# LessonRelated Rates: A Sand Pile

#### MarkFL

##### La Villa Strangiato
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Sand is being dumped from a conveyor belt at a constant rate, forming a conical pile. The ratio of the base radius of this pile to its height is constant. Find the height of the pile as a function of time.

To begin this problem, let's define some constants:

$$a$$ = the rate at which sand is being dumped on the pile, and will have units of volume per time.

$$b$$ = the ratio of the base radius to the height of the pile.

And our variables:

$$V$$ = the volume of the pile at time $$t$$.

$$r$$ = the base radius of the pile.

$$h$$ = the height of the pile.

And so immediately, we may write:

$$\displaystyle \d{V}{t}=a$$

Now, the volume $$V$$ of a cone is given by:

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

We know:

$$\displaystyle b=\frac{r}{h}\implies r=bh$$

Hence:

$$\displaystyle V=\frac{\pi}{3}(bh)^2h=\frac{\pi b^2}{3}h^3$$

Implicitly differentiating with respect to $$t$$, we obtain:

$$\displaystyle \d{V}{t}=\pi b^2h^2\d{h}{t}$$

And so we have:

$$\displaystyle \frac{a}{\pi b^2}=h^2\d{h}{t}$$

Letting $$h_0$$ be the initial height of the pile, and exchanging dummy variables of integration, we may integrate as follows:

$$\displaystyle \frac{a}{\pi b^2}\int_0^t \,du=\int_{h_0}^{h} v^2\,dv$$

Applying the FTOC, there results:

$$\displaystyle \frac{a}{\pi b^2}t=\frac{1}{3}\left(h^3-h_0^3\right)$$

And so, we find:

$$\displaystyle h(t)=\sqrt[3]{\frac{3a}{\pi b^2}t+h_0^3}$$

harpazo and anemone

#### harpazo

##### Pure Mathematics
Is this in any way related to differential equations?

#### MarkFL

##### La Villa Strangiato
Staff member
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Math Helper
Is this in any way related to differential equations?
Yes, we wind up with an IVP in this problem, which we solve via integration.

anemone and harpazo

#### harpazo

##### Pure Mathematics
Yes, we wind up with an IVP in this problem, which we solve via integration.
I guess that related rates is also taught in a differential equations course.

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