I've worked through both parts of this question twice in what I assume is the correct manner, but I'm receiving an unexpected result from part B. The question is as follows:
Sand is dumped such that the shape of the sandpile remains a cone with height equal to twice the radius.
A) If the sand is dumped at the constant rate of 20ft^3/s, find the rate at which the radius is increasing when the height reaches 6 feet.
B) Repeat for a sandpile for which the edge of the sandpile forms an angle of 45° with the horizontal.
Volume of a cone [itex]V=(1/3)πr^2h[/itex]
Where r = radius
Where h = height
The Attempt at a Solution
[itex]dV/dt=d/dt((1/3)πr^2)*h + (1/3)πr^2*(d/dt)(h)[/itex]
[itex]20ft^3/s=(2/3)πr(dr/dt)*h + (1/3)πr^2*(1)[/itex]
If the height = 2*radius, and h=6, r=3
[itex]20=(2/3)π(3)(dr/dt)*(6) + (1/3)π(3^2)[/itex]
[itex]20=(36/3)π(dr/dt) + (9/3)π[/itex]
[itex](20-3π)/12π = dr/dt[/itex]
So when the height of the cone is 6ft, the radius is expanding at about (20-3π)/12π or .2805ft/s.
B) If the edge of the sandpile forms a 45° angle with the horizontal, that means the peak of the cone is 90°, and the height is equal to the radius (since both the bisected angle at the peak and the angle at the edge of the base must be 45°)
[itex]20=(2/3)π(6)(dr/dt)*(6) + (1/3)π(6^2)[/itex]
[itex]20=(72/3)π(dr/dt) + (36/3)π[/itex]
[itex](20-12π)/24π = dr/dt[/itex]
So if the height is equal to the radius, and the height is 6ft, the radius seems to be expanding at about (20-12π)/24π or -.2347ft/s. Since the base of the cone can't be shrinking, I know I must have done something wrong, but I don't see my error.