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Speed of moving particle

  1. Sep 26, 2007 #1
    This is a related rates problem

    The coordinates of a particle moving in the metric xy-plane are differentiable functions of time t with dx/dt = 10m/sec & dy/dt = 5m/sec. How fast is the particle moving away from the origin as it passes through the point (3,-4).

    First used the Pythagorean theorem & found D (distance) from the origin, which was 5. Then I implicitly differentiated the problem d^2=x^2+y^2

    D dD/dt = x dx/dt + y dy/dt (divided out all the 2s). Then I plugged in the rates of change as outlined in the problem.
    D dD/dt = 10x + 5y.

    But at this step I must be misunderstanding something. I tried to plug (3,-4) in for x & y, & 5 in for D. However, according to the book thats incorrect. So what am I misunderstanding? The book shows 5 dD/dt = (5)(10)+(12)(5) but where are the second five in the equation & 12 coming from?
     
  2. jcsd
  3. Sep 26, 2007 #2

    Dick

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    I don't know, because I agree with your solution. The 'book' isn't always right.
     
  4. Sep 26, 2007 #3
    Ok, thank you for taking a look at the problem.

    So it would probably look

    5 dD/dt = (10)(3)+(-4)(5)

    Then just solve? Or because it is distance, would I have to use positive 4 instead of negative? I'm thinking I'd have to use positive 4, but not so sure since I'm just learning all of this.
     
  5. Sep 26, 2007 #4
    No, the negative is correct. In a related rates problem, negatives are very important. If dD/dt wound up negative, the distance would be decreasing

    The only thing I can think of is that your book may want the magnitude of the particle's velocity. dx/dt is the horizontal component and dy/dt is the vertical component. The vector sum will give you a single velocity vector, and the magnitude is the speed. In this case, the point (3,-4) is just extra information and the last part of the question isnt very clear
     
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