Question 1146721
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(1) Formula for the volume of a cone: {{{V = (1/3)(pi)(r^2)(h)}}}<br>
(2) Use the given dimensions of the cone to get the volume formula in terms of a single variable.  Since the problem asks for the rate of change of the depth (height), we want a volume formula in terms of h.<br>
The cone has a depth of 20 and a diameter of 10, so a radius of 5.  So at all times as the cone is filling, the radius is 1/4 of the depth: r = h/4.<br>
{{{V = (1/3)(pi)((h/4)^2)(h) = (1/48)(pi)(h^3)}}}<br>
(3) Find the derivative with respect to time:<br>
{{{dV/dt = (1/48)(pi)(3h^2*(dh/dt)) = (1/16)(pi)(h^2)(dh/dt)}}}<br>
dV/dt is given; solve for dh/dt when h=8:<br>
{{{15 = (1/16)(pi)(8^2)(dh/dt)}}}
{{{15 = 4pi*(dh/dt)}}}
{{{dh/dt = 15/(4pi)}}}<br>
ANSWER: the depth of the water in the tank is changing at a rate of 15/(4pi) feet per minute when the depth is 8 feet.