1. sand is being dumped from a conveyor belt at a rate of 1.2 m³/min and forms a pile in the shape of a cone…

1. sand is being dumped from a conveyor belt at a rate of 1.2 m³/min and 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 growing when the pile is 3 m high? ($v = \\frac{\\pi r^{2}h}{3}$)

1. sand is being dumped from a conveyor belt at a rate of 1.2 m³/min and 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 growing when the pile is 3 m high? ($v = \\frac{\\pi r^{2}h}{3}$)

Answer

Explanation:

Step1: Relate radius and height

Given diameter $d = h$, and $d = 2r$, so $r=\frac{h}{2}$.

Step2: Substitute $r$ into volume formula

Substitute $r = \frac{h}{2}$ into $V=\frac{\pi r^{2}h}{3}$, we get $V=\frac{\pi(\frac{h}{2})^{2}h}{3}=\frac{\pi h^{3}}{12}$.

Step3: Differentiate with respect to time $t$

Differentiate both sides of $V=\frac{\pi h^{3}}{12}$ with respect to $t$ using the chain - rule. $\frac{dV}{dt}=\frac{\pi}{12}\times3h^{2}\frac{dh}{dt}=\frac{\pi h^{2}}{4}\frac{dh}{dt}$.

Step4: Solve for $\frac{dh}{dt}$

We know that $\frac{dV}{dt}=1.2\ m^{3}/min$ and $h = 3m$. Substitute these values into $\frac{dV}{dt}=\frac{\pi h^{2}}{4}\frac{dh}{dt}$. Then $1.2=\frac{\pi\times3^{2}}{4}\frac{dh}{dt}$. First, simplify the right - hand side: $\frac{\pi\times9}{4}\frac{dh}{dt}$. Then solve for $\frac{dh}{dt}$: $\frac{dh}{dt}=\frac{1.2\times4}{9\pi}=\frac{4.8}{9\pi}=\frac{8}{15\pi}\ m/min$.

Answer:

$\frac{8}{15\pi}\ m/min$