Question

Recall that the volume of the a right circular cone with height h and radius …

Answer 1

Given:

The change in volume by time is, (dV/dt) = 10 cubic feet per minute.

The diameter and height of the right circular one are equal, 2r = h.

Height of the pile is, h = 21 ft.

The objective is to find the rate of increase in height of the pile.

Step-by-step explanation:

The general formula of volume of cone is,

`V=(1)/(3)\pi r^2h\text{ . . . . . (1)}`

From the given data, radius of the cone can be calculated as,

`\begin{gathered} 2r=h \\ r=(h)/(2) \end{gathered}`

Substitute the value of r in equation (1).

`\begin{gathered} V=(1)/(3)\pi((h)/(2))^2h \\ V=(1)/(3)\pi((h^2)/(4))^{}h \\ V=(1)/(12)\pi h^3\text{ . . . . . (2)} \end{gathered}`

To find rate of increase in height of the pile:

Now, differentiate equation (2) with respect to time t.

`\begin{gathered} (dV)/(dt)=(\pi)/(12)*(d)/(dt)(h^3) \\ (dV)/(dt)=(\pi)/(12)*3h^2(dh)/(dt) \\ (dV)/(dt)=(\pi)/(4)* h^2(dh)/(dt)\text{ . . . . (3)} \end{gathered}`

Substitute (dV/dt) and h in equation (3).

On plugging the obtained values in eqation (3),

`\begin{gathered} 10=(\pi)/(4)*(21)^2(dh)/(dt) \\ \frac{dh}{\mathrm{d}t}=(10*4)/(21^2*\pi) \\ (dh)/(dt)=0.02887164\ldots.\text{.} \\ (dh)/(dt)\approx0.029\ldots\text{ ft/min} \end{gathered}`

Hence, the rate of increase in height of the pile is 0.029 ft/min.