Hey all, I cant seem to figure out this last question..I have the answer but I cant get the full step by step process. Please help :(

Find the length of the curve: Y = ln x on 1 less than and equal to x less than or equal to square root 3.

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- May 29th 2007, 06:46 AMsikhestLength Of the curve
Hey all, I cant seem to figure out this last question..I have the answer but I cant get the full step by step process. Please help :(

Find the length of the curve: Y = ln x on 1 less than and equal to x less than or equal to square root 3. - May 29th 2007, 06:57 AMJhevon
- May 29th 2007, 07:00 AMsikhest
Hey thanks, All that is left to do is change the bounds and solve it..any ideas on how the bounds are changed?

Thanks - May 29th 2007, 07:44 AMSoroban
Hello, sikhest!

Quote:

Find the length of the curve: .$\displaystyle y = \ln x$ .on $\displaystyle 1 \leq x \leq \sqrt{3}$

If you followed Jhevon's advice: .$\displaystyle y' = \frac{1}{x}$

Then: .$\displaystyle S \;=\;\int^{\sqrt{3}}_1\sqrt{1 + \frac{1}{x^2}}\:dx \;=\; \int^{\sqrt{3}}_1\frac{\sqrt{x^2+1}}{x}\:dx$

Let $\displaystyle x = \tan\theta\quad\Rightarrow\quad dx = \sec^2\!\theta\ d\theta$

. . and we have: .$\displaystyle S \;=\;\int\frac{\sec\theta}{\tan\theta}(\sec^2\!\th eta\ d\theta) \;=\;\int\frac{\sec^3\!\theta}{\tan\theta}\ d\theta $

To change the limits, we have: .$\displaystyle \tan\theta = x$

. . When $\displaystyle x = 1$, we have: .$\displaystyle \tan\theta = 1\quad\Rightarrow\quad \theta = \frac{\pi}{4}$

. . When $\displaystyle x = \sqrt{3}$, we have: .$\displaystyle \tan\theta = \sqrt{3}\quad\Rightarrow\quad\theta = \frac{\pi}{3}$

Therefore, the integral is: .$\displaystyle S \;=\;\int^{\frac{\pi}{3}}_{\frac{\pi}{4}} \frac{\sec^3\!\theta}{\tan\theta}\ d\theta$

*Good luck!*