Example 29: Helicopter on $y = x^2 + 7$⁠, soldier at $(3, 7)$

Basics you need

Solution

1. The distance as a function of $x$

   When the helicopter is at $x$⁠, it is at $(x, x^2 + 7)$⁠. Its distance from $(3, 7)$ is

   $$\begin{aligned} &\sqrt{(x - 3)^2 + (x^2 + 7 - 7)^2} \\ &= \sqrt{(x - 3)^2 + x^4}. \end{aligned}$$
   One number, $x$⁠, describes where the helicopter is, so everything becomes a function of $x$⁠. The $+7$ and $-7$ cancel because the soldier is at the same height as the lowest point of the path.
   Distance formula (Basic 1). Basics: building the function to optimise

2. Work with the squared distance

   Let

   $$f(x) = (x - 3)^2 + x^4, \qquad x \in \R.$$

   The distance is $\sqrt{f(x)}$⁠, and by Basic 2 it is least where $f$ is least.

   Differentiating $f$ is simpler than differentiating its square root, and the answer is the same point. (The book calls this squared distance $f$ and takes the square root at the end.)
   $\sqrt{\ }$ is increasing on $[0, \infty)$ (Basic 2).

3. Differentiate and factorise

   $$\begin{aligned} f'(x) &= 2(x - 3) + 4x^3 \\ &= 4x^3 + 2x - 6 = 2(2x^3 + x - 3). \end{aligned}$$

   $x = 1$ makes $2x^3 + x - 3 = 0$⁠, so $x - 1$ is a factor:

   $$f'(x) = 2(x - 1)(2x^2 + 2x + 3).$$
   To find where $f' = 0$ we need factors. Trying small whole numbers in $2x^3 + x - 3$ quickly finds $x = 1$⁠; dividing by $x - 1$ (or comparing coefficients) gives the quadratic. Check: $(x - 1)(2x^2 + 2x + 3) = 2x^3 + x - 3$⁠.
   Chain rule for $(x-3)^2$⁠; power rule for $x^4$⁠; factor theorem (Basic 3). Basics: factorising and the discriminant

4. The only critical point is $x = 1$

   For the quadratic factor, the discriminant is

   $$2^2 - 4\cdot 2\cdot 3 = 4 - 24 = -20 < 0,$$

   so $2x^2 + 2x + 3 = 0$ has no real roots; since its leading coefficient is positive, $2x^2 + 2x + 3 > 0$ for every $x$⁠. Hence $f'(x) = 0$ only at $x = 1$⁠. $f$ is a polynomial (differentiable everywhere), and $x$ ranges over all of $\R$⁠, so there are no end points to add.

   The book makes the same two remarks: no real roots for the quadratic, and no end points. Knowing that the quadratic factor is always positive (not just non-zero) is what lets us read the sign of $f'$ in the next step.
   Discriminant test (Basic 3). Equivalently, $2x^2 + 2x + 3 = 2\left(x + \frac12\right)^2 + \frac52 > 0$⁠.

5. $x = 1$ gives the least value of $f$ on all of $\R$

   In $f'(x) = 2(x - 1)(2x^2 + 2x + 3)$⁠, the factors $2$ and $2x^2 + 2x + 3$ are always positive, so $f'$ has the sign of $x - 1$⁠:

   • $x < 1$⁠: $f'(x) < 0$⁠, $f$ is decreasing;
   • $x > 1$⁠: $f'(x) > 0$⁠, $f$ is increasing.

   So $f(x) > f(1)$ for every $x \ne 1$⁠, and the least value is

   $$f(1) = (1 - 3)^2 + 1^4 = 4 + 1 = 5.$$
   The book argues differently: with only one critical point, $\sqrt5$ is "either a maximum value or a minimum value", and since $\sqrt{f(0)} = 3 > \sqrt5$⁠, it must be the minimum. That argument needs more care (a single critical point need not be either), while the sign of $f'$ settles it completely. As a further check, $f''(x) = 2 + 12x^2 > 0$⁠, so the second derivative test also gives a local minimum at 1.
   Theorem 1, Section 6.3 (Basic 4); Theorem 4, Section 6.4. Lesson: optimisation problems

6. The nearest distance and the helicopter's position

   The nearest distance is $\sqrt{f(1)} = \sqrt5$ units, when the helicopter is at $(1, 1^2 + 7) = (1, 8)$⁠.

   Geometric check: the slope of the path at $x = 1$ is $\frac{dy}{dx} = 2x = 2$⁠, and the slope of the segment from $(3, 7)$ to $(1, 8)$ is $\frac{8 - 7}{1 - 3} = -\frac12$⁠. Since $2 \times \left(-\frac12\right) = -1$⁠, the shortest segment meets the path at a right angle, as a shortest distance to a smooth curve must.
   Perpendicular lines have slopes with product $-1$⁠.

Every fact used, and where it comes from