Question

A container is the shape of an inverted right circular cone has a radius of 4.00 inches at the top and a height of 5.00 inches. At the instant when the water in the container is 2.00 inches deep, the surface level is falling at the rate of -2.00 in./s. Find the rate at which water is being drained.

Answer 1

The rate at which the water from the container is being drained is 24 inches per second.

Given radius of right circular cone 4 inches .height being 5 inches, height of water is 2 inches and rate at which surface area is falling is 2 inches per second.

Looking at the image we can use similar triangle propert to derive the relationship:

r/R=h/H

where dh/dt=2.

Thus r/5=2/5

r=2  inches

Now from r/R=h/H

we have to  write with initial values of cone and differentiate:

r/5=h/5

5r=5h

differentiating with respect to t

5 dr/dt=5 dh/dt

dh/dt is given as 2

5 dr/dt=5*-2

dr/dt=-2

Volume of cone is 1/3 π$r^{2} h$

We can find the rate at which the water is to be drained by using partial differentiation on the volume equation.

Thus

dv/dt=1/3 π(2rh*dr/dt)+($r^{2}$*dh/dt)

Putting the values which are given and calculated we get

dv/dt=1/3π(2*2*2*2)+(4*2)

=1/3*3.14*(16+8)

=3.14*24/3.14

=24 inches per second

Hence the rate at which the water is drained from the container is 24 inches per second.

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