1. ## Arc 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?

2. ln(sex)?. That's everyone's favorite integral. What is E?.

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

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

$\int_{0}^{\frac{\pi}{4}}\sqrt{1+tan^{2}(x)}dx=\int _{0}^{\frac{\pi}{4}}sec(x)dx$

3. 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)]

4. The derivative of ln(sec(x))=tan(x).

Chain rule:

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

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

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

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

Which we found whittles down to something easy.

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

$\sqrt{(\sqrt{2}-1)^{2}+(\frac{\pi}{4}-0)^{2}}$

5. yeah! thats the E. but that makes more sense now. thank you so much!!