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Related Rates Problem

  1. Jun 30, 2008 #1
    1. The problem statement, all variables and given/known data
    A rocket is launched vertically and is tracked by a radar station located on the ground 5 mi from the launch pad. Suppose that the elevation angle θ of the line of sight to the rocket is increasing at 3° per second when θ=60°. What is the velocity of the rocket at this instant?


    2. Relevant equations
    c2=x2+y2
    tanθ=y/x

    3. The attempt at a solution
    52=c2-y2
    25=c2-y2
    0=2c(dc/dt)-2y(dy/dt)
    0=c(dc/dt) - y(dy/dt)

    dc/dt=3°=π/60

    θ=60°=π/3

    tan π/3=y/5
    5tan π/3=y
    y=3

    c2=52+32
    c2=square root of 34

    Can I just plug in 3 for y, square root of 34 for c, and π/60 for dc/dt
     
  2. jcsd
  3. Jun 30, 2008 #2

    dynamicsolo

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    Homework Helper

    This would be OK up to here,

    But this is not the rate of change of c, but the rate of change of [tex]\theta[/tex].

    For your definitions, x = 5 and y is the altitude of the rocket. So you are looking for dy/dt (since the rocket is going straight up) and you will need to differentiate

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

    (The rate at which the hypotenuse, c, is changing is not needed in this problem.)
     
  4. Jul 2, 2008 #3
    I realized my mistakes. I now have sec2θ dθ/dt = (dy/dt)/5.
    Should I plug in dθ/dt in degrees or radians?
     
  5. Jul 2, 2008 #4

    dynamicsolo

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    Homework Helper

    In mathematical expressions, angles are expressed in radians.
     
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