It says...

Find the length fo the graph and compare it to the straight-line distance between the endpoints of the graph.

f(x)=ln[sex(x)] x E [0,(pi/4)]

how would you do this?

please help

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- Oct 22nd 2007, 11:27 AMrunner07Arc Length question
It says...

Find the length fo the graph and compare it to the straight-line distance between the endpoints of the graph.

f(x)=ln[sex(x)] x E [0,(pi/4)]

how would you do this?

please help - Oct 22nd 2007, 11:34 AMgalactus
ln(

**sex**)?. That's everyone's favorite integral:). What is E?.

Do you mean arc length of $\displaystyle ln(sec(x)), \;\ [0,\frac{\pi}{4}]$

If so, $\displaystyle \frac{d}{dx}[ln(sec(x))=tan(x)$

$\displaystyle \int_{0}^{\frac{\pi}{4}}\sqrt{1+tan^{2}(x)}dx=\int _{0}^{\frac{\pi}{4}}sec(x)dx$ - Oct 22nd 2007, 11:43 AMrunner07
oops. haha sorry about that. about the E, im not sure what it means. thats what it has in my book. it doesnt exactly look like an E though, its more curved than straight. it just says f(x)= ln[sec(x)], x E [0, (pi/4)]

but i think you're right. though when you take the derivative of sec(x) it equals sec(x)tan(x) right? i dont quite understand how you got tanx from ln[sec(x)] - Oct 22nd 2007, 11:58 AMgalactus
The derivative of ln(sec(x))=tan(x).

Chain rule:

$\displaystyle \frac{1}{sec(x)}\cdot\sec(x)tan(x)=tan(x)$

The E may be 'element of'?. Is this it, $\displaystyle \in$?.

If so, they just mean that's the limits of integration.

$\displaystyle \int_{0}^{\frac{\pi}{4}}\sqrt{1+[f'(x)]^{2}}dx$

Which we found whittles down to something easy.

Once you find your solution to the integral, compare it to:

$\displaystyle \sqrt{(\sqrt{2}-1)^{2}+(\frac{\pi}{4}-0)^{2}}$ - Oct 22nd 2007, 06:20 PMrunner07
yeah! thats the E. but that makes more sense now. thank you so much!!