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What times is the bird flying horizontally/vertically?

  1. Mar 27, 2013 #1
    The flight of a bird follows the curve x=t-cos(t), y=3-2sin(t)... where 0<=t<=4pi is the time in seconds.
    a) What times is the bird flying horizontally? find the (x,y) coordinates of the corresponding points.
    b) What times is the bird flying vertically? find the (x,y) coordinates of the corresponding points.

    my work:
    i found dx/dt=1+sin(t)
    dy/dt=-2cos(t)
    and therefore dy/dx=(-2cos(t))/(1+sin(t))

    The bird flies horizontally when dy/dt=0 and dx/dt=/=0, right?
    dy/dt=0 at t=pi/2 and 3pi/2. However, dx/dt=0 when 3pi/2, so would it be just pi/2?

    Same problem for part b. The bird flies vertically when dx/dt=0 and dy/dt=/=0.
    dx/dt=0 at t=3pi/2, but dy/dt equals zero then. What does this mean for the bird's flight?
     
  2. jcsd
  3. Mar 27, 2013 #2

    SammyS

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    Indeed, what does it mean if both dx/dt = 0 and dy/dt = 0 simultaneously ?

    Also, don't forget, the flight is from t = 0 to t = 4π.
     
  4. Mar 27, 2013 #3
    Would it mean that the bird was either stationary or floating in place?

    Ah, that means I need to include every interval of pi/2 up until 8pi/2 correct?
     
  5. Mar 27, 2013 #4

    SammyS

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    Stationary --- momentarily, but stationary.
    Yes.
     
  6. Mar 27, 2013 #5
    Great, thank you!!

    Also, how do you interpret the "find the (x,y) coordinates of the corresponding points"? What do you reckon it's asking for? Just to plug in the different t values into each x and y equation initially given?
     
    Last edited: Mar 27, 2013
  7. Mar 27, 2013 #6

    SammyS

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    Certainly, if dx/dt=0 and dy/dt=0, the bird is stationary, and thus is not flying in any direction.

    To find the location at any particular time, yes, just plug the time into the expressions for x & y .
     
  8. Mar 28, 2013 #7
    ignore this please. I got it :)
     
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