# The smallest distance between any point on a curve and the parabola $y=x^2$ is 1. What is the equation of the curve?

Find the equation of the function $f(x)$, where:

• the optimal distance between any one point $P$ on the curve $y=f(x)$ and the parabola $y=x^2$ is always equal to $1$
• $f(x)>x^2$ for all $x$ (looking for the solution curve above the parabola, not below)

Here is a visual approximation of the construction on Desmos: https://www.desmos.com/calculator/pyomsrazo7

This problem arose when attempting to find the equation of motion of a ball with radius $1$ as it rolls along the concave side of the parabola $y=x^2$. Point $P$ is the center of the ball whose path of motion is the curve $y=f(x)$.

• In the Desmos construction, I use $f(x)=1.8x^2+1$, which is not the solution to the problem. I only use this equation to approximate what the actual solution curve would look like. I don't actually know the solution, hence my asking. – H. Luo Apr 26 '18 at 4:44
• – Michael Hoppe Apr 26 '18 at 5:57
• Move one unit along the normal on every point. This is called an offset curve. – Yves Daoust Apr 26 '18 at 7:00

Let a point on the curve be $(a,f(a)$ and the tangent point on the parabola be $(b,b^2)$. Two conditions must be true for all $a \in \mathbb R$

First, the distance is $1$ $$(b-a)^2 + (b^2-f(a))^2 = 1$$

Second, the slope between the two points must be normal to the tangent slope, which is $2b$ $$\frac{b^2-f(a)}{b-a} = -\frac{1}{2b}$$

Substituting $b^2-f(a) = -\frac{b-a}{2b}$, we get $$(b-a)^2 = \frac{4b^2}{4b^2+1}$$

Since $(a,f(a))$ is always on the interior of the parabola, we have $b-a > 0$ if $b > 0$, and $b-a < 0$ if $b < 0$. Therefore $b-a$ should have the same sign as $b$ and we can simplify

$$b-a = \frac{2b}{\sqrt{4b^2+1}}$$

and

$$b^2 - f(a) = -\frac{1}{\sqrt{4b^2+1}}$$

From here, you have the solution in parametric form \begin{align} x &= b - \frac{2b}{\sqrt{4b^2+1}} \\ y &= b^2 + \frac{1}{\sqrt{4b^2+1}} \end{align}

Here is a visualization. Two "bends" occur when $b = \pm \frac12$

Also note that this does not represent one continuous motion in your original proposed problem, as the ball is too large and would get stuck at the peak, where the tangent points are $b = \pm \frac{\sqrt{3}}{2}$

EDIT: A natural question you might ask is, what radius does the ball need to be to not get stuck at the peak? To find out, redo the problem with a radius of $r>0$ to get the parametric solution

\begin{align} x &= b - \frac{2br}{\sqrt{4b^2+1}} \\ y &= b^2 + \frac{r}{\sqrt{4b^2+1}} \end{align}

A simple fix is to restrict $r$ so we have only one tangent point at $x=0$. Solving $x=0$ gives $$(4b^2 + 1 - 4r^2)b^2 = 0$$

The only solution needs to be $b=0$, which means $$1-4r^2 \ge 0$$

Therefore $r \le \frac12$

Here's a solution curve with $r=0.4$

Actually, $r=\frac12$ does not result in a smooth curve, so you may not want to include it. But $\forall r < \frac12$ is valid

EDIT 2: You can also obtain an "outer" solution by switching the signs

\begin{align} x &= b + \frac{2br}{\sqrt{4b^2+1}} \\ y &= b^2 - \frac{r}{\sqrt{4b^2+1}} \end{align}

This solution has no restriction on $r$ (the curve is always smooth no matter how large the radius)

• +1. Note that the question "What radius does the ball need to be to not get stuck?" has the answer "Any radius smaller than the parabola's minimum radius of curvature". The radius of curvature is the reciprocal of curvature and thus can be computed as $$\frac{1}{k} = \frac{(1+(y^\prime)^2)^{3/2}}{|y^{\prime\prime}|}$$ which for $y=x^2$ becomes $\frac12(1+4x^2)^{3/2}$. This is clearly (and intuitively) minimized when $x=0$, where the parabola is most bendy, so that the minimum radius of curvature is $1/2$. – Blue Apr 26 '18 at 6:07

Parabola $P: (t,t^2)$.

Tangent to $P$ at $(t,t^2)$: $(y-t^2) = (\frac{\mathrm{d}}{\mathrm{d}t} t^2)(x-t)$, so $y = 2t x - t + t^2$. All we really need is that the slope is $2t$.

Normal to $P$ at $(t,t^2)$: This has slope $\frac{-1}{2t}$, the negative reciprocal of the slope of the tangent line.

Points one unit from $P$: From the point $(t,t^2)$ we go one unit along the line through that point with slope $-1/2t$. This is $(t,t^2) \pm \frac{1}{\sqrt{1+4t^2}}(2t,-1)$. Checking signs for $t$ near $0$, we want $(t,t^2) - \frac{1}{\sqrt{1+4t^2}}(2t,-1)$.

There's a small hiccup near $x = 0$, where the point along the normal is inside the envelope of circles. We need to excise this piece. We should find the three values of $t$ giving $0$ in the first coordinate.

Solving $t - \frac{2t}{\sqrt{1+4t^2}} = 0$: We have $t(1 - \frac{2}{\sqrt{1+4t^2}}) = 0$, so either $t=0$ or $\sqrt{1+4t^2} = 2$, $1+4 t^2 = 4$, $t^2 = 3/4$, so $t = \pm\sqrt{3}/2$.

Therefore, the curve you want is given parametrically by $(t,t^2) - \frac{1}{\sqrt{1+4t^2}}(2t,-1)$ for $t \in (-\infty, -\sqrt{3}/2] \cup [\sqrt{3}/2, \infty)$.

I have no reason to believe we can eliminate $t$.

Note that your ball is experiencing discontinuous "lifting" forces as $t$ jumps from $-\sqrt{3}/2$ to $\sqrt{3}/2$ as it goes from being in contact with one point on the parabola, to two widely separated points, then back to one point. ("Lifting" in the sense that this is the point providing resistance to the ball just sinking through the floor as if it were insubstantial.)

Actually, we can eliminate $t$, giving the equation implicitly: $$16 x^6 + (5-4y)^2(y^2-1)+x^4(16 y^2 - 40 y - 47)+x^2(-32y^3+6y+28) = 0$$ Given that the full red curve above is not a function, in fact there are $x$s near $0$ with three choices of $y$, we can't do better than implicit solution of degree $3$ in $y$ (and we haven't done quite that well -- we have degree $4$ in $y$).

Elimination method: Set $u^2 = 1+4t^2$ so that the original two equations are $ux = ut -2t, uy = ut^2 +2$, then eliminate $u$ and $t$ from these three equations.

doing this with parametric equations.

$y-x^2 = 0\\ \nabla (y-x^2) = (-2x, 1)$

This is vector is perpendicular to the curve.

Normalize it.

$(\frac {-2x}{\sqrt {1+4x^2}},\frac {1}{\sqrt {1+4x^2}})$

A parametric equation for your curve would be:

$x = t - \frac {2t}{\sqrt {1+4t^2}}\\ y = t^2+ \frac {1}{\sqrt {1+4t^2}}$

Why another solution? I guess it presents a more convenient parametrization.

Say the curve is $f(x,y) = 0$, and we want point on the normal to the curve, at distance $\delta$, in the direction of $f>0$. The point $(x,y)$ on curve gets shifted by $\frac{\delta}{\|\operatorname{grad} f(x,y)\|}\cdot \operatorname{grad} f(x,y)$. In our case, the curve $C\colon f=0$ can be parametrized by $t\mapsto (t,t^2)$ so get a parametrization for $C_{\delta}$ $$(t, t^2) + \delta ( \frac{-2 t}{\sqrt{1+ 4 t^2}}, \frac{1}{\sqrt{1+ 4 t^2}})$$

Let $t = \frac{\tan \theta}{2}$. We get another parametrization $$(\frac{\tan \theta}{2}, \frac{\tan^2 \theta}{4}) + \rho ( - \sin \theta, \cos \theta)$$

The curves $C_{\delta}$ are not smooth for $\rho\ge \frac{1}{2}$ and smooth for $\rho < \frac{1}{2}$. $C_{\frac{1}{2}}$ has near the point $(0, \frac{1}{2})$ the parametrization $$(\frac{\theta^3}{4} + O(\theta^5), \frac{1}{2} + \frac{3 \theta^4}{16} + O(\theta^6))$$, so near $(0, \frac{1}{2})$ it apears as the the graph of a function $s\mapsto \frac{1}{2} + c\cdot s^{4/3}$, $C^1$ but not $C^2$ around $0$.

We want $C_1$ parametrized by $$C(\theta)=(\frac{\tan \theta}{2}- \sin \theta, \frac{\tan^2\theta}{4} + \cos \theta)$$

The curve was sketched in the other answers. The first coordinate is $0$ for $\theta = -\frac{\pi}{3}, 0, \frac{\pi}{3}$. So the curve of optimal distance $1$ is parametrized as $$\tilde C(\theta)= \begin{cases} C(\theta-\frac{\pi}{3})\ \ \theta\in (-\frac{\pi}{6}, 0]\\C(\theta+\frac{\pi}{3} ) \ \ \theta\in [0,\frac{\pi}{6}) \end{cases}$$

Let $P$ be the convex region $y\ge x^2$. We are interested in different regions $$P_\delta=\{x \ | d(x, P^c) \ge \delta$$ for $\delta\ge 0$ and $$P_{\delta} = \{x \ | \ d(x,P) \le |\delta| \}$$ for $\delta\le 0$. All of these sets are convex, which is valid in general.

Indeed, take two points $x$, $x'$ in $P_{\delta}$, $\delta\ge 0$. The balls of radiu $\rho$ centered at $x$, $x'$ are contained in $P$, so their convex hull is. We see now that the distance from any point of the segment $xx'$ to the complement of this convex hull is $> \delta$, so $[x,x']\subset P_{\delta}$.

Let now $\delta<0$, $x$, $x'$ in $P_{\delta}$, that is $d(x,y)$, $d(x',y')\le \delta$ for some points $y$, $y'$ in $P$. But for any $\lambda\in [0,1]$ we have $$d(\lambda x + (1-\lambda) x', \lambda y + (1-\lambda) y') \le \lambda d(x,y) + (1-\lambda)d(x', y')$$ and so $y''\in P_{\delta}$.