# Integrating the tractrix equation

• Jul 1st 2012, 02:25 PM
phys251
Integrating the tractrix equation
I'm starting out with the following:

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

Rearranging terms, I get:

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

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?
• Jul 1st 2012, 02:49 PM
pickslides
Re: Integrating the tractrix equation
Substitute $\displaystyle y = 10\sin u \implies \frac{dy}{du} = 10 \cos u$
• Jul 1st 2012, 03:02 PM
Reckoner
Re: Integrating the tractrix equation
$\displaystyle \frac{dy}{dx}=\frac{-y}{\sqrt{10^2-y^2}}$

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

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

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

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

Open if you're still stuck:

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

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

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

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

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

$\displaystyle \Rightarrow x + 10\ln|y| + \sqrt{100-y^2} - 10\ln\left(10 + \sqrt{100-y^2}\right) + C=0$
• Jul 1st 2012, 06:25 PM
phys251
Re: Integrating the tractrix equation
Quote:

Originally Posted by Reckoner
$\displaystyle \frac{dy}{dx}=\frac{-y}{\sqrt{10^2-y^2}}$

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

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

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

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

Open if you're still stuck:

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

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

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

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

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

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

Isn't step 4 under the spoiler supposed to be $\displaystyle x + 10\ln\left|\csc y - \cot y\right| + 10\cos\theta + C=0$?

EDIT: NM, I guess I forgot that $\displaystyle \ln\left|\csc y - \cot y\right| = -\ln\left|\csc y + \cot y\right|$.
• Jul 1st 2012, 06:45 PM
Reckoner
Re: Integrating the tractrix equation
Quote:

Originally Posted by phys251
NM, I guess I forgot that $\displaystyle \ln\left|\csc y - \cot y\right| = -\ln\left|\csc y + \cot y\right|$.

Yes, their equivalence follows from the identity $\displaystyle \csc^2\theta - \cot^2\theta = 1.$

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