Not your typical car problem

The instantaneous distance of the car from the origin (shown in violet), can be represented by $y = \sqrt ( x^2 + sin^2 (\frac{\pi}{6}x))$ .

By virtue of related rates, we see that

$\large \frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}$

However, instead of $\frac{dx}{dt}$ , we are given with the rate of the car along the graph of $y = sin(\frac{\pi}{6} x)$ . If we let this given rate as $\frac{dL}{dt}$ , then we can derive a relationship that

$\large \frac{dx}{dt}=\frac{dx}{dL}\cdot \frac{dL}{dt}$ , simplifying our original equation to

$\large \frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dL}\cdot \frac{dL}{dt}$

Now, we will have to find $\frac{dx}{dL}$ . Note that $L =\large \int \:\sqrt{1+\left(y'\right)^2}dx$

thus, $\large \frac {dL}{dx} = \sqrt {1+ \frac{\pi^2}{36}cos(\frac{\pi}{6}x)}$ .

It follows that

$\large \frac {dx}{dL} = \frac{1}{\sqrt {1+ \frac{\pi^2}{36}cos(\frac{\pi}{6}x)}}$ .

Now, we have $\large \frac {dx}{dL}$ and $\frac {dL}{dt}$ . We can easily get the derivative $\frac {dy}{dx}$ and get

$\large\frac{6x+\pi \sin \left(\frac{\pi x}{6}\right)\cos \left(\frac{\pi x}{6}\right)}{6\sqrt{x^2+\sin ^2\left(\frac{\pi x}{6}\right)}}$ .

So,

$\large \frac {dy}{dt} = \frac {dy}{dx} \cdot \frac {dx}{dL} \cdot \frac {dL}{dt}$ and get $\frac{6x+\pi \sin \left(\frac{\pi x}{6}\right)\cos \left(\frac{\pi x}{6}\right)}{6\sqrt{x^2+\sin ^2\left(\frac{\pi x}{6}\right)}} \cdot \frac{1}{\sqrt {1+ \frac{pi^2}{36}cos(\frac{pi}{6}x)} }\cdot \frac { 9x^2}{3x^2 - 1} \sqrt{1+ \frac {\pi^2}{36}cos^{2} \frac{\pi}{6}x}$

Simplifying that, we get

$\frac {dy}{dx} = \large \frac{6x+\pi \:\sin \:\left(\frac{\pi \:x}{6}\right)\cos \:\left(\frac{\pi \:x}{6}\right)}{6\sqrt{x^2+\sin \:^2\left(\frac{\pi \:x}{6}\right)}}\cdot \frac{9x^2}{3x^2-1}$ .

We are asked to find out $\lim _{x\to \infty }\left(\frac{dy}{dx}\right)$ .

And here, by use of squeeze theorem, L'Hopital's rule and other related theorems,

$\large \lim _{x\to \infty }\left(\frac{6x+\pi \sin \:\:\left(\frac{\pi \:\:x}{6}\right)\cos \left(\frac{\pi \:\:x}{6}\right)}{6\sqrt{x^2+\sin ^2\left(\frac{\pi \:\:x}{6}\right)}}\cdot \frac{9x^2}{3x^2-1}\:\right) = \boxed {3}$