Question

sand falls from an overhead bin and accumulates in a conical pile with a radius that is always two times its height. suppose the height of the pile increases at a rate of 1 cm/s when the pile is 15 cm high. at what rate is the sand leaving the bin? the sand is leaving the bin at a rate of . (type an exact answer, using π as needed.)

Explanation:

Step1: Write the volume formula for a cone

The volume formula of a cone is $V=\frac{1}{3}\pi r^{2}h$. Given that $r = 2h$, we substitute $r$ into the volume formula: $V=\frac{1}{3}\pi(2h)^{2}h=\frac{4}{3}\pi h^{3}$.

Step2: Differentiate the volume formula with respect to time $t$

Using the chain - rule, $\frac{dV}{dt}=4\pi h^{2}\frac{dh}{dt}$.

Step3: Substitute the given values of $h$ and $\frac{dh}{dt}$

We are given that $h = 15$ cm and $\frac{dh}{dt}=1$ cm/s. Substituting these values into the equation $\frac{dV}{dt}=4\pi h^{2}\frac{dh}{dt}$, we get $\frac{dV}{dt}=4\pi\times(15)^{2}\times1$.

Step4: Calculate the value of $\frac{dV}{dt}$

$\frac{dV}{dt}=4\pi\times225\times1 = 900\pi$ $cm^{3}/s$. This is the rate at which sand is entering the pile. Since the sand is leaving the bin at a rate equal to the rate at which it is entering the pile (to maintain the described situation), the rate at which the sand is leaving the bin is $900\pi$ $cm^{3}/s$.

Answer:

$900\pi$