Integrating the tractrix equation

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July 1st 2012, 02:25 PM
phys251

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?
July 1st 2012, 02:49 PM
pickslides

Re: Integrating the tractrix equation

Substitute $y = 10\sin u \implies \frac{dy}{du} = 10 \cos u$
July 1st 2012, 03:02 PM
Reckoner

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:

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

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

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

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

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

$\Rightarrow x + 10\ln|y| + \sqrt{100-y^2} - 10\ln\left(10 + \sqrt{100-y^2}\right) + C=0$
July 1st 2012, 06:25 PM
phys251

Re: Integrating the tractrix equation

Quote:

Originally Posted by Reckoner

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

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

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 $\ln\left|\csc y - \cot y\right| = -\ln\left|\csc y + \cot y\right|$ .
July 1st 2012, 06:45 PM
Reckoner

Re: Integrating the tractrix equation

Quote:

Originally Posted by phys251

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.