Why does a distance and its square reach their minimum at the same point?

There is a question in my calculus textbook that asks to find a point on the parabola $y^2 = 2x$ that is closest to point $(1,4)$.

They want us to first use the distance formula, but then proceeded to square it leaving us with $d^2 = (y^2/2 - 1)^2 + (y-4)^2$.

I am following the math up to this point, where I get lost is when the textbook states "You should convince yourself that the minimum of $d^2$ occurs at the same point as the minimum of $d$, but $d^2$ is easier to work with".

I don't understand how that can be true at all, $d$ is a distance between a fixed point and another point on the graph $y^2 = 2x$. $d^2$ must be that distance scaled by itself, how can the minimum be the same then? I've worked out the math for $d$ and $d^2$ and I do get the same point but I don't understand why.

Solution 1:

If you want to figure out which of $\sqrt{2}$ and $\sqrt{3}$ is the smallest, you can do this simply by comparing $2$ and $3$ instead. More generally, if you want to make a square root as small as possible, you can do this by making what's under the square root as small as possible.

Solution 2:

This has nothing to do with derivatives, nor epsilons; it's pure logic.

If you have a function $f:\>P\to{\mathbb R}_{\geq0}$ defined on some set $P$ (like a parabola in the plane) and a strictly increasing function ${\rm sqr}:\>{\mathbb R}_{\geq0}\to{\mathbb R}_{\geq0}$ then the function $$g:={\rm sqr}\circ f:\quad P\to{\mathbb R}_{\geq0},\qquad x\mapsto{\rm sqr}\bigl(f(x)\bigr)$$ satisfies $$g(x)>g(y)\quad\Leftrightarrow\quad f(x)>f(y)\qquad\qquad\forall x,\>y\in P\ .$$ It follows that the functions $f$ and $g$ take their extrema at the same points of $P$.

Solution 3:

If $x \ge 0$ and $y \ge 0$ then $x \le y \iff x^2 \le y^2$.

So as distances are greater or equal to 0, a distance is smaller or equal to another if and only its square is smaller or equal to the other's square.

So as $\min distance$ (in terms of $x$) is the smallest possible value, it will occur at the same $x$ where the $\min distance^2$, the smallest possible value of the squares, occurs.

That's all there is to it.

Solution 4:

This isn't a rigorous explanation, but one that will hopefully intuitively make sense.

Suppose that you are minimizing the distance between some curve $f(x)$ and a point $(x_0,y_0)$. We might not know what that minimum distance is, but we know that it is some value $d$, where $d\in\mathbb{R}$. Since we are assuming that $d$ is the minimum value, we know that all other points on $f(x)$ will be further away from $(x_0,y_0)$ and thus their distances from the point will be $d+\epsilon$, where $\epsilon\in\mathbb{R^+}$.

Now we will compare our minimum distance $d$ with the other distances $d+\epsilon$ by squaring each.

$$d^2=d^2$$ $$(d+\epsilon)^2=d^2+2d\epsilon+\epsilon^2$$

Since $\epsilon$ is a positive value, we can say for certain that $d^2<d^2+2d\epsilon+\epsilon^2$, and therefore working with the square of the distance will still result in finding the minimum distance.