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Wheat falls from an overhead bin and accumulates in a conical pile, so that the radius of the base of the cone is always twice the height of the cone. If the wheat falls at a rate of $3$ m$^3$/min, how fast is the height of the wheat pile changing when the diameter of the pile is $20$ meters?

Here is what I tried. Let $r =$ radius, $h =$ height, $v =$ volume, and $t =$ time. $$r = 2h.$$ $$v = \frac13\pi r^2h =\pi (2h)^2h/3 = 4\pi h^3/3.$$ $$\frac{\mathrm d v}{\mathrm dt} = \frac{\mathrm d}{\mathrm dt}(4\pi h^3/3).$$ $$\frac{\mathrm d v}{\mathrm dt} = 4\pi h^2 \frac{\mathrm d h}{\mathrm dt}.$$ When diameter $= 10$, height $= 2.5$. $$3 = 4\pi(2.5)^2 \frac{\mathrm d h}{\mathrm dt}$$ $$3 = 25\pi\frac{\mathrm d h}{\mathrm dt}.$$ $$\frac{\mathrm d h}{\mathrm dt} =\frac3{25\pi} = 0.038\text{ m/min.}$$

Ian
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jack
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1 Answers1

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Basically you have $$\frac{dh}{dt} = \frac{dv}{dt} \times \frac{dh}{dv}$$

and you are given that $\frac{dv}{dt} = 3$ and you want to find $\frac{dh}{dt}$ so your goal is to find $\frac{dh}{dv}$ and then use the above equation to find your $\frac{dh}{dt}$

$$v=\frac{4\pi h^3}{3}$$ like you calculated and so $$\frac{dv}{dh} = 4 \pi h^2$$

but remember you want $\frac{dh}{dv}$ so you simply take $$\frac{1}{\frac{dv}{dh}}$$ and so $$\frac{dh}{dv} = \frac{1}{ 4 \pi h^2}$$

and now you are basically done so when $d= 20$ you have $h = 5$ and $$\frac{dh}{dv} = \frac{1}{4 \pi (5)^2}$$ so your $$\frac{dh}{dt} = 3 \times \frac{1}{4 \pi (5)^2}$$m/min

alkabary
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