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Homework Help: Derivatives of Trigonometric Functions

  1. Dec 30, 2007 #1
    1.A kite 40 m above the ground moves horizontally at the rate of 3 m/s. At what rate is the angle between the string and the horizontal decreasing when 80 m of string has been let out. Answer is 0.02 m/s



    2. What I did was:

    -Drew a triangle as prescribed above
    -I found the unknown length of the ground at that instaneous time.
    -I found the derivative of the vertical height.
    -Then I took derivative of sin theta which gave:
    cos theta (d theta/t) = ry'-yr'(becomes 0 since the string is 80 m all the time) all divided by r^2
    Then I divide by cos theta to get the derivitive the angle.


    Thanks very much
     
  2. jcsd
  3. Dec 30, 2007 #2

    HallsofIvy

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    I'm not sure why you have included y and y'. You are told "A kite 40 m above the ground moves horizontally" so y= 40 and y'= 0. (You didn't actually say that y represented the height above the ground (which you should have) but I assume that from the fact that you say [itex]sin(\theta)= y/r)[/itex] and I assume (because, again, you didnt say that) that r is the hypotenuse, the length of the kites string. Since you are told "moves horizontally at the rate of 3 m/s", it might be better to use [itex]tan(\theta)= x/40[/itex], where x is the horizontal distance from directly over head. Then [itex]sec^2(\theta)(d\theta/dt)= x'/40= 3/40[/itex]. Use the fact that, at the moment in question, r= 80 and y= 40 to determine both x and [itex]\theta[/itex] at that moment.
     
  4. Dec 30, 2007 #3

    dynamicsolo

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    Incidentally, if the question is asking about the rate at which the angle is changing, the answer would have to be in radians/sec (or perhaps degrees/sec), rather than a linear velocity such as meters/sec.

    [EDIT: Just finished working this out. The given answer is rounded-off to one significant figure, but would give 0.02 rad/sec. The "mathematically exact" answer is 3/160 rad/sec. Now you get to show why...]
     
    Last edited: Dec 30, 2007
  5. Dec 30, 2007 #4
    thanks :) that helped!
     
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