Question 650834
Water is being pumped into a conical tank at the rate  of 100 ft3/min. The height of the tank is 20 ft and its radius is 5ft. How fast is the water level rising when the water height is 10ft?
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The volume of a cone, {{{V = (pi/3)r^2*h}}}
We can express r in terms of h: r = h/4
So we have an equation for the volume in terms of the height only:
{{{V = (pi/3)(h/4)^2*h = (pi/48)h^3}}}
The rate of change of volume, {{{dV/dt = (dV/dh)(dh/dt) = 100 ft^3/min}}}
{{{dV/dh = (pi/16)h^2}}}
The rate of change of the water height, {{{dh/dt}}}, at h=10 is:
{{{dh/dt = (100)/((pi/16)*10^2) = 16/pi}}}
So the rate of rise at h=10 ft = 5.093 ft/min