0
$\begingroup$

An airplane is flying at a constant speed at a constant altitude of $3$ km in a straight line that will take it directly over an observer at a ground level.

At a given instant the observer noted that that the angle $\theta$ from the ground to the plane is $\frac{\pi}{3}$ radians and is increasing at $\frac{1}{60}$ radians per second. Find the speed, in km/h, at which the airplane is moving towards the observer.

In my calculations, I found that the horizontal distance between the guy and the airplane is $1.732$ km. So $$\frac{d\theta}{dt} = \frac{1}{60} = \frac{d\theta}{ds} \cdot \frac{ds}{dt}$$

Is this the right way of doing it?

P.S. Also, it would be great if someone would teach me how to use formulas instead of text in this website, as I am new here.

$\endgroup$
  • $\begingroup$ Regarding the use of formulas, check out meta.math.stackexchange.com/questions/5020/…. I've edited your answer to add this. You can view it by either right clicking something and choosing "Show Math As">"TeX Commands" or choosing the "edit" option for the post and looking at the markup. $\endgroup$ – RandomUser May 23 '14 at 20:14
  • $\begingroup$ @RandomUser Thank you! $\endgroup$ – Badalyan May 23 '14 at 22:35
0
$\begingroup$

I have the following. Yes the initial horizontal distance is $s=3*\cot(\pi/3)=\sqrt{3}$. Now $s-ds=3*\cot(\pi/3+d\theta)$. So $$ {ds\over dt}=-3*\cot(\theta)^\prime {d\theta \over dt}=-3*(\cot(\theta)^2-1) {d\theta \over dt} $$ So if you are looking for the initial speed then it will be $$ {ds\over dt}=-3(1/3-1)/60=1/30 km/sec=120 km/h $$ I hope I did not make any mistake here.

$\endgroup$

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.