Sand of Time

Calculus Level 4

An hour glass is constructed from rotating the graphs $y = \frac{x^{2}}{3}$ and $y = -\frac{x^{2}}{3}$ around the y-axis within the domain of $[-6, 6]$ as shown in the picture above.

The upper inverted bulb of the hour glass has a circular base with radius of $6$ cm. and is $12$ cm. high, and the sand, filled in the upper bulb, is dripping into the lower bulb at a constant rate, falling all down in an hour.

What is the rate of decrease in height of the upper bulb in cm./minute when half an hour has passed?

Give your answer to the nearest 3 decimal places.

The answer is 0.141.

1 solution

Worranat Pakornrat
Mar 29, 2016

The volume of revolution = $\int_{0}^{12} \pi(x^2) \ \mathrm{d}y$ = $\int_{0}^{12} \pi(3y)\ \mathrm{d}y$ = $3\pi[\dfrac{y^2}{2}|_0^{12}$ ] = $216\pi$ .

Then $\dfrac{dV}{dh} = \dfrac{\frac{dV}{dt}}{\frac{dh}{dt}}$ = $\dfrac{dV}{dy} = 3\pi(y)$ .

$\dfrac{dV}{dt}$ is constant and equals to $\dfrac{216\pi}{60}$ .

And when half an hour has passed, volume of $108\pi$ stays upward, and volume = $3\pi[(\dfrac{y^2}{2}] = 108\pi$ . Then $y = \sqrt{72}$ .

Thus, at $y = \sqrt{72}$ , $\dfrac{dV}{dy} = 3\pi(\sqrt{72})$

$\dfrac{dh}{dt} = \dfrac{\frac{dV}{dt}}{\frac{dV}{dh}} = \dfrac{\frac{216\pi}{60}}{3\pi(\sqrt{72})} = \dfrac{1}{5\sqrt{2}} = \dfrac{\sqrt{2}}{10}$ .

That would be approximately $0.141$ .

Sand of Time

The answer is 0.141.

1 solution

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