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Related rates

  1. Jun 12, 2010 #1
    1. The problem statement, all variables and given/known data
    A rocket has launched straight up, and its altitude is h = 10t2
    feet after t seconds. You are on the ground 1000 feet from the launch site. The line
    of sight from you to the rocket makes an angle θ with the horizontal. By how many
    Radians per second is θ changing ten seconds after the launch?

    It's a problem form MIT OpenCourseWare, and I know the solution is given, but I did it my way and (suprisingly) it doesn't work. What did I wrong?
    The answer is 1/10

    It's not a homework question, but I didn't want to spoil your Mathematics section :)
    2. Relevant equations
    [tex]\frac{dh}{dt}=20t[/tex]
    3. The attempt at a solution
    [tex]\theta=arctan\frac{h}{1000}[/tex]
    [tex]\frac{d}{dt}(\theta=arctan\frac{h}{1000})[/tex]
    [tex]\frac{d\theta}{dt}=\frac{1}{1+\frac{h^{2}}{1000^2}}\frac{dh}{dt}[/tex]
    which is 100 ;/

    Thanks for any help :)
     
  2. jcsd
  3. Jun 12, 2010 #2

    Mark44

    Staff: Mentor

    No, [itex]d \theta/dt[/itex] is a variable quantity. At t = 10 sec. dh/dt = 200 ft/sec, h = 1000 ft, and theta = pi/4. By your calculation, d(theta)/dt is 100 when t = 10 sec.

    Your mistake is in your derivative of arctan(h/1000). You forgot to use the chain rule, so your value for the derivative is too large by a factor of 1000.
     
  4. Jun 13, 2010 #3
    I'm so stupid ! the last part (dh/dt) should be:
    [tex]\frac{d}{dt}(\frac{h}{1000})=\frac{1}{1000}\frac{dh}{dt}[/tex]
    Right ?
    And thank you for help :) I'd been already really upset; I thought the whole solution was wrong and didn't know why.

    And of course "100" was only when t=10s. I just hadn't written it.
     
  5. Jun 13, 2010 #4

    Mark44

    Staff: Mentor

    Yes, that's right.
     
  6. Jun 13, 2010 #5
    thanks :)
     
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