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Related Rates Problem: Angle of Elevation

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data
    "An airplane flies at an altitude of 5 miles toward a point directly over an observer. The speed of the plane is 600 miles per hour. Find the rate at which the angle of elevation [tex]\theta[/tex] is changing when the angle is 30[tex]\circ[/tex]"

    • x = ground distance of plane from the observer.
    • [tex]\theta[/tex] = angle of elevation from observer to plane.

    • altitude of plane is 5 miles.
    • [tex]\frac{dx}{dt}[/tex] = -600mi/h (because the plane is travelling towards observer, distance between them decreases).

    2. Relevant equations

    the situation creates a right triangle with base x and height 5 mi; hypotenuse is not necessary in this problem. Angle of elevation is [tex]\theta[/tex], so the base equation for this problem is:

    tan [tex]\theta[/tex] = [tex]\frac{5}{x}[/tex]

    3. The attempt at a solution
    I first found what x was, since I would need x to solve my problem:

    • tan [tex]\theta[/tex] = [tex]\frac{5}{x}[/tex]

    • tan 30[tex]\circ[/tex] = [tex]\frac{5}{x}[/tex]

    • x = 5[tex]\sqrt{3}[/tex]

    Then, I found the derivative of both sides of the base equation, so my work looks like this:

    • tan [tex]\theta[/tex] = [tex]\frac{5}{x}[/tex]

    • sec2 [tex]\theta[/tex] [tex]\frac{d\theta}{dt}[/tex] = 5 (-x-2) [tex]\frac{dx}{dt}[/tex]

    • [tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{-5cos^2\theta}{x^2}[/tex] [tex]\frac{dx}{dt}[/tex]

    plugging in, the equation would turn into:

    • [tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{-5cos^230}{(5\sqrt{3})^2}[/tex] * -600 = 30

    however, the answer the book gave was 1/2 radians, and also, if you look at the way the units interact in the equation:

    [tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{miles}{miles^2}[/tex] * (miles/hour) = hour-1

    the answer I had was in "nothing per hour" when the answer should be in "radians per hour".

    What did I do wrong in this problem? Thanks in advance!
  2. jcsd
  3. Sep 20, 2009 #2
    You and I are getting the same answer, so either we both are wrong, or your book is wrong.

    Radians is a dimensionless unit, so "nothing per hour" and "radians per hour" are the same.
  4. Sep 20, 2009 #3
    yes, thanks for the confirmation!

    Yet, I also forgot to note this: when I convert the 30 (supposedly degrees, because I was working in degrees) to radians, I got .5236 which rounds to approximately 1/2 radians. However, since I got the answer in "radians per hour", I wasn't supposed to convert, which brings me to another question:

    If it doesn't matter which format (radians or degrees) you use for the angle (because tan of any format always becomes a ratio without any units), why does the change in angle always stay 30?? isn't it supposed to change according to whatever unit you use in the first place?
  5. Sep 20, 2009 #4
    Because the derivative of tan x is only sec^2 x if you are using radians. If you use degrees, you must use the chain rule which gives a factor of pi/180 or 180/pi.
  6. Sep 20, 2009 #5
    I don't think I understand. When you take the derivative of tan x don't you always get sec2 x?
  7. Sep 20, 2009 #6
    No, you were really working in radians, you just wrote "30" instead of pi/6, but your computations were really in radians.

    Your answer was 30 radians per hour.


    Explanation of derivatives in degrees: Sketch a graph of a sinusoid function with amplitude 1 and period 360, i.e. put tickmarks on the x-axis at 180 and 360, with the x intercepts at 0, 180, 360, etc. There is no way the slope at the origin is 1.

    This function is y=sin(pi x/180). Its derivative wrt x is cos(pi x/180) * pi/180. The derivative at x=0 is pi/180.

    You don't need to know this to do your problem. Just work in radians.
  8. Sep 21, 2009 #7
    Aha. 30 radians per hour is 1/2 radian per minute. I bet that's what they did.
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