Question 1106186
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The conical tank has a radius of 3m and a height (depth) of 8m.  The ratio of the radius of the surface of the water to the depth of the water will always be 3:8.  So<br>
{{{r = (3/8)h}}}; {{{h = (8/3)r}}}<br>
(a) The volume remaining at any time is
{{{V = (1/3)(pi)(r^2)(h) = (1/3)(pi)(r^2)((8/3)r) = (8/9)(pi)(r^3)}}}<br>
{{{V(r) = (8/9)(pi)(r^3)}}}
(b) We want to find the rate at which the radius of the surface of the water is changing (dr/dt) when the depth is 5m.  We are given that the volume of water in the tank is decreasing at a rate of 0.2 m^3/minute: dV/dt = -0.2.<br>
{{{dV/dt = (dV/dr)*(dr/dt)}}}
{{{dr/dt = (dV/dt)/(dV/dr)}}}<br>
We know dV/dt; and from part (a) we can find dV/dr to be {{{(8/3)(pi)(r^2)}}}<br>
So<br>
{{{dr/dt = (-0.2)/((8/3)(pi)(r^2))}}}<br>
When the depth is 5m, the radius is (3/8)*5 = (15/8)m.  So dr/dt when the depth is 5 is<br>
{{{dr/dt = (-0.2)/((8/3)(pi)((15/8)^2))}}}
{{{dr/dt = (-1/5)/((75/8)(pi))}}}
{{{dr/dt = (-8)/(375(pi))}}}