Problem An airplane is flying at a constant speed at a constant altitude of $10$ km in a straight line directly over an observer. At a given moment the observer notes that the angle of elevation $\theta$ to the plane is $54^\circ$ and is increasing at $1^\circ$ per second. find the speed, in kilometres per hour, at which the airplane is moving towards the observer.

I'm working on a related rates problem, and the equation that i'm using to relate the two variables is $$\tan \theta= \frac{10}{x}.$$ so I can differentiate this, and after simplifying, I end up with $$\frac{dx}{dt}=-\frac{1}{10} \cdot x^2\sec^2 \theta\frac{d\theta}{dt}.$$ But if I rearrange the original equation so that it's $x=\frac{10}{\tan \theta}$, I get $$\frac{dx}{dt}=(-10\csc^2 \theta) \frac{d\theta}{dt},$$ which is different from the other derivative. Is there something quirky about rearranging the equation, or am I just blindly messing something up?

• Where does the variable $t$ come into play? Is $x$ a function of $t$? Please post the problem statement (the exercise you are working on). You're missing something, or your differentiation makes no sense. Jul 2 '18 at 22:42
• Welcome to MSE! Could you provide a little more context, like the original statement of the problem? Also, your question will be much easier to read (and therefore much more likely to get answered!) if you use MathJax formatting to format the math in the question: math.meta.stackexchange.com/questions/5020/…. (Kind of a long tutorial, but the basics are close to the top, and the whole thing is well worth reading if you have the time.) Jul 2 '18 at 22:44
• ok yeah sorry my bad, the problem goes like this: An airplane is flying at a constant speed at a constant altitude of 10 km in a straight line directly over an observer. At a given moment the observer notes that the angle of elevation θ to the plane is 54° and is increasing at 1° per second. find the speed, in kilometres per hour, at which the airplane is moving towards the observer. Jul 2 '18 at 22:53
• Hint: if you substitute $x = \frac{10}{\tan \theta}$ into the first derivative, what do you get? Jul 2 '18 at 23:07
• omgggg i can't believe i didn't see that before, thank u so much! Jul 2 '18 at 23:21

Though the two derivatives may seem different from one another, they are actually the same! This is because $$x^2=\left(\frac{10}{\tan\theta}\right)^2=\frac{100}{\tan^2\theta}=100\frac{\cos^2\theta}{\sin^2\theta}=100\frac{\csc^2\theta}{\sec^2\theta}.$$
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