Question

Gravel is being dumped from a conveyor belt at a rate of 10 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 9 ft high? (Round your answer to two decimal places.)

Answer 1

Answer: The height of the pile increasing is increasing at $0.16\frac{ft}{min}$

Explanation:

Given

Rate at which gravel is being dumped , $\frac{\mathrm{d} V}{\mathrm{d} t}=10\frac{ft^{3}}{min}$

=>Rate of increase of volume of cone=$\frac{\mathrm{d} V}{\mathrm{d} t}=10\frac{ft^{3}}{min}$

If height of the cone at any instant is h then the diameter of cone is also h

Volume of cone , $V=\frac{\pi r^{2}h}{3}=\frac{\pi h^{3}}{4\times 3}=\frac{\pi h^{3}}{12}$

Now differentiate both sides w.r.t time(t)

$\frac{\mathrm{d} V}{\mathrm{d} t}=\frac{\pi h^{2}}{4}\frac{\mathrm{d} h}{\mathrm{d} t}$

Therefore at h = 9 ft

$10=\frac{\pi \times 9^{2}}{4}\times \frac{\mathrm{d} h}{\mathrm{d} t}$

=>$\frac{\mathrm{d} h}{\mathrm{d} t}=0.16\frac{ft}{min}$

Thus the height of the pile increasing is increasing at $0.16\frac{ft}{min}$