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Math Help - Integrating the tractrix equation

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    Integrating the tractrix equation

    I'm starting out with the following:

    \frac{dy}{dx}=\frac{-y}{\sqrt{10^2-y^2}}

    Rearranging terms, I get:

    \frac{\sqrt{10^2-y^2}}ydy=-dx}

    Someone remind me how to integrate that left side? I know I could turn it all into tan(theta), but then what would I do with dy?
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  2. #2
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    Re: Integrating the tractrix equation

    Substitute y = 10\sin u \implies \frac{dy}{du} = 10 \cos u
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    Re: Integrating the tractrix equation

    \frac{dy}{dx}=\frac{-y}{\sqrt{10^2-y^2}}

    \Rightarrow\int\frac{\sqrt{10^2-y^2}}y\,dy=-\int dx

    Use trig substitution. Let \textstyle y = 10\sin\theta,\ -\frac\pi2\leq\theta\leq\frac\pi2. Then dy = 10\cos\theta\,d\theta and we have

    \int\frac{\sqrt{10^2-y^2}}y\,dy=-x

    \Rightarrow\int\frac{\sqrt{10^2-10^2\sin^2\theta}}{10\sin\theta}\left(10\cos\theta \right)\,d\theta=-x

    Open if you're still stuck:

    Spoiler:
    \Rightarrow x+10\int\frac{\cos^2\theta}{\sin\theta}\,d\theta=0

    \Rightarrow x+10\int\frac{1-\sin^2\theta}{\sin\theta}\,d\theta=0

    \Rightarrow x+10\int\left(\csc\theta - \sin\theta\right)\,d\theta=0

    \Rightarrow x - 10\ln\left|\csc\theta + \cot\theta\right| + 10\cos\theta + C=0

    \Rightarrow x - 10\ln\left|\frac{10}y + \frac{\sqrt{100-y^2}}y\right| + \sqrt{100-y^2} + C=0

    \Rightarrow x + 10\ln|y| + \sqrt{100-y^2} - 10\ln\left(10 + \sqrt{100-y^2}\right) + C=0
    Last edited by Reckoner; July 1st 2012 at 06:47 PM.
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    Re: Integrating the tractrix equation

    Quote Originally Posted by Reckoner View Post
    \frac{dy}{dx}=\frac{-y}{\sqrt{10^2-y^2}}

    \Rightarrow\int\frac{\sqrt{10^2-y^2}}y\,dy=-\int dx

    Use trig substitution. Let \textstyle y = 10\sin\theta,\ -\frac\pi2\leq\theta\leq\frac\pi2. Then dy = 10\cos\theta\,d\theta and we have

    \int\frac{\sqrt{10^2-y^2}}y\,dy=-x

    \Rightarrow\int\frac{\sqrt{10^2-10^2\sin^2\theta}}{10\sin\theta}\left(10\cos\theta \right)\,d\theta=-x

    Open if you're still stuck:

    Spoiler:
    \Rightarrow x+10\int\frac{\cos^2\theta}{\sin\theta}\,d\theta=0

    \Rightarrow x+10\int\frac{1-\sin^2\theta}{\sin\theta}\,d\theta=0

    \Rightarrow x+10\int\left(\csc\theta - \sin\theta\right)\,d\theta=0

    \Rightarrow x - 10\ln\left|\csc y + \cot y\right| + 10\cos\theta + C=0

    \Rightarrow x - 10\ln\left|\frac{10}y + \frac{\sqrt{100-y^2}}y\right| + \sqrt{100-y^2} + C=0

    \Rightarrow x + 10\ln|y| + \sqrt{100-y^2} - 10\ln\left(10 + \sqrt{100-y^2}\right) + C=0
    Isn't step 4 under the spoiler supposed to be x + 10\ln\left|\csc y - \cot y\right| + 10\cos\theta + C=0?

    EDIT: NM, I guess I forgot that \ln\left|\csc y - \cot y\right| = -\ln\left|\csc y + \cot y\right|.
    Last edited by phys251; July 1st 2012 at 06:33 PM.
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    MHF Contributor Reckoner's Avatar
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    Re: Integrating the tractrix equation

    Quote Originally Posted by phys251 View Post
    NM, I guess I forgot that \ln\left|\csc y - \cot y\right| = -\ln\left|\csc y + \cot y\right|.
    Yes, their equivalence follows from the identity \csc^2\theta - \cot^2\theta = 1.

    Those particular y's were supposed to be \theta's, however (that was before I back-substituted), so that was still sloppy on my part. I'll change that now.
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