Let

`h=`current water level

`r=`radius at current water level

`A=`section area at current water level

`v=`speed at which water level is rising at `h`

Since height of cup is 20cm and radius is 10cm it follows that

`20:10=h:r => r=x/2`

Since our section is a circle with radius `r`

`A=r^2pi=(h/2)^2pi=(x^2pi)/4 cm^2`

Now to get `v` we simply devide rate at which cup is being filled and area of section `A.`

`v=((16pi)/3)/((h^2pi)/4)=64/3h^2`cm/s

So speed at which water level is rising when water level is `h=5cm` is

`v=64/(3cdot5^2)=0.85dot(3)`cm/s