Rolling a Parabola

Let $(c,c^2)$ denote the point on the graph of $y=x^2$ which then rolls to the point $(a,0)$ .

The act of rolling the graph on the $x$ -axis implies the following:

• The arc from $(0,0)$ to $(c,c^2)$ on the graph of $y=x^2$ traces out the segment from $(0,0)$ to $(a,0)$ on the $x$ -axis. Using this fact along with the arclength formula, we find a formula for $a$ : $a = \int_0^c \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx = \int_0^c \sqrt{1 + 4x^2}\,dx = \frac{1}{2}c\sqrt{1+4c^2}+\frac{1}{4}\sinh^{-1}(2c).$
• The tangent line to the graph of $y=x^2$ at $(c,c^2)$ rotates clockwise $\psi^{\circ}$ to become a horizontal line. Since the slope of the tangent line is $2c$ , the angle it makes with the horizontal, and therefore the angle the entire parabola must rotate, is given by $\psi^\circ = \tan^{-1}(2c).$

To find the equation of the graph after we roll it, we do the following, in order:

• Translate the point $(c,c^2)$ to the origin $(0,0)$ .
• Rotate the graph clockwise by $\psi^\circ$ about the origin.
• Translate the origin to the point $(a,0)$ .

Note that the result will be a rotated graph such that the original point $(c,c^2)$ moves to the point $(a,0)$ , which is exactly the same result as rolling the graph. Here are the results of each step:

• Replace $x$ with $x+c$ and replace $y$ with $y+c^2$ : $y+c^2 = (x+c)^2$
• Replace $x$ with $x\cos\psi^\circ - y\sin\psi^\circ$ and replace $y$ with $y\cos\psi^\circ + x\sin\psi^\circ$ : $y\cos\psi^\circ + x\sin\psi^\circ+c^2 = (x\cos\psi^\circ - y\sin\psi^\circ+c)^2$ However, since we know $\psi^\circ = \tan^{-1}(2c)$ we can find $\sin\psi^\circ = \frac{2c}{\sqrt{4c^2+1}}$ and $\cos\psi^\circ = \frac{1}{\sqrt{4c^2+1}}$ and the whole thing simplifies to $y^2 - y\left(\frac{x}{c}+\frac{(4c^2+1)^{3/2}}{4c^2}\right) + \frac{x^2}{4c^2}=0.$
• Replace $x$ with $x-a$ : $y^2 - y\left(\frac{x-a}{c}+\frac{(4c^2+1)^{3/2}}{4c^2}\right) + \frac{(x-a)^2}{4c^2}=0.$

By definition, this is the equation of the rolled parabola. Therefore, it has a vertical tangent at the point $(0,b)$ , and algebraically, this just means that setting $x=0$ in the equation will result in a quadratic with a repeated root at $y=b$ . Following this, we set $x=0$ : $y^2 - y\left(-\frac{a}{c}+\frac{(4c^2+1)^{3/2}}{4c^2}\right) + \frac{a^2}{4c^2}=0.$ Then since it has a repeated root at $y=b$ (and from the geometric picture, we know $b>0$ ), it follows that $b = \sqrt{\frac{a^2}{4c^2}} = \frac{a}{2c}.$ Additionally, since the root is repeated , the discriminant of the quadratic equals zero: $\left(-\frac{a}{c}+\frac{(4c^2+1)^{3/2}}{4c^2}\right)^2 - 4\cdot \frac{a^2}{4c^2} = 0$ which simplifies to $a = \frac{(4c^2+1)^{3/2}}{8c}.$

That's basically all we need! We have found that $a = \frac{(4c^2+1)^{3/2}}{8c}, \qquad b = \frac{a}{2c} = \frac{(4c^2+1)^{3/2}}{16c^2}, \qquad \psi^\circ = \tan^{-1}(2c)$ and in order to solve for $c$ , we can set the two equations we found for $a$ equal to one another: $\frac{(4c^2+1)^{3/2}}{8c} = \frac{1}{2}c\sqrt{1+4c^2}+\frac{1}{4}\sinh^{-1}(2c),$ which simplifies to $\sqrt{4c^2+1} = 2c\sinh^{-1}(2c).$

And.... well, that's as far as we can go algebraically. We probably could use some argument with the Lambert W function to find an exact form, but I honestly didn't bother with that. Instead, we can just solve it numerically to find $c \approx 0.754439780769159964454942244080289286847$

Putting this into the equations for the other quantities: $a \approx 0.98275200809309315883722546197544930437 \\ b \approx 0.65131242621589113026534167674042822633 \\ \psi \approx\, 56.4658351274523481063059796450721801295$ and putting it together: $\boxed{10a + 10b + \psi \approx 72.8064794705421909973316510322309554}$

We're given the parabola $y = x^2$ . We want to roll it such that it is always tangent to the x-axis. To that end, we have two coordinate frames: $Oxy$ which is the fixed reference frame, and $O'x'y'$ which is rigidly attached to the rolling parabola and rolls with it. To relate the two frames, let $r'$ be the position vector in $O'x'y'$ of any point, and $r$ be the position vector of the same point in $O x y$ . Note that frame $O'x'y'$ is rotated by an angle $\psi^{\circ}$ clockwise, with respect to frame $Oxy$ . Further, the contact point has coordinates ${r_1}' = (x_1, {x_1}^2)$ in frame $O'x'y'$ with a corresponding coordinate vector in $Oxy$ of $r_1 = ( L, 0 )$ . Hence, from the first observation, we have

$r = R r' + d$

where

$R = \begin{bmatrix} \cos \psi^{\circ} && \sin \psi^{\circ} \\ -\sin \psi^{\circ} && \cos \psi^{\circ} \end{bmatrix}$

Substituting the contact point, results in,

$r_1 = R {r_1}' + d$

Substracting, and re-arranging, we get,

$r = R (r' - {r_1}') + r_1$

Now we need to specify angle $\psi$ and $L$ in terms of $x_1$ . It is straight forward to find that,

$\psi = \tan^{-1} (2 x_1)$

and

$L = \displaystyle \int_{0}^{x_1} \sqrt{ 1 + 4 t^2 } dt = \frac{1}{4} ( \sec \psi \tan \psi + \ln | \sec \psi + \tan \psi | )$

The other point of interest on the parabola, is the point ${r_2}' = (x_2, {x_2}^2)$ where the tangent is perpendicular to the tangent at $(x_1, {x_1}^2)$ . It can be seen that $x_2 = \dfrac{-1}{2} \cot \psi$ .

The corresponding absolute position vector in $Oxy$ is

$r_2 = R ( {r_2}' - {r_1}' ) + r_1$

In particular, the x-coordinate of $r_2$ is

${r_2}_x = x^* = L + \cos(\psi) (x_2 - x_1) + \sin(\psi) ({x_2}^2 - {x_1}^2 )$

Note that $L$ , $x_1$ and $x_2$ are functions of the angle $\psi$ as defined above.

Finally, we want to solve for $\psi$ that will result in $x^* = 0$ . That is, we want to find the root of $x^*(\psi)$ . This can be done using Newton's method, as the derivative is readily available.

Doing so, results in

$\psi^{\circ} = 56.4658351274523^{\circ}$

$a = L = 0.982752008093093$

$b = {r_2}_y = 0.651312426215891$

This makes the answer,

$10 a + 10 b + \psi = 72.8064794705422$