Thread: 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.





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?:

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:



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

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 $\displaystyle \frac\pi4$?

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

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 $\displaystyle r$?

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

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.

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


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


Thank you so much!

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.

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

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:

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

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

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

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.

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