Finding the length of a curve in multivariable calculus. Sin/Cos identities problem..

**sparacinoj** · February 19th 2009, 01:01 PM

Original problem:

Find the length of the curve.

I have gotten this far:

$L = \int_{0}^{\pi/4} \sqrt{16sin^2(4t)+16cos^2(4t)+16tan^2(t)}dt$

but now I'm stuck because I don't perfectly recall the sin/cos/tan identities... I know that if I can get rid of the $sin^2$ and $cos^2$ I can change $tan^2+1$ to $sec^2$ , which I can get the square root of and solve for... but what identities do I need to use? Thanks for any help.

**Reckoner** · February 19th 2009, 01:35 PM

Originally Posted by sparacinoj

Original problem:

Find the length of the curve.

I have gotten this far:

$L = \int_{0}^{\pi/4} \sqrt{16sin^2(4t)+16cos^2(4t)+16tan^2(t)}dt$

but now I'm stuck because I don't perfectly recall the sin/cos/tan identities... I know that if I can get rid of the $sin^2$ and $cos^2$ I can change $tan^2+1$ to $sec^2$ , which I can get the square root of and solve for... but what identities do I need to use? Thanks for any help.

A simple Google search for trig identities could have solved your problem faster than a post on a message board, but here you go:

$16\sin^24t+16\cos^24t+16\tan^2t$

$=16(\sin^24t+\cos^24t+\tan^2t)$

$=16(1+\tan^2t)$ (Pythagorean identity: $\sin^2x+\cos^2x = 1$ )

$=16\sec^2t$ (and again)

**sparacinoj** · February 19th 2009, 02:32 PM

Well, I did not know that you could still have a constant before x in the identity

, such as 4 as in my problem, and still have it equal 1.

I thought you had to use another complex identity to get rid of the constant before the variable...

HOWEVER, with that said, after taking the square root I get:

$L = \int_{0}^{\pi/4} 4sectdt$

the integral of $sec(x)$ is $ln |sec x + tan x|$ , so my final answer is:

L =

which as you can see, has gotten me the big "X"!

Any ideas?

**Reckoner** · February 19th 2009, 04:22 PM

Originally Posted by sparacinoj

Any ideas?

Everything looks good to me. Perhaps you are expected to evaluate the secants and tangents? We have

$\int_0^{\pi/4}\sqrt{16\sec^2t}\,dt$

$=4\int_0^{\pi/4}|\sec t|\,dt$

$=4\int_0^{\pi/4}\sec t\,dt$ (since $\sec t>0\text{ for }t\in[0,\pi/4]$ )

$=4\left[\ln|\sec t + \tan t|\right]_0^{\pi/4}$

$=4\left(\ln\left\lvert\sec\frac\pi4+\tan\frac\pi4\ right\rvert-\ln|\sec0+\tan0|\right)$

$=4\left(\ln\left\lvert\sqrt2+1\right\rvert-\ln|1+0|\right)$

$=4\left(\ln(\sqrt2+1)-0\right)=4\ln(\sqrt2+1)$

**sparacinoj** · February 19th 2009, 04:27 PM

Wow....

Usually it will allow you to just have the general answer without punching it into the calculator, so I never thought of evaluating the secants and tangents...

But sure enough, I put in that answer and it gave me a checkmark. I guess their program isn't completely perfect yet.

Thanks a lot, Reckoner. I had spent around an hour and a half doing the problem over and over to make sure I integrated properly, and it ends up being a server-side problem... Oh well...

Thanks again~

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