Question 1127032
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The formula for the volume of a cone is<br>
{{{V = (1/3)(pi)(r^2)(h)}}}<br>
The diameter of the base is 4, so the radius r is 2; the height h is 6.  So h is 3 times r.<br>
As the tank fills, the height (depth of the water) is always 3 times the radius (of the surface of the water).  So h = 3r, or r = h/3.<br>
We are given the rate at which the water level is rising (dh/dt); so we want our volume formula to be in terms of h alone.  So<br>
{{{V = (1/3)(pi)((h/3)^2)(h) = ((pi)h^3)/27}}}<br>
Find the derivative:<br>
{{{dV/dt = (((pi)h^2)/9)(dh/dt)}}}<br>
The given height is 2m, which is 200cm; dh/dt is given as 20cm/min.  Evaluate the derivative with those values.<br>
{{{dV/dt = (((pi)(200^2))/9)(20) = 279252}}} to the nearest whole number.<br>
The water volume is increasing at a rate of 279,252 cm^3/min while 10,500 cm^3/min is leaking out; that means the rate at which the water is being pumped into the tank is 279,252+10,500 cm^3/min, or 289,752 cm^3/min.