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 (in ft/min) is the height of the pile increasing when the pile is 12 ft high? (round your answer to two decimal places.)

1. -/1 points

a kite 100 ft above the ground moves horizontally at a speed of 6 ft/s. at what rate (in rad/s) is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?

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 always 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, $\frac{dV}{dt}=\frac{dV}{dh}\cdot\frac{dh}{dt}$. Differentiate $V=\frac{1}{12}\pi h^{3}$ with respect to $h$: $\frac{dV}{dh}=\frac{1}{4}\pi h^{2}$. Then $\frac{dV}{dt}=\frac{1}{4}\pi h^{2}\frac{dh}{dt}$.

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

We know that $\frac{dV}{dt}=35$ ft³/min and $h = 12$ 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 $h = 12$ into the formula for $\frac{dh}{dt}$: $\frac{dh}{dt}=\frac{4\times35}{\pi\times12^{2}}=\frac{140}{144\pi}=\frac{35}{36\pi}\approx0.31$ ft/min.

Answer:

$0.31$ ft/min