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Related Rates Question

  1. Nov 30, 2007 #1
    A plane flying with a constant speed of 4 km/min passes over a ground radar station at an altitude of 14 km and climbs at an angle of 40 degrees. At what rate, in km/min is the distance from the plane to the radar station increasing 3 minutes later?



    The only equation I am using is the cosine law:
    c^2=a^2+b^2-2abCos(Angle)

    a=14
    b=(i am not sure if its 4 or not)

    I have been doing this:
    c^2= 14^2+ b^2- 2(14)b cos(130)


    But I can't seem to get the answer, Please help me out
     
  2. jcsd
  3. Dec 1, 2007 #2
    When you say "climbs at an angle of 40 degrees," I assume you mean with respect to the ground. If this is the case, then you'll have a right triangle where the vertical displacement is 14km, call it "a," the plane's horizontal displacement is 14/tan(40), call it "b," and the plane's displacement with respect to the radar station is 14/ sin(40), call this "c." The right triangle relationship gives us:

    [tex] a^2 + b^2 = c^2 [/tex].

    You need to differentiate both sides of this equation with respect to time. When you do this, you'll have the three displacements stated above, and you'll have three rates of change (one rate of change with respect to each variable), one of which you'll be solving for. To find the two known rates of change (as opposed to the one you're solving for), you'll want to use the fact that the acceleration of the plane is zero both vertically and horizontally, and thus you can use the equation distance = rate*time. After this, you should only have the one unknown to solve for.
     
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