# Shifrin Differential Geometry Exercise $1.2.27$ -- A Differential Equation For Bikes

## The Question

Suppose the front wheel of a bicycle follows the arclength-parametrized plane curve $$\vec{\alpha}$$. Determine the path $$\vec{\beta}$$ of the rear wheel, $$1$$ unit away. As the hint explains, the goal is a differential equation involving $$\theta$$, the angle of the front wheel with the axle of the bike, and $$\kappa$$, the curvature of $$\vec{\alpha}$$.

This question is very interesting to me, and I haven't seen a solution written up anywhere. This is sort of shocking to me as it seems like it should be a very relevant problem for e.g. autonomous driving.

## What I've Tried

I've only been able to make minimal progress.

The hint tells us to write $$\vec{\alpha} - \vec{\beta}$$ in terms of $$\theta$$, $$\vec{T}$$ (i.e. $$\vec{\alpha}'$$), and $$\vec{N}$$ (i.e. $$\frac{\vec{\alpha}''}{\kappa}$$). We obviously have $$\| \vec{\alpha} - \vec{\beta}\|^2 = 1$$. Differentiating, we obtain $$(\vec{\alpha}' - \vec{\beta}') \cdot (\vec{\alpha} - \vec{\beta}) = 0$$ That is: $$(\vec{T} - \vec{\beta}') \cdot (\vec{\alpha} - \vec{\beta}) = 0$$ Differentiating again, we obtain $$(\kappa \vec{N} - \vec{\beta}'') \cdot (\vec{\alpha} - \vec{\beta}) + (\vec{T} - \vec{\beta}') \cdot (\vec{T} - \vec{\beta}') = 0$$ Now it seems to me we should have $$\vec{T} \cdot \vec{\beta}' = \|\vec{\beta}'\| \cos \theta$$ So we can expand $$(\vec{T} - \vec{\beta}') \cdot (\vec{T} - \vec{\beta}') = 1 - 2\|\beta'\| \cos\theta + \|\vec{\beta}'\|^2$$ And that's as far as I've gotten. It seems like I'm going about this all wrong. In particular, I have no idea what to do with the derivatives of $$\vec{\beta}$$. If only $$\vec{\beta}$$ were arclength parametrized I feel like I could make some progress, but I don't think there is any reason it should be. The only thing I can think is that we should have $$\vec{\alpha} - \vec{\beta} = \lambda \vec{\beta}'$$ for some $$\lambda$$ that could depend on the arclength of $$\vec{\alpha}$$. I didn't push too far in this direction, though, since it required introducing yet another unknown.

I thought of yet another line of attack. Since $$\|\vec{\alpha} - \vec{\beta}\| = 1$$, we can say that $$\|\vec{\alpha} - \vec{\beta}\|$$ is just $$\vec{T}$$ rotated by $$\theta$$, i.e. $$\vec{\alpha} - \vec{\beta} = \cos (\theta) \vec{T} + \sin(\theta) \vec{N}$$ With this expression I've gone as far as the hint suggests, but I don't see what to do next.

What am I missing?

If this post summons Ted Shifrin, and he'd rather answers to his textbook questions not be given out, I'd be happy to delete this question and post it as a reference request instead. I really am shocked I haven't been able to find this problem written about anywhere. I'm guessing it's because I'm bad at searching the literature, not because it actually hasn't been written about.

• I haven't seen this written up anywhere, either. I will tell you to follow your second idea. There's always one thing that is obvious to do in differential geometry. Do it. And obviously there's another obvious thing to do once you've done that. Every single problem in this section requires that. And, yes, I'd rather not have a solution posted, but I'm not opposed to giving a bit more of a hint if you do the obvious things and get stuck. Remember that you should not expect to be able to solve the differential equation in general. Feb 19 at 23:44
• @TedShifrin Thank you for the encouragement. I saw several directions to go in with this problem and none seemed very promising. By "second idea" do you mean $\vec{\alpha}' - \vec{\beta}' = \lambda \vec{\beta}'$ or $\vec{\alpha} - \vec{\beta} = \cos(\theta) \vec{T} - \sin(\theta) \vec{N}$ or some combination thereof? Anyway, very strange that this easy to state and highly applicable problem is so scarcely written about. The best I could find is a physicsforum thread, in which the replies greatly simplify the problem and call it solved. Do the pairs $\vec{\alpha}, \vec{\beta}$ even have a name? Feb 20 at 2:22
• Yes, the trigonometric one. Then, as I suggested, you must apply the first rule of differential geometry. Differentiate. What you have to think about is what you know about $\beta’$ from the physics of the situation. Feb 20 at 2:25
• Is the physics saying $\|\vec{\beta}'\| = \|\vec{\alpha}'\| = 1$? That seems possible to me, but I can't convince myself that it's true. The question amounts to whether or not the front wheel and the back wheel spin at the same speed. Again, seems plausible, but I can't make a convincing argument. Feb 20 at 4:17
• Yes, thinking about it some more it's obviously false (a tight turn quickly disproves it). I think I get it now. I found a way to come up with two expressions for $\|\vec{\beta}'\|$ in terms of $\kappa$ and $\theta$. Feb 20 at 4:39

## 1 Answer

With help from Ted Shifrin, I was able to solve the problem. Here's a sketch of the solution.

The first observation is that $$\|\vec{\alpha} - \vec{\beta}\| = 1$$ since the bike has length $$1$$. Since $$\vec{\alpha}$$ is arc-length parametrized, $$\|\vec{T}\| = \|\vec{\alpha}\| = 1$$, so $$\vec{T}$$ and $$\vec{\alpha} - \vec{\beta}$$ must be related by a rotation. The angle of this rotation is exactly the angle $$\theta$$ that the front wheel makes with the frame of the bike (up to a sign depending on how you define things). Applying the rotation to $$\vec{T}$$ gives you an expression for $$\vec{\alpha} - \vec{\beta}$$ in the $$\{\vec{T} , \vec{N}\}$$ basis in terms of $$\theta$$. Differentiate this expression for $$\vec{\alpha}- \vec{\beta}$$ to obtain an expression for $$\vec{\beta}'$$ in the $$\{\vec{T}, \vec{N}\}$$ basis in terms of $$\theta$$ and $$\kappa$$.

On the other hand, $$\vec{\beta}'$$ must point in the same direction as $$\vec{\alpha} - \vec{\beta}$$ since the rear wheel is fixed to point in the same direction as the frame of the bike. This allows us to write $$\vec{\beta}' = \lambda(\vec{\alpha} - \vec{\beta})$$ from which it easily follows that $$\lambda = \vec{\beta}' \cdot (\vec{\alpha} - \vec{\beta})$$ We have expressions for both vectors on the RHS in the $$\{\vec{T}, \vec{N}\}$$ basis, so we can easily calculate this dot product in terms of $$\theta$$ and $$\kappa$$.

Knowing $$\lambda$$, we have yet another expression for $$\vec{\beta}'$$. Since $$\{\vec{T}, \vec{N}\}$$ is a basis, we may equate the components of each expression. After just a little algebra, you should obtain $$\kappa = \theta ' + \sin\theta$$ an expression for $$\kappa$$ entirely in terms of $$\theta$$ and its derivatives.

• Wonderful. This is a very nice problem with a surprisingly elegant solution. Feb 22 at 15:29