Calculating the Speed of a Sand Pile's Height Increase

superjen
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Sand is falling into a cone whose height is always 1/2 the diameter of the base. If the sand is falling at a rate of 1/2 cubic meters per min. How fast is the height increasin when the pile is 3 meters high?

so far I've drawn my diagram and got this far.

Given:
h = 1/2 the base
dv/dt = +1/2m3/min

Needs:
dh/dt

the equation i was going to use is
V = 1/12pih3

then i get

dv/dt = (1/2pi)(3h2)dh/dt

i don't know how to get the h?
any help?
 
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You're asked to find dh/dt when the height is 3 meters...so h=3:wink:

But there is a problem with your starting equation V = (1/12)pih3...If the height is always half the diameter, then r=4h not h/4
 
ok, that was stupid of me,
i don't even know why i was having trouble with that now >.<
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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