The Physicist and the Particle

Let $x$ be the x-coordinate of the particle. We are given that $\dfrac{dx}{dt} = 1$ . We also know that the particle's position is given by $(x, \sqrt{x})$ .

Let $A(x, \sqrt{x})$ be an arbitrary point on the particle's path such that $x < 16$ , as shown above. Point $B(16, 0)$ is the physicist's position, which does not move. Let $C$ be the foot of the perpendicular dropped from $A$ onto the x-axis. We find that $AC = \sqrt{x}$ , $BC = 16 - x$ , and $AB = \sqrt{x + (16 - x)^2}$ .

Let $\theta = \angle ABC$ . We want to find $\dfrac{d\theta}{dt}$ when $x = 4$ . We can use trigonometric relationships to write

$\tan \theta = \dfrac{AC}{BC} = \dfrac{\sqrt{x}}{16 - x} \\ \sec \theta = \dfrac{AB}{BC} = \dfrac{\sqrt{x + (16 - x)^2}}{16 - x}.$

Taking the derivative with respect to $t$ on both sides of $\tan \theta = \dfrac{\sqrt{x}}{16 - x}$ yields

$\begin{aligned} \dfrac{d}{dt} [ \tan \theta ] &= \dfrac{d}{dt} \left [ \dfrac{\sqrt{x}}{16 - x} \right ] \\ \sec^2 \theta \dfrac{d\theta}{dt} &= \dfrac{(16 - x) \left ( \frac{1}{2\sqrt{x}} \right) \frac{dx}{dt} - \sqrt{x} \left ( -\frac{dx}{dt} \right)}{(16 - x)^2} \\ \left ( \dfrac{\sqrt{x + (16 - x)^2}}{16 - x} \right )^2 \dfrac{d\theta}{dt} &= \dfrac{\frac{16 - x}{2 \sqrt{x}} + \sqrt{x}}{(16 - x)^2} \dfrac{dx}{dt} \\ \dfrac{d\theta}{dt} &= \dfrac{16 + x}{2\sqrt{x}(x + (16 - x)^2)} \dfrac{dx}{dt}. \end{aligned}$

When $x = 4$ ,

$\begin{aligned} \left. \dfrac{d\theta}{dt} \right |_{x = 4} &= \dfrac{16 + 4}{2\sqrt{4}(4 + (16 - 4)^2)} \dfrac{dx}{dt} \\ &= \dfrac{20}{2(2)(4 + 12^2)}(1) \\ &= \dfrac{5}{148}. \end{aligned}$

This means that the physicist's head is turning at a rate of $\dfrac{5}{148}$ radians per second, so $p + q = 5 + 148 = \boxed{153}.$