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Math Help - How do I start this problem?

  1. #1
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    How do I start this problem?

    What is the velocity vector of a particle traveling to the right
    along the hyperbola y = 1/x with constant speed 5cm/s when the
    particle's location is (2, 1)?

    I'm not sure where to start or what formula(s) I should use.
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  2. #2
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    Quote Originally Posted by ZeroVector View Post
    What is the velocity vector of a particle traveling to the right
    along the hyperbola y = 1/x with constant speed 5cm/s when the
    particle's location is (2, 1)?

    I'm not sure where to start or what formula(s) I should use.
    Start with the speed is 5 cm/s so the square of the speed:

    v^2=\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}  {dt}\right)^2=25

    Now you know that y=1/x so:

    \dfrac{dy}{dt}=-\;\dfrac{1}{x^2}\dfrac{dx}{dt}

    etc..

    CB
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  3. #3
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    Quote Originally Posted by ZeroVector View Post
    What is the velocity vector of a particle traveling to the right
    along the hyperbola y = 1/x with constant speed 5cm/s when the
    particle's location is (2, 1)?

    I'm not sure where to start or what formula(s) I should use.
    (2, 1) does not lie on y = 1/x.

    And what coordinate system is to be used? Intrinsic coordinates?
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  4. #4
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    Sorry, I was mindlessly altering it so that I could (try and) reproduce it by myself. The coordinates my book gives is (2, 1/2).

    Edit: I'm still sort of confused. What do I do with dy/t = -1/x^2 dx/dt? And (dx/dt)^2 + (dy/dt)^2 = 25?
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  5. #5
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    Quote Originally Posted by ZeroVector View Post
    Sorry, I was mindlessly altering it so that I could (try and) reproduce it by myself. The coordinates my book gives is (2, 1/2).

    Edit: I'm still sort of confused. What do I do with dy/t = -1/x^2 dx/dt? And (dx/dt)^2 + (dy/dt)^2 = 25?
    Refer to my edit of CB's reply:
    Quote Originally Posted by CaptainBlack View Post
    [snip]
    v^2=\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}  {dt}\right)^2=25 .... (1)

    Now you know that y=1/x so:

    \dfrac{dy}{dt}=-\;\dfrac{1}{x^2}\dfrac{dx}{dt} .... (2)
    Substitute x = 2 into (2). Then substitute the result into (1) and solve for dx/dt. Substitute your value of dx/dt into (2) and solve for dy/dt. Use the values to get the velocity vector.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    Refer to my edit of CB's reply:

    Substitute x = 2 into (2). Then substitute the result into (1) and solve for dx/dt. Substitute your value of dx/dt into (2) and solve for dy/dt. Use the values to get the velocity vector.
    Doesn't the Bible say something about not numbering the equations of your posts!?

    CB
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    Doesn't the Bible say something about not numbering the equations of your posts!?

    CB
    I think the Book of Numbers says somewhere that it's OK ....
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  8. #8
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    Just to make sure, is the velocity vector:
    v = < SQRT(400/17), -0.25 * SQRT(400/17) >
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  9. #9
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    Quote Originally Posted by ZeroVector View Post
    Just to make sure, is the velocity vector:
    v = < SQRT(400/17), -0.25 * SQRT(400/17) >
    since \sqrt{400}=20 might I suggest you try simplifying the above (otherwise it looks OK)

    CB
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