# Length Of the curve

• May 29th 2007, 07:46 AM
sikhest
Length 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, 07:57 AM
Jhevon
Quote:

Originally Posted by sikhest
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.

use the formula: $s = \int_{1}^{ \sqrt {3}} \sqrt {1 + (y')^2}dx$

where $s$ is the length of the curve on the desired interval and $y'$ is the derivative of $y$ with respect to $x$
• May 29th 2007, 08:00 AM
sikhest
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, 08:44 AM
Soroban
Hello, sikhest!

Quote:

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

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

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

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

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

To change the limits, we have: . $\tan\theta = x$
. . When $x = 1$, we have: . $\tan\theta = 1\quad\Rightarrow\quad \theta = \frac{\pi}{4}$
. . When $x = \sqrt{3}$, we have: . $\tan\theta = \sqrt{3}\quad\Rightarrow\quad\theta = \frac{\pi}{3}$

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

Good luck!