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

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    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:

    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.
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    Quote Originally Posted by sparacinoj View Post
    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)
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    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?
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    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by sparacinoj View Post
    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)
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    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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