Related rates and derivatives word problem. Need help clearing a few things up.

Hey all, thanks in advance for any help you can offer.

I have completed most of the problem here, I just need some help in the right direction! Most of this problem will be trig, but I posted this in the calculus forum because it is calculus level. I will go ahead and post the whole problem in the form of an image to make things easier, with my questions at the bottom.

http://i.imgur.com/Ko20L.png

My questions are from 46.1.2 and on. I have determined that the radius at the top of the cup is twelve, and I used the concept of similar triangles to express r as h.

Could somebody please look this over?:

http://i.imgur.com/BlZfo.png

**Is this correct so far?**

Now is where my questions come in. 46.1.3 is driving me crazy. **How do I find dh/dt? ** Here is my work so far:

http://i.imgur.com/yWA15.png

__Is what I found for dv/dt correct? If it is, how do I go about finding dh/dt? __

Thanks for any help you can offer, and sorry for the wall of text!

-dan

Re: Related rates and derivatives word problem. Need help clearing a few things up.

Quote:

Originally Posted by

**dan147** My questions are from 46.1.2 and on. I have determined that the radius at the top of the cup is twelve, and I used the concept of similar triangles to express r as h.

The radius of the top of the cup is not twelve. Why are you assuming that the angle is $\displaystyle \frac\pi4$?

Re: Related rates and derivatives word problem. Need help clearing a few things up.

Because they didn't give me enough information. How do I get the angle?

Re: Related rates and derivatives word problem. Need help clearing a few things up.

Quote:

Originally Posted by

**dan147** Because they didn't give me enough information. How do I get the angle?

You don't really need the angle. You know that the volume of a right circular cone is $\displaystyle V=\frac13\pi r^2h$. You are told that the volume of the cup is $\displaystyle V = 36\pi\ \mathrm{cm}^3$ and the height is $\displaystyle h = 12\ \mathrm{cm}$. Can you solve for $\displaystyle r$?

Re: Related rates and derivatives word problem. Need help clearing a few things up.

Whoops! You sir are absolutely right! I found that 3 = r. Is the rest of this right?

http://i.imgur.com/g6otJ.png

Re: Related rates and derivatives word problem. Need help clearing a few things up.

Quote:

Originally Posted by

**dan147** I found that 3 = r. Is the rest of this right?

Well, I don't think you calculated $\displaystyle \arctan\frac14$ correctly. But again, you really don't need to find the angle. As the problem states, we have similar triangles. So

$\displaystyle \frac r3 =\frac h{12}$

$\displaystyle \Rightarrow r=\frac h4$

This is our "relation equation," as the problem calls it.

Re: Related rates and derivatives word problem. Need help clearing a few things up.

Ok, I think I am starting to get it. Thank you! How does this look?

http://i.imgur.com/SDJHw.png

Now we get down to derivatives. How do I get dh/dt? I think this is correct for dv/dt, except for missing information:

http://i.imgur.com/aLOHE.png

Thank you so much!

Re: Related rates and derivatives word problem. Need help clearing a few things up.

Quote:

Originally Posted by

**dan147** Ok, I think I am starting to get it. Thank you! How does this look?

Looks good. Now solve the equation for $\displaystyle \frac{dh}{dt}$. Note that we already know $\displaystyle \frac{dV}{dt}$ from the problem statement, and we can determine what $\displaystyle h$ is at the time when there is $\displaystyle 8\pi\ \mathrm{cm}^3$ of slushy left.

Re: Related rates and derivatives word problem. Need help clearing a few things up.

Ok, I feel like I am really close. How do I get h? This doesn't seem right...

http://i.imgur.com/5PAv8.png

Re: Related rates and derivatives word problem. Need help clearing a few things up.

Quote:

Originally Posted by

**dan147** Ok, I feel like I am really close. How do I get h? This doesn't seem right...

$\displaystyle h^3 = \frac{48V}\pi$

$\displaystyle \Rightarrow h^3 = \frac{48(8\pi)}\pi = 384$

$\displaystyle \Rightarrow h = \sqrt[3]{384} = 4\sqrt[3]6$

And we also have $\displaystyle \frac{dV}{dt} = -0.1$.

Now substitute:

$\displaystyle \frac{dh}{dt} = \frac{16}{\pi h^2}\left(\frac{dV}{dt}\right) = -\frac{1.6}{16\pi\sqrt[3]{36}}$

$\displaystyle = -\frac1{10\pi\sqrt[3]{36}}$

Re: Related rates and derivatives word problem. Need help clearing a few things up.

How does this look? The final answer doesn't really look right to me...

http://i.imgur.com/QqcSZ.png

Re: Related rates and derivatives word problem. Need help clearing a few things up.

Quote:

Originally Posted by

**dan147** How does this look? The final answer doesn't really look right to me...

That's what I got. So at that particular moment, the depth of the slushy is decreasing by about 0.010 cm/s.

Re: Related rates and derivatives word problem. Need help clearing a few things up.

What a problem! Thank you so much for your help, I really appreciate it. I thanked each one of your posts.