A paper cup has the shape of a cone with height 10 cm and radius 3 cm (at the top. if water is poured into the cup at a rate of 2cm3/s, how fast is the water level rising when the water is 5 cm deep?

Let

h: height of the water

r: radius of the circular top of the water

V: the volume of water in the cup.

We have:

r/h = 3/10

So,

r = (3/10) * h

the volume of a cone is:

V = (1/3) * π*r^2*h

Rewriting:

V (t) = (1/3) * π * ((3/10) * h (t)) ^2*h (t)

V (t) = (3π/100) * h (t) ^3

Using implicit differentiation:

V' (t) = (9π/100) * h (t) ^2*h' (t)

Clearing h' (t)

h' (t) = V' (t) / ((9π/100) * h (t) ^2)

the rate of change of volume is V' (t) = 2 cm3/s when h (t) = 5 cm.

substituting:

h' (t) = 8 / (9π) cm/s

Answer:

the water level is rising at a rate of:

h' (t) = 8 / (9π) cm/s