Question 1103812
<br>
The volume of sand in the pile at any time is<br>
{{{V=(1/3)(pi)(r^2)(h)}}}<br>
We are given that dV/dt is 5 (m^3/sec); we need to find dh/dt at the moment the height h is 2.<br>
Since we need to find dh/dt, we need to have the volume formula in terms of h only.  We can do that using the given information that the height of the pile is always equal to the diameter.<br>
If the diameter is equal to the height, then the radius is half the height; r = h/2.  Then the volume formula is<br>
{{{V=(1/3)(pi)((h/2)^2)(h) = (1/12)(pi)(h^3)}}}<br>
Now we need to differentiate this formula to find the relationship between dh/dt and dV/dt.<br>
(NOTE: Many beginning calculus students try to substitute the given value h=2 prior to this point.  However, that gives a constant for the volume; and the derivative of a constant is 0 -- so we won't be able to finish the problem....)<br>
{{{dV/dt = (1/4)(pi)(h^2)(dh/dt)}}}<br>
We could plug in the given value h=2 at this point, along with the given value of 5 for dV/dt, to find the answer to the problem.  But let's hold off on substituting the given value for h; perhaps later on we will be asked to find dh/dt when the height is some value other than 2.<br>
So we are getting a formula for dh/dt in terms of dV/dt and h, so that we can, if required, find dh/dt for several different values of h.<br>
{{{dh/dt = (4*(dV/dt))/((pi)(h^2)) = 20/((pi)(h^2))}}}<br>
Now we can plug in the given value of h to find dh/dt when the height is 2:<br>
{{{dh/dt = 20/(4pi) = 5/pi = 1.59}}}<br>
To 2 decimal places, dh/dt = 1.59 when h = 2.