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One of my students arrived at a correct solution to a straightforward related rates problem, but I can't understand their method...

Problem: an airplane flies at a constant altitude of 10km and speed 0.2km/s in the direction of a stationary camera on the ground. The camera is trained upward at an angle $\theta$, aimed directly at the airplane. How quickly does $\theta$ change when the distance between the camera and airplane is 15km?

(see figure [1])

My student's solution:

\begin{align} &\theta = \frac{10}{x} = 10x^{-1} \\ &\frac{d\theta}{dt}=-10x^{-2} \frac{dx}{dt} \\ &\frac{d\theta}{dt}=\frac{(-10)(-0.2)}{15^2}=0.008888\dots . \end{align}

Did the student stumble into a reasonable application of a small angle approximation (or something like that), or is the answer a coincidence?

I am particularly baffled because, as written, $\frac{dx}{dt}$ is not -0.2 (that value is the speed over the ground, not the speed at which the hypotenuse diminishes, because the airplane flies at a constant altitude).

Xander Henderson
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Michael Megliola
  • 61

2 Answers2

2

I think it's a coincidence.

The correct solution is to write $$\sin(\theta)=\frac{10}{x} \\ \cos(\theta)\dot\theta=\frac{-10}{x^2}\dot x$$

If we let $a$ be the ground distance from the observer to the airplane (i.e the adjacent side of the triangle) we can write $$x=\sqrt{a^2+10^2}~~,~~\cos\theta=\frac{a}{x}$$ And hence $$\frac{a}{x}\dot\theta=\frac{-10}{x^2}\frac{a\dot a}{x}$$ The $a/x$ cancels from both sides so $$\dot\theta=\frac{-10}{x^2}\dot a$$

Your student, as far as I can tell, made two mistakes:

(1): They thought $\theta=10/x$

(2): They misinterpreted the question and thought $\dot x=-0.2$. (As did I!)

However, in this case, the two mistakes miraculously cancelled each other out, due precisely to the fact that the $\cos\theta=a/x$ cancels on both sides in the above equation.

Another way to see this is to write $$\theta=\operatorname{arcsin}(10/x) \\ \dot\theta=\frac{1}{\sqrt{1-(10/x)^2}}\frac{-10}{x^2}\dot x \\ \dot\theta=\frac{x}{\sqrt{x^2-10^2}}\frac{-10}{x^2}\frac{a\dot a}{x} \\ \dot\theta=\frac{x}{a}\frac{-10}{x^2}\frac{a\dot a}{x} \\ \dot \theta=\frac{-10}{x^2}\dot a$$

So we can see that the equation $\dot\theta=-\frac{10}{x^2}\cdot\text{something}$ is in fact correct. Your student happened to pick the right $\text{something}$, despite their misunderstanding.

K.defaoite
  • 13,890
-2

EDIT: THE FOLLOWING IS INCORRECT, since $\dot x\neq -0.2.$

I don't think the answer of $\frac{-10\cdot(-0.2)}{15^2}$ is correct in the first place.

See $$\sin\theta=\frac{10}{x} \\ \cos(\theta)~\dot{\theta}=\frac{-10}{x^2}\dot{x}$$ When $x=15$ , $\cos(\theta)=\frac{\sqrt{15^2-10^2}}{15}=\frac{\sqrt{125}}{15}$, since the length of the adjacent side of the triangle is $\sqrt{x^2-10^2}$. Thus $$\frac{\sqrt{125}}{15}\dot{\theta}=\frac{-10}{15^2}\cdot (-0.2) \\ \implies \dot{\theta}=\frac{2}{15\cdot\sqrt{125}}\approx 0.0119...$$

Please correct me if I made any mistakes

K.defaoite
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