MHB Sand Pile Growth Rate: 10 Seconds After Falling

[jiasyuen]

Sand falls on to horizontal ground at the rate of $$9m^3$$ per minute and forms a heap in the shape of a right circular cone with vertical angle $$60^{\circ}$$. Show that 10 seconds after the sand begins to fall, the rate at which the radius of the base of the pile is increasing is $$3^{\frac{1}{2}}(\frac{4}{\pi})^{\frac{1}{3}}m$$ per minute.

 

[MarkFL]

First, what is the volume of a right circular cone?

 

[jiasyuen]

$$\frac{1}{3}\pi r^2h$$

 

[MarkFL]

$$\frac{1}{3}\pi r^2h$$

Correct, we have:

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

Now, given that the angle of repose is $60^{\circ}$, can you express the height $h$ as a function of $r$?

 

[jiasyuen]

I've no idea. Can you show me? Thanks

 

[MarkFL]

Consider the following diagram:

View attachment 3989

Can you think of a way to relate $h$ and $r$?

 

[jiasyuen]

$$\tan 30=\frac{r}{h}$$

$$\frac{\sqrt{3}}{3}=\frac{r}{h}$$

$$r=\frac{h}{sqrt{3}}$$

Am I correct? How to proceed?

 

[MarkFL]

You are correct, although I would have written:

$$\tan\left(60^{\circ}\right)=\frac{h}{r}$$

$$h=\sqrt{3}r$$

So, substitute this into your formula for the volume of the cone so that you have the volume as a function of $r$ only. The reason we want only $r$ is because we are asked to find $$\d{r}{t}$$.

Then differentiate both sides of the formula with respect to time $t$...what do you get?

 

[jiasyuen]

$$v=\frac{\sqrt{3}}{3}\pi r^3$$

How to proceed from here? How to 'differentiate both sides of the formula with respect to time t' ?

 

[MarkFL]

$$v=\frac{\sqrt{3}}{3}\pi r^3$$

How to proceed from here? How to 'differentiate both sides of the formula with respect to time t' ?

On the right side, you will need to use the power and chain rules...what do you get?

 

[jiasyuen]

Do you mean this $$\frac{dv}{dr}=\sqrt{3}\pi r^2$$?

 

[MarkFL]

Do you mean this $$\frac{dv}{dr}=\sqrt{3}\pi r^2$$?

Close, but you have omitted one crucial piece, resulting from the application of the chain rule...you should have:

$$\d{V}{t}=\sqrt{3}\pi r^2\d{r}{t}$$

Now, in the problem statement, you are given information regarding $$\d{V}{t}$$...that is, how fast the volume of the pile of sand is changing with respect to time...can you use this to make a substitution?

 

[jiasyuen]

$$\frac{dr}{dt}=\frac{9}{\sqrt{3}\pi r^2}$$

How to proceed?

 

[MarkFL]

Okay, you need to know the value of $r$ when $t=10$...any ideas on how you can find this?

 

[jiasyuen]

I've no idea.

 

[MarkFL]

If the volume of the pile is increasing at a rate of $9\text{ m}^3$ per second, what is the volume after 10 seconds? Can you use the relationship you have between $V$ and $r$ to then find $r$?

 

[jiasyuen]

I'm stupid. Stuck at here :(

 

[MarkFL]

The volume of the pile can be written as:

$$V(t)=9t$$

Hence:

$$V(10)=9(10)=90$$

And so we must have:

$$90=\frac{\sqrt{3}\pi}{3}r^3$$

So, solve that for $r$, and then plug the result for $r$ into your value of $$\d{r}{t}$$ which is a function of $r$. After you simplify, you should get the desired result. :D

 

[jiasyuen]

$$\frac{9}{\sqrt{3}\pi\left [ (\frac{90\sqrt{3}}{\pi})^\frac{1}{3} \right ]^2}$$. How to simplify it into $$3^{\frac{1}{2}}(\frac{4}{\pi})^{\frac{1}{3}}m$$ ?

 

[MarkFL]

We neglected the fact that the time rate of change of the volume of the pile is given in meters cubed per minute, and we are to find the rate of change after 10 seconds, or 1/6 of a minute, and so we want:

$$V\left(\frac{1}{6}\right)=9\left(\frac{1}{6}\right)=\frac{3}{2}$$

Hence:

$$\frac{3}{2}=\frac{\pi}{\sqrt{3}}r^3\implies r=\left(\frac{3\sqrt{3}}{2\pi}\right)^{\frac{1}{3}}$$

Now see what you get. :D

 

[jiasyuen]

Teach me how to simplify.
$$\frac{9}{\sqrt{3}\pi \left [ (\frac{3\sqrt{3}}{2\pi})^\frac{1}{3} \right ]^2}$$

 

[MarkFL]

$$\frac{9}{\sqrt{3}\pi \left [\left(\dfrac{3\sqrt{3}}{2\pi}\right)^\frac{1}{3} \right ]^2}=\frac{3^2\cdot4^{\frac{1}3{}}}{3^{\frac{3}{2}}\cdot\pi^{\frac{1}{3}}}=3^{\frac{1}{2}}\left(\frac{4}{\pi}\right)^{\frac{1}{3}}$$