Minimum distance between curves

Question

We define $$f: \mathbb{R} \to \mathbb{R^3}, f(x)=\begin{pmatrix} cos(\pi \cdot x) \\ sin(\pi\cdot x) \\ 1-x^2 \end{pmatrix}$$

Problem: Calculate the minimum distance to the point of orign

My ideas:

The distance is definded by: $$d(x)=\sqrt{|(cos(π⋅x)-0)^2|+|(sin(π⋅x)-0)^2|+|((1-x^2)-0)^2}|$$ $$d(x)=\sqrt{|(cos(π⋅x))^2|+|(sin(π⋅x))^2|+|(1-x^2)^2}|$$

The minimum distance is the extrema of the first derivation, so I have to calculate $$d'(x)=0$$

As solution I get x=-1 or x=0 or x=1 (Wolfram Alpha)

But I'm not actually sure absout the solution and I don't know how to solve it without calculator.

Thanks in advance!

Answer 1

You can continue, knowing that place where maximum of its square occurs is also the same.

$$ d^2(x) = y (x) = 1 + (1-x^2)^2 = 2 - 2 x^2 + x^4 $$

Derivative =0,

$$ 4x = 4 x^3,\ \quad x(1-x)(1+x) = 0 $$

gives WA result. Next derivative finds where max/min occur for above cubic.

Answer 2

You're on the right track, you can minimize distance squared though. Try minimizing $$ d^2(x)=f(x)=(\cos\pi x)^2+(\sin \pi x)^2+(1-x^2)^2=1+(1-x^2)^2 $$ Then we have, for a minimum $$ f'(x)=-4x(1-x^2)=x^3-x=0\implies x=0,x=\pm1 $$ and $$ f''(x)=12x^2-4 $$ Meaning we have a maximum at $x=0$ and minima at $x=\pm 1$

Answer 3

A geometric rather than an analytical approach would be the following:

The x and y values of the curve describe a circular motion (as the free variable $t$ increases, not the substition $x \rightarrow t$ here), projected on the x-y-plane, this is a circle around $(0,0)$.

All the points on a circle around the origin have the same distance to the origin in the x-y-plane. Hence, only the z-component $1-t^2$ is of imprortance to the distance. As a function of $t$, the z-component is a parabola which is closest to $0$ (in fact, equal to $0$) at the points $t=\pm1$.