[tex]h^{\prime}=\frac{4}{1089\pi}ft\text{ /min}[/tex]
STEP - BY - STEP EXPLANATION
What to find?
dh/dt
Given that;
At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 4 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude.
Since we have that the rate is 4 cubic per minute, then dv/dt = 4 (since v is the volume of the cone at time t.)
The formula for the volume of a cone is
[tex]V=\frac{1}{3}\pi r^2h---------------(1)[/tex]Where r is the radius and h is the height.
We have that, the diameter of the cone is approximately three times the altitude.
That is;
diameter = 3h
But, d = 2r
⇒2r = 3h
⇒ r = 3h/2
Now, we substitute r=3h/2 into equation (1).
[tex]V=\frac{1}{3}\pi(\frac{3h}{2})^2h[/tex][tex]=\frac{1}{3}\pi(\frac{9h^2}{4})h[/tex][tex]V=\frac{3}{4}\pi h^3[/tex]Now, differentiate the above with respect to t.
[tex]\frac{dv}{dt}=\frac{3}{4}\pi3h^2\frac{dh}{dt}[/tex]Simplify .
[tex]\frac{dv}{dt}=\frac{9}{4}\pi h^2\frac{dh}{dt}[/tex]Make dh/dt subject of formula.
[tex]\frac{dh}{dt}=\frac{dv}{dt}\times\frac{4}{9\pi h^2}----------(2)[/tex]Recall that, dv/dt = 4 and h=22
Substituting the values into equation (2), we have;
[tex]\frac{dh}{dt}=4\times\frac{4}{9\pi(22)^2}[/tex][tex]=\frac{16}{9\pi\text{ (484)}}[/tex][tex]=\frac{4}{9\pi\text{ (121)}}[/tex][tex]=\frac{4}{1089\pi}[/tex]Therefore, h' = dh/dt = 4/1089π ft/min.
