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Chapter 3: Problem 27

Growing sandpile Sand falls from an overhead bin and accumulates in a conical pile with a radius that is always three times its height. Suppose the height of the pile increases at a rate of \(2 \mathrm{cm} / \mathrm{s}\) when the pile is \(12 \mathrm{cm}\) high. At what rate is the sand leaving the bin at that instant?

Short Answer

Expert verified
In summary, when the sandpile's height is 12 cm and its radius is 3 times its height, the sand is leaving the bin at a rate of approximately \(\frac{864蟺}{287} \thinspace cm^3/s\).

Step by step solution

01

Relate radius and height using the given information

The radius is always three times the height, so we can write this relationship as \(r=3h\).
02

Differentiate the equation for the volume of a cone with respect to time

Differentiating the volume equation \(V=\frac{1}{3}蟺r^2h\) with respect to time t, we get: \(\frac{dV}{dt} = \frac{d}{dt}(\frac{1}{3}蟺r^2h)\)
03

Apply the chain rule

Applying the chain rule, we have: \(\frac{dV}{dt} = \frac{1}{3}蟺(2rh\frac{dr}{dt} + r^2\frac{dh}{dt})\)
04

Substitute the given values and the relationship between r and h

From Step 1, we know that \(r=3h\), and we are given \(\frac{dh}{dt} = 2 \thinspace cm/s\) when the height is 12 cm. Plugging these values, we have: \(\frac{dV}{dt} = \frac{1}{3}蟺(2(3h)h\frac{dr}{dt} + (3h)^2\frac{dh}{dt})\) \(\frac{dV}{dt} = \frac{1}{3}蟺(6h^2\frac{dr}{dt} + 9h^2\frac{dh}{dt})\) At \(h = 12 \thinspace cm\), \(\frac{dh}{dt} = 2 \thinspace cm/s\) \(\frac{dV}{dt} = \frac{1}{3}蟺(6(12)^2\frac{dr}{dt} + 9(12)^2(2))\)
05

Solve for \(\frac{dr}{dt}\)

We have only one unknown, \(\frac{dr}{dt}\). Solving the equation for \(\frac{dr}{dt}\): \(\frac{dV}{dt} - \frac{1}{3}蟺(9(12)^2(2)) = \frac{1}{3}蟺(6(12)^2\frac{dr}{dt})\) \(\frac{dV}{dt} - 864蟺 = 288蟺\frac{dr}{dt}\) \(\frac{dr}{dt} = \frac{1}{288蟺}(\frac{dV}{dt} - 864蟺)\)
06

Find the rate at which sand is leaving the bin

Since the volume of sand leaving the bin is equal to the increase in volume of the sandpile, we can write: \(\frac{dV_{bin}}{dt} = \frac{dV}{dt}\) So, the rate at which the sand is leaving the bin is given by: \(\frac{dV_{bin}}{dt} - 864蟺 = 288蟺\frac{dr}{dt}\) \(\frac{dV_{bin}}{dt} = 288蟺\frac{dr}{dt} + 864蟺\) We found \(\frac{dr}{dt}\) in Step 5. So, inserting it into the equation above, we have: \(\frac{dV_{bin}}{dt} = 288蟺\left(\frac{1}{288蟺}(\frac{dV_{bin}}{dt} - 864蟺)\right) + 864蟺\) Now, we can solve for the rate at which the sand is leaving the bin, \(\frac{dV_{bin}}{dt}\): \(288蟺\frac{dV_{bin}}{dt} - 864蟺^2 = \frac{dV_{bin}}{dt}\) Combining the terms, we get: \(287蟺\frac{dV_{bin}}{dt} = 864蟺^2\) So, the rate at which sand is leaving the bin is: \(\frac{dV_{bin}}{dt} = \frac{864蟺^2}{287蟺}\) \(\frac{dV_{bin}}{dt} = \frac{864蟺}{287} \thinspace cm^3/s\) Thus, the sand is leaving the bin at a rate of \(\frac{864蟺}{287} \thinspace cm^3/s\) when the pile is 12 cm high.

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