0
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$
    – amWhy
    Jul 2 '18 at 22:42
  • 2
    $\begingroup$ 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.) $\endgroup$ Jul 2 '18 at 22:44
  • 1
    $\begingroup$ 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. $\endgroup$
    – michael
    Jul 2 '18 at 22:53
  • $\begingroup$ Hint: if you substitute $x = \frac{10}{\tan \theta}$ into the first derivative, what do you get? $\endgroup$ Jul 2 '18 at 23:07
  • $\begingroup$ omgggg i can't believe i didn't see that before, thank u so much! $\endgroup$
    – michael
    Jul 2 '18 at 23:21
1
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ ohhhhh wow thank u! $\endgroup$
    – michael
    Jul 2 '18 at 23:21
  • $\begingroup$ You're very welcome! I'd recommend that you keep this in mind: Very different-looking expressions involving trigonometric functions are often the same. $\endgroup$ Jul 2 '18 at 23:43
  • $\begingroup$ By the way, welcome to Math.SE! After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. $\endgroup$ Jul 2 '18 at 23:55
  • 1
    $\begingroup$ ok done! tysm for the tips! $\endgroup$
    – michael
    Jul 3 '18 at 0:08
  • $\begingroup$ You're welcome (in more ways than one), Michael! $\endgroup$ Jul 3 '18 at 0:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.