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

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

Original problem:

Find the length of the curve.

I have gotten this far:

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

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 $\displaystyle sin^2$ and $\displaystyle cos^2$ I can change $\displaystyle tan^2+1$ to $\displaystyle sec^2$, which I can get the square root of and solve for... but what identities do I need to use? Thanks for any help.

2. Originally Posted by sparacinoj
Original problem:

Find the length of the curve.

I have gotten this far:

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

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 $\displaystyle sin^2$ and $\displaystyle cos^2$ I can change $\displaystyle tan^2+1$ to $\displaystyle sec^2$, 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:

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

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

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

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

3. 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:

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

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

L =

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

Any ideas?

4. Originally Posted by sparacinoj
Any ideas?
Everything looks good to me. Perhaps you are expected to evaluate the secants and tangents? We have

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

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

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

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

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

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

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

5. 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~