I am trying to 'draw' a two-dimensional path in the shape of a semi circle with thickness d in the xy-plane. The way I would like to do this is to have two parabolas, $f(x) = ax^2$ and $g(x)= bx^2 +cx + d$, and have the semi circle be defined by the area in between the parabolas. I've attached a terse picture if this is unclear:
I have tried a vector approach:
$\vec{F} = x\hat{x}+ax^2\hat{y}$ -- vector form of $f(x)$
$\vec{G} = x\hat{x}+(bx^2+cx+d)\hat{y}$ -- vector form of $g(x)$
Assume $a$ is known. I've imposed the following constraints:
i) $|\vec{F}-\vec{G}| = d \ \ \forall \ \ x$
ii)$f(x_0)=g(x_0) \to ax_0^2=bx_0^2 +cx_0+d$
With these constraints, i) ensures constant path width, I believe, and ii) ensures the ends of the semi circle to be 'flat' in the y direction
Clearly the constant in $g(x)$ is $d$, as it must be to satisfy i) for $x=0$. Further algebra in finding $b,c$ has not yielded me any apparent results after plotting - I'm finding that b and c are functions of x that result in $g(x)$ being non-quadratic.
Can anyone help me out? What did I do wrong? Is this even possible?
It's a theorem that two curves whose distance from each other is a nonzero constant cannot both be parabolas with $a \ne 0$, so you're never going to be able to solve your equations.
In fact, the fact that you couldn't solve them is more or less a proof of this fact.
If you'd like a proof, I can provide details, but I'm guessing you just want to know that you can stop working on this approach.
Proof that the offset curve for (at least one) parabola is not a parabola:
Look at $y = x^2$. At a point $P = (a, a^2)$ of this curve, the tangent vector is $(1, 2a)$, so the normal vector is $n = (-2a, 1)$, which has squared length $1 + 4a^2$. Thus the point $P + rn$, where $r = \frac{1}{\sqrt{1 + 4a^2}}$, is distance 1 from the parabola. The coordinates of this point are $$ (a, a^2) + r(-2a, 1) = (a(1-2r), a^2 - r) $$ Calling these $X$ and $Y$, the question becomes "are there constants $A,B,C$ with $AX^2 + BX + C = Y$?"
For $a = 0$, we have $(X, Y) = (0, 1)$.
For $a = 1$, we have $(X, Y) = ((1-2\sqrt{5}), 1-\sqrt{5})$.
For $a = -1$, we have $(X, Y) = (-(1-2\sqrt{5}), 1-\sqrt{5})$.
The first of these tells us that if there are values $A,B,C$ as needed, then $C = 1$. The second and third say that $$ A(1-2\sqrt{5})^2 + B(1-2\sqrt{5}) + C = A(-(1-2\sqrt{5})^2) + B(-(1+2\sqrt{5})) + C, $$ hence that $2B(1-2\sqrt{5}) = 0$, so $B = 0$.
But when added, they say that $$ A(1-2\sqrt{5})^2 + C + A(-(1-2\sqrt{5})^2) + C = 2(1-\sqrt{5}) $$ Dividing by two, we get $$ Au + C = u, $$ where $u = 1 - 2\sqrt{5}$.
But $C = 1$, so $$ A(u-1) = 1\\ A = \frac{1}{u-1}= \frac{1}{2\sqrt{5}}. $$
Thus the offset curve, if it's a quadratic must be $$ (**) Y = A\frac{1}{2\sqrt{5}} X^2 + 1. $$
Now look at $a = 2$; this gives us the point $$ (X, Y) = (a(1-2r), a^2 - r) = (2(1-2\sqrt{17}), 4 - \sqrt{17}).$$
If we plug these into equation (**), we don't get an equality, however, so we must conclude that the only quadratic consistent with the values for $a = -1, 0, 1$ is inconsistent with the value for $a = 2$. Hence there's no quadratic whose graph contains all distance-1 offset points for $y = x^2$.
There's a cuter visual proof, which is that if you just draw the offset curve for distance 1.5 (I think), it turns out to have a self-intersection; such a curve is clearly not the graph of any quadratic!