Question

gravel is being dumped from a conveyor belt at a rate of 35 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 15 ft high? round the result to the nearest hundredth.
a conveyor belt is shown with gravel being dumped off the end in conical shape.
the correct answer is not listed.
0.27 ft/min
0.20 ft/min
0.33 ft/min
0.14 ft/min

Explanation:

Step1: Write the volume formula for a cone

The volume formula for a cone is $V=\frac{1}{3}\pi r^{2}h$. Given that the base - diameter $d$ and height $h$ are equal, and $d = 2r$, so $2r=h$ or $r=\frac{h}{2}$. Substitute $r=\frac{h}{2}$ into the volume formula: $V=\frac{1}{3}\pi(\frac{h}{2})^{2}h=\frac{1}{12}\pi h^{3}$.

Step2: Differentiate both sides with respect to time $t$

Using the chain - rule, $\frac{dV}{dt}=\frac{dV}{dh}\cdot\frac{dh}{dt}$. Differentiating $V = \frac{1}{12}\pi h^{3}$ with respect to $h$ gives $\frac{dV}{dh}=\frac{1}{4}\pi h^{2}$. So, $\frac{dV}{dt}=\frac{1}{4}\pi h^{2}\frac{dh}{dt}$.

Step3: Solve for $\frac{dh}{dt}$

We know that $\frac{dV}{dt}=35$ ft³/min and $h = 15$ ft. Rearranging the equation $\frac{dV}{dt}=\frac{1}{4}\pi h^{2}\frac{dh}{dt}$ for $\frac{dh}{dt}$, we get $\frac{dh}{dt}=\frac{4\frac{dV}{dt}}{\pi h^{2}}$.

Step4: Substitute the given values

Substitute $\frac{dV}{dt}=35$ and $h = 15$ into the formula for $\frac{dh}{dt}$: $\frac{dh}{dt}=\frac{4\times35}{\pi\times15^{2}}=\frac{140}{225\pi}=\frac{28}{45\pi}\approx0.20$ ft/min.

Answer:

$0.20$ ft/min