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.
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.