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Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!

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Carl-E:
So close! 

The chain rule gives d/dt[f(g(t))] = f'(g(t))* g'(t), so when you take the derivative of sqrt(A/pi), you should be getting

{1/[2sqrt(A)*pi]}*{dA/dt}.  The derivative ofthe inner function is multiplied onto the outside of the composition - you have it inside with the inner function A. 

By the way, one of the nicer tricks with related rate problems is that you don't need to solve for the variable first... you could leave it as

A = pi*r2

and take the derivative as it stands, getting

dA/dt = 2*pi*r*dr/dt

then plug in the values and solve for dr/dt.  It gives the same result, and is a little easier to work with (no square roots and such). 


Prost! 

Jace:
Well, I first tried to use the original equation, but then I had dr/dt and r as unknown variables when trying to solve, and realized that I needed to have A in there as my known values were A and dA/dt.

It probably didn't help that I was on my 9th hour of work when trying to do that problem. At least I was messing up doing the derivative (which is a simple thing to figure out the error in)

Carl-E:
But you do have r.  The circle's area is 9 mi2, so the radius is sqrt(9/pi). 

I know, the 9th hour is not a pretty one...

de_la_Nae:
I used to be decent at math, but I haven't had to think in these terms in quite a while. My eyes are getting crossed just looking at the equations. D;

de_la_Nae:
Also I hate/love all of you for those earlier puns and wordplay.  :psyduck:

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