The rate at which the height of the pile is changing when the height is 2 ft. is 0.5659 ft./min. Computed using differentiation.
The sand is falling onto a conical pile, thus we get the pile in the shape of a cone, whose volume is given as:
V = (1/3)πr²h,
where V represents the volume of the cone, r represents its radius, and h represents its height.
In the question, we are given that the rate of piling is 16 ft.³/min, which implies that:
dV/dt = 16 ... (i)
We are also given that the diameter is three times the altitude (height), which can be shown as:
2r = 3h,
or, r = (3/2)h.
Thus, the volume can now be shown as:
V = (1/3)π{(3/2)h}²h,
or, V = (3πh³)/4.
Differentiating this with respect to time, we get:
dV/dt = (3π/4)(3h²)dh/dt.
Now, we need to calculate the rate of change of height, when the height is 2 feet, thus we take h = 2
Using dV/dt = 16 from (i), we can write:
16 = (3π/4)(3(2)²)dh/dt.,
or, 16 = 9π(dh/dt),
or, dh/dt = 16/9π = 0.5659 ft./min.
Thus, the rate at which the height of the pile is changing when the height is 2 ft. is 0.5659 ft./min.
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