Example 23: Shortest distance from $(0, c)$ to $y = x^2$

Basics you need

Solution

1. A general point of the parabola

   Let $(h, k)$ be any point on $y = x^2$⁠. Then $k = h^2$⁠, so $k \ge 0$⁠, and every $k \ge 0$ occurs (at $h = \pm\sqrt{k}$⁠).

   To minimise over "all points of the parabola" we need a single variable that runs over all of them. The height $k$ does that, with allowed range $k \ge 0$⁠. The two points $(\sqrt{k}, k)$ and $(-\sqrt{k}, k)$ are mirror images in the $y$-axis, and $(0, c)$ is on that axis, so both are at the same distance from it: using $k$ loses nothing.
   A square is never negative: $k = h^2 \ge 0$⁠. Basics: interval notation

2. The distance as a function of $k$

   $$\begin{aligned} D &= \sqrt{(h - 0)^2 + (k - c)^2} \\ &= \sqrt{h^2 + (k - c)^2}. \end{aligned}$$

   Replace $h^2$ by $k$⁠:

   $$D(k) = \sqrt{k + (k - c)^2}, \qquad k \ge 0.$$
   This is the book's equation (1) followed by the substitution $h^2 = k$⁠. Now $D$ depends on one variable only, and calculus can be used.
   Distance formula (Basic 1) with $(x_1, y_1) = (0, c)$ and $(x_2, y_2) = (h, k)$⁠.

3. Differentiate by the chain rule

   Inner function $u = k + (k-c)^2$⁠, with $\frac{du}{dk} = 1 + 2(k - c)$⁠; outer function $\sqrt{u}$⁠. So

   $$D'(k) = \frac{1 + 2(k - c)}{2\sqrt{k + (k - c)^2}}.$$

   The denominator is positive for every $k \ge 0$⁠: $k + (k-c)^2$ could be 0 only if $k = 0$ and $k = c$⁠, but $c \ge \frac{1}{2}$⁠.

   We need $D'$ to find the critical points and to read off where $D$ falls and rises. Because the denominator is positive, the sign of $D'(k)$ is simply the sign of the numerator $2(k - c) + 1$ (Basic 5). That makes the rest easy.
   Chain rule: $\frac{d}{dk}\sqrt{u} = \frac{1}{2\sqrt{u}}\,\frac{du}{dk}$⁠; and $\frac{d}{dk}(k-c)^2 = 2(k-c)$⁠. Basics: chain rule

4. The critical point, and why $c \ge \frac{1}{2}$ matters

   $$D'(k) = 0 \iff 2(k - c) + 1 = 0 \iff k = \frac{2c - 1}{2}.$$

   Call this $k_0 = c - \frac{1}{2}$⁠. Since $c \ge \frac{1}{2}$⁠, $k_0 \ge 0$⁠: it is an allowed height.

   This is where the condition $c \ge \frac{1}{2}$ is used. If $c$ were smaller than $\frac{1}{2}$⁠, $k_0$ would be negative, a height the parabola never reaches, and the method would break (Step 7 says what happens then). The upper limit $c \le 5$ plays no part in the working; the answer holds for every $c \ge \frac{1}{2}$⁠.
   Solving a linear equation: $2k - 2c + 1 = 0$ gives $2k = 2c - 1$⁠. Basics: solving equations

5. First derivative test, and why it gives the least value

   The numerator is $2(k - c) + 1 = 2(k - k_0)$⁠. So

   • for $0 \le k < k_0$⁠: $D'(k) < 0$⁠, and $D$ is decreasing;
   • for $k > k_0$⁠: $D'(k) > 0$⁠, and $D$ is increasing.

   When $k_0 > 0$⁠, $D'$ changes sign from negative to positive at $k_0$⁠, so by the first derivative test $k_0$ is a point of local minima. More than that: $D$ decreases all the way from $k = 0$ to $k_0$ and increases for ever after, so $D(k) > D(k_0)$ for every other allowed $k$⁠. The least distance occurs at $k = k_0$⁠. (When $c = \frac{1}{2}$⁠, $k_0 = 0$ is the end of the range and $D$ is increasing on all of $k \ge 0$⁠, so again the least value is at $k_0$⁠.)

   The book uses the first derivative test here and, in its Note, explains why: it is easier and shorter than computing $D''$ for this square root. The sign pattern "$-$ then $+$" over the whole range is what turns a local minimum into the shortest distance overall.
   Theorem 3(ii), Section 6.4, and Theorem 1, Section 6.3. Lesson: first and second derivative tests

6. The shortest distance

   At $k_0 = \frac{2c-1}{2}$ we have $k_0 - c = -\frac{1}{2}$⁠, so

   $$\begin{aligned} D(k_0) &= \sqrt{\tfrac{2c - 1}{2} + \left(-\tfrac{1}{2}\right)^2} \\ &= \sqrt{\tfrac{4c - 2 + 1}{4}} = \frac{\sqrt{4c - 1}}{2}. \end{aligned}$$

   The nearest points of the parabola are $\left(\pm\sqrt{c - \frac{1}{2}},\ c - \frac{1}{2}\right)$⁠: one on each side of the axis (they coincide at the vertex when $c = \frac{1}{2}$⁠).

   Putting $k_0 - c = -\frac{1}{2}$ in first keeps the arithmetic short. Writing $\frac{2c-1}{2} = \frac{4c-2}{4}$ puts everything over 4, whose square root is 2.
   $\sqrt{\frac{a}{4}} = \frac{\sqrt{a}}{2}$ for $a \ge 0$⁠; here $a = 4c - 1 \ge 1$⁠.

7. A check without calculus, and the case $c < \frac{1}{2}$

   Completing the square inside the root:

   $$\begin{aligned} k + (k-c)^2 &= k^2 - (2c-1)k + c^2 \\ &= \left(k - k_0\right)^2 + c - \tfrac{1}{4}. \end{aligned}$$

   The square is least (zero) at $k = k_0$⁠, giving $D^2 = c - \frac{1}{4} = \frac{4c-1}{4}$⁠, the same answer. If instead $0 < c < \frac{1}{2}$ (not part of this question), $k_0 < 0$ is not allowed, $D$ increases on all of $k \ge 0$⁠, and the shortest distance is $D(0) = c$⁠, to the vertex. At $c = \frac{1}{2}$ the two formulas agree: $\frac{\sqrt{1}}{2} = \frac{1}{2}$⁠.

   An independent route to the same number is the best check that the calculus is right. It also explains the restriction on $c$ in the question.
   $k_0^2 = \left(c - \frac12\right)^2 = c^2 - c + \frac14$⁠, so $c^2 - k_0^2 = c - \frac14$⁠.

Every fact used, and where it comes from