Minimum distance between parabolas

Question

Is there a simple way to find the minimum distance between two parabolas?

For example, between y=-0.1(x-17)² + 8.6 {7.726 < x < 19.134} and y=-0.12(x-17.5)² + 6.2 {10.313 < x < 18.829}

See graph of the two parabolas here (the exponential function simply determines the restricted domain for the parabolas)

Context: I have to ensure that there is a minimum of 1.8 units between these two parabolas, to ensure the "path" is at least 1.8 metres wide always.

Attempt: I thought about trying to find the closest distance visually then finding the normal and subbing in for x and y at that specific point, but that doesn't seem very scientific and I'm a bit stuck.

Thanks in advance

Answer 1

Hint:

Minimize

$$(x'-x'')^2+((a'x^2+b'x+c')-(a''x''^2+b''x''+c''))^2.$$

The stationary points are given by

$$(x'-x'')+(2a'x'+b')((a'x'^2+b'x'+c')-(a''x''^2+b''x''+c''))=0,\\(x''-x')+(2a''x''+b'')((a'x'^2+b'x'+c')-(a''x''^2+b''x''+c''))=0.$$

By adding the two equations and as the parabolas do not intersect in the given domain, we have one linear equation

$$2a'x'+b'+2a''x''+b''=0.$$

We can eliminate $x''$ by drawing it from the latter and plugging it in one of the formers. This results in a cubic equation, having one or three real solutions.

But that's not all. As the curves are bounded, the minimum might occur at an endpoint. From the figure we can see that there are two candidate endpoints: the left endpoint of the lower parabola, and the right endpoint of the upper one.

Knowing $x'$ or $x''$, the stationarity equations are still cubic.

Answer 2

In general, the shortest line between two curves is the one that is normal to both of them, unless this line is at the boundary of the curves.

Let $y_1(x)=a_1(x-b_1)^2+c_1$ and $y_2(x)=a_2(x-b_2)^2+c_2$.

Then the tangents at $x$ are $y_i'(x)=2a_i(x-b_i)$ and the normal lines are $$\mathbf{r}_i^N(x,t)=\begin{pmatrix}x\\y_i(x)\end{pmatrix}+t\begin{pmatrix}1\\-1/y_i'(x)\end{pmatrix}$$

The requirements on $x_i,t$ are that $$\mathbf{r}_1^N(x_1,t)=\begin{pmatrix}x_2\\y_2(x_2)\end{pmatrix},\quad -1/y_1'(x_1)=-1/y_2'(x_2)$$ These three equations, when solved, give $$x_1+t=x_2,\\ a_1(x_1-b_1)^2+c_1-\frac{t}{2a_1(x_1-b_1)}=a_2(x_2-b_2)^2+c_2,\\ a_1(x_1-b_1)=a_2(x_2-b_2)$$

Solving gives $$x_1=8.23107, x_2=10.1926, t=1.96149,\\ x_1=27.4119, x_2=26.1765, t=-1.23531,\\ x_1=15.3571, x_2=16.1309, t=0.77382$$

The first and third are within the given ranges; of these the distance of 2.26 is minimised by the first.

Answer 3

This is not an answer. I just want to show you the graphical solution to your problem. This diagram was drawn to scale in GeoGebra. For easy viewing please download it to your computer.

The restricted domain is $EGKJ$. The absolute minimum of the non-restricted area is $MF$. The absolute minimum for the restricted are is $JR$. Since this is equal to 2.1569, a gap of 1.8 is guaranteed. The other distances are given for comparison only.

For the non-restricted case, there are two global minimums on either sides of the axis of the inner parabola (e.g. $MF$ and $QH$). One of them is always the absolute minimum. There is a construction to determine these two values. However, I could not prove it yet. Nevertheless, I hope to post an answer at a later date.

If neither of the above mentioned minimums occurs in the restricted area, then, the minimum inside this area is to be found at one of the boundaries as shown in the diagram. But, the minimum is not always the boundary line. In this diagram $JK$ is the boundary line, but the minimum is $JR$.

Can you please share your results with us, if you have already solved the problem?