# Math Help - Length Of the curve

1. ## 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.

2. 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 $x$

3. Hey thanks, All that is left to do is change the bounds and solve it..any ideas on how the bounds are changed?

Thanks

4. Hello, sikhest!

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!