Suppose water is running out of a conical
funnel at the speed of 2 in3/sec. The radius of the base of
the funnel is 5 in and the height 15 in. Find the rate at which the water
level is dropping when it is 2 in from the top.
Let r be the radius and h the height of the surface of
the water at time t, and let v be the volume of the water in the funnel.
By similar triangles, r/5 = h/15. Therefore, r = h/3.
Therefore the volume of the water in the funnel is
v = πr2h/3
= πh3/27
Differentiate both side of the above equation.
dv/dt = (πh2/9)(dh/dt)
The water runs out off the funnel at speed of 2 in3/sec,
so dv/dt = -2. Thus,
-2 = (πh2/9)(dh/dt)
Hence,
dh/dt = -18/πh2
in2/sec |