Dragged

We will call the distance the bottom of the pen moves $x$ , the distance the top of the pen moves $p$ , the angle of the pen $\theta$ , and the initial angle of the pen $\alpha$ .

When we move the bottom of the pen $dx$ , it will move the tip $dp$ (parallel to the pen) and rotate the bottom of the pen $d\theta$ as shown. We are interested in finding x, so we can use $\theta$ as our parameter to get $\large x=\int_{\theta=\alpha}^{\frac{\pi}{2}} dx$ Using the diagram, this is equal to $\large \int_{\theta=\alpha}^{\frac{\pi}{2}} \csc(\theta) d \theta$ All we have to do now is find $\alpha$ . Because the tip of the pen moves a unit distance, $\large 1=p= \int_{\theta=\alpha}^{\frac{\pi}{2}} dp=\int_{\theta=\alpha}^{\frac{\pi}{2}} \cot(\theta) d \theta$ Solving the second integral for $\alpha$ ( $\alpha=\arcsin(1/e)$ ), and plugging that into the first integral gives us $x=\ln\big(e + \sqrt{e^2-1}\big)$

Suppose that the lower end of the rod has travelled a distance $X$ at the moment that the rod makes and angle of $\theta$ with the horizontal, and that the initial value of $\theta$ is $\alpha$ . The coordinates of the upper end of the rod are $(X + \cos\theta,\sin\theta)$ and we shall assume that the roughness of the table means that the upper end of the rod always moves in a direction that is parallel to the rod. Thus we must have $\left(\begin{array}{c} X' - \sin\theta \\ \cos\theta\end{array}\right) \; = \; \frac{ds}{d\theta} \left(\begin{array}{c} \cos\theta \\ \sin\theta \end{array} \right)$ where $s$ is the distance travelled by the upper end of the rod. Thus $\frac{ds}{d\theta} \; = \; \frac{X' - \sin\theta}{\cos\theta} \; = \; \frac{\cos\theta}{\sin\theta}$ and hence $\frac{dX}{d\theta} \; = \; \mathrm{cosec}\,\theta \hspace{2cm} \frac{ds}{d\theta} \; = \; \cot\theta$ Thus the distances travelled by the two ends of the rod, during the motion $\alpha \le \theta \le \tfrac12\pi$ are $X \; = \; \Big[-\ln(\mathrm{cosec}\,\theta + \cot\theta)\Big]_\alpha^{\frac12\pi} \; = \; \ln(\mathrm{cosec}\,\alpha + \cot\alpha) \hspace{2cm} 1 \; = \; s \; = \; \Big[\ln(\sin\theta)\Big]_\alpha^{\frac12\pi} \; = \; \ln(\mathrm{cosec}\,\alpha)$ Thus $\mathrm{cosec}\,\alpha = e$ , so that $\cot\alpha = \sqrt{e^2-1}$ , and so $X = \boxed{\ln\big(e + \sqrt{e^2-1}\big)}$ .

It is a little messy but I just wanted to make sure I didn't commit any mistakes.

The answer is in the question: Thanks to the dotted lines the creator provided, the answer should be approximately equal to 41/25, which is 1.64. The correct answer is 1.657454454... I know this solution is not appealing, but it gets the job done! :)

I recognized this curve as a tractrix, which I came across in an undergrad course on ODEs. Funny what sticks even after 40+ years. So I looked up the solution on Wolfram (I did solve it way back then), but it took me a while to realize that we wanted the dog walker's distance, not the horizontal distance traveled by the dog. If the dog reference is confusing, apparently it is common in German to call it a "hundekurve" link text . In my ODE book the dog was replaced by a small boat on smooth water. That was a nice way to physically satisfy the tangency condition.