1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Angle of Elevation Related Rates help

  1. Sep 27, 2012 #1

    B3NR4Y

    User Avatar
    Gold Member

    I need help with this problem in my calculus book.
    An airplane at an altitude of 10,000 feet is flying at a constant speed on a line which will take it directly over an observer on the ground. If, at a given instant, the observer notes that the angle of elevation of the airplane is 60 degrees and is increasing at a rate of one degree a second, find the speed of the airplane. (Hint: tangent of theta is equal to sine of theta over cosine of theta)

    I've worked it and had tangent of theta equals "x" over 10,000. And took the derivative of each side to get secant squared of theta equals one over 10,000 times dx/dt. I solve for dx/dt with my information and get 40,000 feet per second. However the boom says 10,000*pi over 135.

    I don't get where the pi comes from.

    Sorry for no LaTeX, I'm on my phone.
     
  2. jcsd
  3. Sep 27, 2012 #2
    The full derivative of the angle term must also include the derivative of the angle, i.e., the angular speed. Whose value is given in degrees per second, but it must be converted to the radian measure.
     
  4. Sep 27, 2012 #3

    B3NR4Y

    User Avatar
    Gold Member

    Thank you. The reason I didn't include it was because it was 1 degree a second, but what you're saying is I should convert all angles to radian? I'll try that...
     
  5. Sep 27, 2012 #4

    B3NR4Y

    User Avatar
    Gold Member

    I still can't seem to get it. Can someone tell me how the book got that answer? I don't know why it would give me the hint it did if I don't really need it.
     
  6. Sep 27, 2012 #5
    What is the formula you got?
     
  7. Sep 29, 2012 #6

    B3NR4Y

    User Avatar
    Gold Member

    If you want to know what hint they gave: The tangent of theta being equal to the sine of theta over the cosine of theta.

    If you want to know the answer I got: (40,000*pi)/180
    I know it isn't right , so I didn't feel the need to simplify. The formula I get for the rate x changes with respect to time is:
    dx/dt = pi/180 * sec^2 (pi/3)*10,000
     
  8. Sep 29, 2012 #7

    B3NR4Y

    User Avatar
    Gold Member

    \begin{equation}
    \frac{dx}{dt} = \frac{\pi sec^{2}(\frac{\pi}{3}) 10,000}{180}
    \end{equation}

    \begin{equation}
    tan \theta =\frac{x}{10000}
    \end{equation}
    \begin{equation}
    \frac{d}{dt} (tan \theta) = \frac{d}{dt} (\frac{x}{10000})
    \end{equation}
    \begin{equation}
    sec^{2} \theta \frac{d\theta}{dt} = \frac{1}{10000} \frac{dx}{dt}
    \end{equation}
    \begin{equation}
    10,000 (sec^{2} (\frac{\pi}{3}) \frac{\pi}{180}) = \frac{dx}{dt}
    \end{equation}
    \begin{equation}
    10,000 (\frac{4\pi}{180})
    \end{equation}
    \begin{equation}
    \frac{40,000\pi}{180}
    \end{equation}
    \begin{equation}
    \frac{2,000\pi}{9}
    \end{equation}

    Where have I steered wrong
     
    Last edited: Sep 29, 2012
  9. Sep 29, 2012 #8

    B3NR4Y

    User Avatar
    Gold Member

    Oh, and the hint they gave was

    \begin{equation}
    tan\theta = \frac{sin\theta}{cos\theta}
    \end{equation}
     
  10. Sep 29, 2012 #9
    ## \tan \theta = \frac x {10000} ## is wrong. The correct formula is ## \tan \theta = \frac {10000} x ##.
     
  11. Sep 29, 2012 #10

    B3NR4Y

    User Avatar
    Gold Member

    So
    \begin{equation}
    sec^{2} \theta \frac{d\theta}{dt} = -\frac{10,000}{x^{2}} \frac{dx}{dt}
    \end{equation}
     
  12. Sep 29, 2012 #11
    Yes. Eliminate x from the expression and then find v. Alternatively, you could use cot x = x/10000.
     
  13. Sep 29, 2012 #12

    B3NR4Y

    User Avatar
    Gold Member

    Thanks :D I got the correct answer. You are a calculus god, have the ceremonial loin cloth.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Angle of Elevation Related Rates help
Loading...