need explaining

**viet** · January 7th 2008, 05:56 PM

$L = \int^{2\pi}_{0}\sqrt{(-sin t)^2 + (cos t)^2 + (\frac{1}{\pi})^2} dt$

$= \int^{2\pi}_{0}\sqrt{sin^2t+cos^2t+(\frac{1}{\pi^2 })} dt$

this is an example from the book it skipped a step, i dont know how they got the next step:

$= \int^{2\pi}_{0}\sqrt{1+\frac{1}{\pi^2}}dt$

$= 2\pi\sqrt{1+\frac{1}{\pi^2}}$ is the arc length

**Jhevon** · January 7th 2008, 06:12 PM

Originally Posted by viet

$L = \int^{2\pi}_{0}\sqrt{(-sin t)^2 + (cos t)^2 + (\frac{1}{\pi})^2} dt$

$= \int^{2\pi}_{0}\sqrt{sin^2t+cos^2t+(\frac{1}{\pi^2 })} dt$

this is an example from the book it skipped a step, i dont know how they got the next step:

$= \int^{2\pi}_{0}\sqrt{1+\frac{1}{\pi^2}}dt$

$= 2\pi\sqrt{1+\frac{1}{\pi^2}}$ is the arc length

i hope that by calc 2 you would know that $\sin^2 x + \cos^2 x = 1$

also, $\sqrt{1 + \frac 1{\pi^2}}$ is a constant, so the integral is just $\left[ \sqrt{1 + \frac 1{\pi^2}} \right]t$ and the rest follows

**viet** · January 7th 2008, 06:16 PM

its been awhile since i took calc2, thanks for the reminder Jhevon

**Jhevon** · January 7th 2008, 06:17 PM

Originally Posted by viet

its been awhile since i took calc2, thanks for the reminder Jhevon

really? what math are you doing now? (well, to be fair, you should have known that identity long before calc 2 even)

**ThePerfectHacker** · January 7th 2008, 06:31 PM

Originally Posted by Jhevon

really? what math are you doing now?

Probably a Civil Engineering course. Maybe calculating length of certain cables.

**viet** · January 7th 2008, 06:54 PM

Originally Posted by ThePerfectHacker

Probably a Civil Engineering course. Maybe calculating length of certain cables.

lol nah im not smart enough to be taking that course, im taking calc3 online. was reading the book and got a lil lost on one of the example.

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