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I had a thought. Let say you're driving along a parabola shaped road (with equation y = x^2 as viewed from the sky). You're driving from (-10, 100) driving along to (10, 100) at a constant speed so the length of curve you have driven along per unit time is constant.

What happens to the steering wheel as you drive along?

We can be fairly sure that it is always between left and straight. As you approach the infinities it would be travelling straight and since the derivative at x = 0 is 0, I feel the same must be true there. But what happens around x = 0, especially as a function of x or time t, where t = 0 when x = 0?

Was wondering if anyone had heard of this problem before or had any insights. Apologies if this is not in the right layout.

Leonhard Euler
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1 Answers1

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The radius of curvature is a helpful concept here. It measures how sharply the car has to be turned. The reciprocal of it is the curvature defined as

$$\kappa=\frac{d\theta}{ds},$$

where $d\theta$ is the change of direction of the tangent and $ds$ is the distance traveled. So the curvature measures the angle turned per unit distance traveled. Given the curve $y=x^2$, we have

$$\frac{dy}{dx}=\tan\theta=2x.$$

Therefore

$$d\theta=d\arctan(2x)=\frac{2dx}{1+4x^2},$$

and

$$ds=\sqrt{dx^2+dy^2}=\sqrt{1+4x^2}\,dx.$$

Then the curvature as a function of $x$ is given by

$$\kappa=\frac{d\theta}{ds}=\frac{2}{(1+4x^2)^{3/2}}.$$

The sharpest turn is at $x=0$. The direction of the turn is always a left, as one drives the car from $(-10,100)$ to $(10,100)$. Since the speed $v=ds/dt$ is constant, the angle turned per unit time, i.e., the angular velocity $\omega=d\theta/dt$, is proportional to the curvature $\kappa$. They satisfy $\omega=\kappa v$. If you want to do physics here, the normal acceleration $\,a_\perp=\omega v=\kappa v^2=v^2/\rho$, where $\,\rho=1/\kappa\,$ is the radius of curvature. The tangential acceleration $\,a_\parallel=0\,$ because the speed $\,v=\mathrm{const}$.

Zhuoran He
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