Calc IV: Functions of Several Variables (Chain Rule)

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April 24th 2008, 06:32 AM
googoogaga

Calc IV: Functions of Several Variables (Chain Rule)

Hello,
I am having a hard time solving this problem. I have no clue. None whatsover. Thanks in advance for your help.

Volume & Surface Area
The radius of a right circular cone is increasing at a rate of 6 inches per minute, and the height is decreasing at a rate of 4 inches per minute. What are the rates of change of the volume and surface area when radius is 12 inches and height is 36 inches?
April 24th 2008, 06:39 AM
Moo

Hello,

$V=(\pi r^2)\frac h3=\frac{\pi}{3} r^2 h$

Where r is the radius and h the height.

You're looking for $\frac{dV}{dt}$

By using the multiplication rule, you have :

$\frac{dV}{dt}=\frac{\pi}{3} \left( \frac{dh}{dt} r^2+2h \frac{dr}{dt} r \right)$

From the text, we know that $\frac{dh}{dt}=-4, \ \frac{dr}{dt}=6, \ r=12, \ h=36$

So it should give you :

$\frac{dV}{dt}=\frac{\pi}{3} (-4*144+2*36*6*12)=\dots$
April 24th 2008, 06:41 AM
XIII13Thirteen

I believe your formula for the cone is $\frac{1}{3}pi r^2h$ for the cone. Would you just not take the derivative as you insert your variables. I didn't know there was a Cal IV I thought they only went to III (just an aside to myself :D)
April 24th 2008, 06:43 AM
XIII13Thirteen

Yeah, listen to Moo, as he understands volume formula much better than myself
April 24th 2008, 06:47 AM
Moo

For the area, it's the sum of the area of the base and the lateral area.

The area of the base is given by $\pi r^2$

The lateral area is given by $\pi r \sqrt{h^2+r^2}$ (I can try to explain it, but later...)

Hence $A=\pi r^2+\pi r \sqrt{h^2+r^2}$

Once again, you're looking for $\frac{dA}{dt}$

Can you do it ?
April 24th 2008, 08:41 AM
googoogaga

hello,
I am not so sure about the second part (lateral area) and the dA/dt what is the dA/dt? can you explain in details please because I am little confused on the second part Thanks!
April 24th 2008, 08:44 AM
Moo

It means "the derivative of A in respect with t".
April 24th 2008, 09:56 AM
googoogaga

How do you have take the derivative , I forgot whether I use the chain rule or
other rules, if so how do I do it?
April 24th 2008, 12:08 PM
XIII13Thirteen

Quote:

Originally Posted by googoogaga

How do you have take the derivative , I forgot whether I use the chain rule or
other rules, if so how do I do it?

Are you needing reminders on how to take a derivative in general or how to take a partial derivative? With partial derivatives you want to take the derivative in the respect to the variable in question. An example would be like

$x^2+y^2+3$ in respect of x would be $2x$

Check out some rules online if your still a bit confused on it
April 28th 2008, 08:22 PM
googoogaga

(Worried)I am so sorry for the late thanks I just got so caught up with school work, you know it's time for exams. Thanks again! You clarified it!
April 28th 2008, 11:32 PM
Moo

Quote:

You clarified it!

You know, this is the most important thing... Sincerely