what is the value of X when Q is the biggest?

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- Nov 28th 2006, 06:56 AMcamherokidTrigonometric Trianle
what is the value of X when Q is the biggest?

- Nov 28th 2006, 07:02 AMThePerfectHacker
What is Q?

(You cetainly do not mean the Q continuum). - Nov 28th 2006, 10:40 PMearboth
Hello,

I assume that you want to know for which value of x has the angle $\displaystyle \theta$ a maximum.

From your sketch you see:

$\displaystyle \theta(x)=\arctan\left(\frac{420}{x}\right)-\arctan\left(\frac{380}{x}\right)$

You'll get an extremum if the first derivative of this function equals zero:

$\displaystyle \theta'(x)=\frac{1}{1+\left(\frac{420}{x} \right)^2}\cdot \left(\frac{-420}{x}\right)-\frac{1}{1+\left(\frac{380}{x} \right)^2}\cdot \left(\frac{-380}{x}\right)$

After a few and simple :D transformations you'll get:

$\displaystyle \theta'(x)=\frac{40·(159600 - x^2)}{(x^2 + 144400)·(x^2 + 176400)}$

That means: $\displaystyle \theta'(x)=0\text{ if }159600 - x^2=0$. Solve for x. I've got:

$\displaystyle x= 399.4996... \approx399.5 $

There is of course a negative solution of x but thisvalue isn't very realistic with your problem.

EB - Nov 29th 2006, 05:14 AMcamherokid
Thanks you Earboth...for the solution.

I knew how to get to the Tan formula...

but I just did not what to do with it.

$\displaystyle \theta$ is the angle...but I wrote Q. I don't know

where I can find that sign.

Great Job...! - Nov 29th 2006, 06:02 AMearboth
Hello,

maybe this will help a little bit further: In this forum you can find a link to the LaTeX-tutorial.

(If I find the link in time I'll add it here)

Have a nice time!

EB

Edit: Well, Ihope I actually did find the right link: http://www.mathhelpforum.com/math-he...rial-latex.pdf