Question

Sand falls from an overhead bin and accumulates in a conical pile with a radius that is always twotwo times its height. Suppose the height of the pile increases at a rate of 33 cm divided by scm/s when the pile is 1010 cm high. At what rate is the sand leaving the bin at that​ instant?

Answer 1

Answer:

-423 m³/s

Explanation:

Volume of a cone is:

V = ⅓ π r² h

Given r = 2h:

V = ⅓ π (2h)² h

V = ⁴/₃ π h³

Taking derivative with respect to time:

dV/dt = 4π h² dh/dt

Given h = 1010 cm and dh/dt = 33 cm/s:

dV/dt = 4π (1010 cm)² (33 cm/s)

dV/dt ≈ 4.23×10⁸ cm³/s

dV/dt ≈ 423 m³/s

The pile is growing at 423 m³/s, so the bin is draining at -423 m³/s.

Answer 2

The rate at which the sand is leaving the bin at that instant is $423\times 10^{6} cm^{3}/s$.

Given

It is given that the radius of a conical bin is two times its height and at the instant when the height of the bin is $1010cm$, the height of the pile increases at a rate of $33 cm/s$.

Volume of the bin

The formula for the volume of a cone is given as,

$V=\frac{1}{3}\pi r^{2}h$

Substitute $r=2h$ as per the question

$V=\frac{4}{3}\pi h^{3}$

This is the volume of the conical bin.

Rate of change in the volume of the bin

To find the rate of change in the volume of the bin, differentiate the expression for volume w.r.t. time using the chain rule as follows,

$\frac{dV}{dt}=\frac{dV}{dh}\times \frac{dh}{dt}$

$\frac{dV}{dt}=\frac{4}{3}\pi (3h^{2}) \times \frac{dh}{dt}\\\\\frac{dV}{dt}=4\pi h^{2} \times \frac{dh}{dt}$

Now, according to the question, at $h=1010cm$, $\frac{dh}{dt}=33$.

Substituting these values, the rate at which the sand is leaving the bin is,

$\frac{dV}{dt}=4\pi (1010)^{2} \times 33\\\frac{dV}{dt}=423,025,503.90235\\\frac{dV}{dt}=423\times10^{6}cm^{3}/s$

So, the rate at which the sand leaves the conical bin at the given instant is $423\times10^{6}cm^{3}/s$

To learn more about the rate of change of a quantity, refer

https://brainly.com/question/20705459

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