# Thread: Related Rates Problem...don't know what to do, help.

1. ## Related Rates Problem...don't know what to do, help.

I don't know what to do at all...I had an idea, but I got stuck, BIG TIME. Please show me what to do step by step, Thanks.

Sand falls from a hopper at a rate of 0.1 cubic meters per hour and forms a conical pile beneath. If the side of the cone makes an angle Pi/6 radians with the vertical, find the rate at which the height of the cone increases. Give your answer in terms of "h," the height of the pile in meters.

Thanks for any help.

2. Originally Posted by ilsj6
I don't know what to do at all...I had an idea, but I got stuck, BIG TIME. Please show me what to do step by step, Thanks.

Sand falls from a hopper at a rate of 0.1 cubic meters per hour and forms a conical pile beneath. If the side of the cone makes an angle Pi/6 radians with the vertical, find the rate at which the height of the cone increases. Give your answer in terms of "h," the height of the pile in meters.

Thanks for any help.
sketch a diagram ... half the cone forms a 30-60-90 triangle

ratio between r and h is ...

$\displaystyle \frac{r}{h} = \frac{1}{\sqrt{3}}$

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

volume of a cone ...

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

$\displaystyle V = \frac{\pi}{3} \left(\frac{h}{\sqrt{3}}\right)^2 h = \frac{\pi}{9}h^3$

take the time derivative of the last equation, then determine $\displaystyle \frac{dh}{dt}$ in terms of h.

3. So basically, my answer would be:

h' = sqrt{v'9/3Pi}

?

and since v' = 0.1, I would just compute a numerical answer, which would be approximately 0.3090?

4. Originally Posted by ilsj6
So basically, my answer would be:

h = SQRT(v'9/3Pi)

?

and since v' = 0.1, I would just compute a numerical answer, which would be approximately 0.3090?
no ... you're supposed to find $\displaystyle \frac{dh}{dt}$ in terms of h.

5. Originally Posted by skeeter
no ... you're supposed to find $\displaystyle \frac{dh}{dt}$ in terms of h.

I don't understand. Please elaborate...my Calculus Professor never explained this.

I know that the derivative of the last equation is:

V' = (3Pih^2)/9 = (Pih^2)/3

6. We know the volume of the cone is $\displaystyle \frac{1}{3}\pi r^{2}h$. Now, we know that there must be a constant relationship between the sides of the right triangle formed by any height $\displaystyle h$ and any radius $\displaystyle r$, since the angle between the vertical (the height) of the cone and the side of the cone is always $\displaystyle \frac{\pi}{6}=30^{\circ}$. It's a 30-60-90 triangle! Here is a diagram of the triangle created:

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

Let's plug that into the cone volume equation:

$\displaystyle V=\frac{1}{9}\pi h^{3}$. Now differentiate with respect to $\displaystyle t$:

$\displaystyle \frac{dV}{dt}=\frac{1}{3}\pi h^{2}\frac{dh}{dt}$

$\displaystyle \frac{dV}{dt}=0.1$, so plug that in and just solve for $\displaystyle \frac{dh}{dt}$, which will be in terms of $\displaystyle h$.

7. Originally Posted by Pinkk
We know the volume of the cone is $\displaystyle \frac{1}{3}\pi r^{2}h$. Now, we know that there must be a constant relationship between the sides of the right triangle formed by any height $\displaystyle h$ and any radius $\displaystyle r$, since the angle between the vertical (the height) of the cone and the side of the cone is always $\displaystyle \frac{\pi}{6}=30^{\circ}$. It's a 30-60-90 triangle! Here is a diagram of the triangle created:

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

Let's plug that into the cone volume equation:

$\displaystyle V=\frac{1}{9}\pi h^{3}$. Now differentiate with respect to $\displaystyle t$:

$\displaystyle \frac{dV}{dt}=\frac{1}{3}\pi h^{2}\frac{dh}{dt}$

$\displaystyle \frac{dV}{dt}=0.1$, so plug that in and just solve for $\displaystyle \frac{dh}{dt}$, which will be in terms of $\displaystyle h$.

So I would have this as my answer then?:

(0.1)/(Pih^2) = dh/dt

Is this correct?

8. It would be $\displaystyle \frac{.3}{\pi h^{2}}=\frac{dh}{dt}$. That is for any height $\displaystyle h$. If the question did not give to a specific height or radius at which to measure $\displaystyle \frac{dh}{dt}$, then your answer is just the general formula that I wrote above. Now, if you were told a specific $\displaystyle h$, you would plug that into the equation. If you were given a specific $\displaystyle r$, you would convert it into an $\displaystyle h$ value and then plug that into the equation. Remember, $\displaystyle h$ and $\displaystyle r$ are constantly changing; they are variables, not constants.

9. Originally Posted by Pinkk
It would be $\displaystyle \frac{.3}{\pi h^{2}}=\frac{dh}{dt}$. That is for any height $\displaystyle h$. If the question did not give to a specific height or radius at which to measure $\displaystyle \frac{dh}{dt}$, then your answer is just the general formula that I wrote above. Now, if you were told a specific $\displaystyle h$, you would plug that into the equation. If you were given a specific $\displaystyle r$, you would convert it into an $\displaystyle h$ value and then plug that into the equation. Remember, $\displaystyle h$ and $\displaystyle r$ are constantly changing; they are variables, not constants.

Thanks, I re-worked the dh/dt part and got what you wrote. Now if I wanted to find the rate at which the radius increases and write my answer in terms of "h", would I just do the same work, but plug in h=rsqrt3 for the volume equation?

Just trying to learn every type of question, so that I am prepared for any quiz.

10. Yes, you would substitute $\displaystyle h=r\sqrt 3$, differentiate with respect to $\displaystyle t$ and solve for $\displaystyle \frac{dr}{dt}$.

11. I am going to work out the radius version by myself, let me know if this is correct.

h = rSqrt3

V = 1/3Pir^2(rSqrt3)

= (Pi*r^3*Sqrt3)/3

V' = (9*Pi*Sqrt3*r^2)/9

= (Pi*Sqrt3*r^2)*(dr/dt)

Solving for dr/dt:

dr/dt = 0.1/(Pi*Sqrt3*r^2)

That is my answer for dr/dt in terms of "h."

Is this correct? I hope so.

Thanks.

12. I don't know why you suddenly switched to 9, but your work seems to check out.

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

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

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

$\displaystyle \frac{0.1}{\sqrt{3}\pi r^{2}} = \frac{dr}{dt}$

And this equation is in terms of $\displaystyle r$; you eliminated the $\displaystyle h$ term with the substitution $\displaystyle h=r\sqrt{3}$.

13. For the 9, I had this as the derivative:

(3*(3*Pi*Sqrt3*r^2))/(3^2)

so I just multiplied out the 3s to get the 9s, which canceled out.

Thank you SO MUCH for everything. I understand these problems much better now.

One last question, since my answer is in terms of "r," how would I put it in terms of "h?"

I'm done after this.

14. Just rewrite your $\displaystyle r$ term as $\displaystyle \frac{h}{\sqrt{3}}$, which would eventually simplify to:

$\displaystyle \frac{dr}{dt}=\frac{.3}{\sqrt{3}\pi h^{2}}$

You may notice that $\displaystyle \frac{dr}{dt}=\frac{1}{\sqrt{3}}\frac{dh}{dt}$ or $\displaystyle \sqrt{3}\frac{dr}{dt}=\frac{dh}{dt}$. Remember how $\displaystyle h=r\sqrt{3}$. The same ratio applies with their respective derivatives.

15. Originally Posted by Pinkk
Just rewrite your $\displaystyle r$ term as $\displaystyle \frac{h}{\sqrt{3}}$, which would eventually simplify to:

$\displaystyle \frac{dr}{dt}=\frac{.3}{\sqrt{3}\pi h^{2}}$

You may notice that $\displaystyle \frac{dr}{dt}=\frac{1}{\sqrt{3}}\frac{dh}{dt}$ or $\displaystyle \sqrt{3}\frac{dr}{dt}=\frac{dh}{dt}$. Remember how $\displaystyle h=r\sqrt{3}$. The same ratio applies with their respective derivatives.
I figured that.

Thanks again!