Answer: The height of the pile increasing is increasing at [tex]0.16\frac{ft}{min}[/tex]
Explanation:
Given
Rate at which gravel is being dumped , [tex]\frac{\mathrm{d} V}{\mathrm{d} t}=10\frac{ft^{3}}{min}[/tex]
=>Rate of increase of volume of cone=[tex]\frac{\mathrm{d} V}{\mathrm{d} t}=10\frac{ft^{3}}{min}[/tex]
If height of the cone at any instant is h then the diameter of cone is also h
Volume of cone , [tex]V=\frac{\pi r^{2}h}{3}=\frac{\pi h^{3}}{4\times 3}=\frac{\pi h^{3}}{12}[/tex]
Now differentiate both sides w.r.t time(t)
[tex]\frac{\mathrm{d} V}{\mathrm{d} t}=\frac{\pi h^{2}}{4}\frac{\mathrm{d} h}{\mathrm{d} t}[/tex]
Therefore at h = 9 ft
[tex]10=\frac{\pi \times 9^{2}}{4}\times \frac{\mathrm{d} h}{\mathrm{d} t}[/tex]
=>[tex]\frac{\mathrm{d} h}{\mathrm{d} t}=0.16\frac{ft}{min}[/tex]
Thus the height of the pile increasing is increasing at [tex]0.16\frac{ft}{min}[/tex]
