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.

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[tex]r^{2} h[/tex]

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)+([tex]r^{2}[/tex]*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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