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Related Rate Problem involving Theta

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data

    A balloon rises at the rate of 8 ft/s from a point on the ground 60 feet from an observer. Find the rate of change of the angle of elevation when the balloon is 25 feet above the ground.

    2. Relevant equations

    n/a

    3. The attempt at a solution

    i tried several times and erased...
    but...
    I tried d/dt 'ing tanX = h/60 and solving for dx/dt but that gave me a number way off from the answer.
    I get (secX)^2 * dx/dt = 8/60 (because dh/dt = 8)

    The answer is supposed to be about .114 radians/second.

    EDIT: I solved it myself, finally...
     
    Last edited: Nov 9, 2008
  2. jcsd
  3. Nov 9, 2008 #2
    I got it myself. Finally....
     
  4. Nov 9, 2008 #3
    For these related rate questions you shouldnt sub in the numbers given to you until after you differentiate

    suppose the distance between observer and the point on the ground is x and the height of hte balloon and is y, then

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

    now differentiate with respect to time
    [tex]\frac{d}{dt}(\tan\theta)=\frac{d}{dt}\left(\frac{y}{x}\right)[/tex]

    when you differentiate theta and y keep in mind that you have differentiate implicitly, so you will end up with terms like dy/dt or [itex]d\theta/dt[/itex]
    also x is a constant so you need not both about differentiating x with respect to t.

    Hope this helps
     
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