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Math Help - can anyone help me with this problem !

  1. #1
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    can anyone help me with this problem !

    Question is:

    A particle is moving along the curve whose equation is:

    xy^3 = 8(1+y^2)/5

    and the x-coordinate is increasing at a rate of 6 units/s when the particle is at the point (1, 2).

    1. At what rate is the y-coordinate changing at that instant

    2. Is the particle rising or falling at that instant?

    Could you explain me with answers plzz

    Thank you..
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  2. #2
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    Did you try to cross-multiply from the initial equation? Rather, 5(xy^3) = 8(1+y^2) ?
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  3. #3
    MHF Contributor

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    Quote Originally Posted by mookie View Post
    Question is:

    A particle is moving along the curve whose equation is:

    xy^3 = 8(1+y^2)/5

    and the x-coordinate is increasing at a rate of 6 units/s when the particle is at the point (1, 2).

    1. At what rate is the y-coordinate changing at that instant
    In other words, you want to find dy/dt. Differentiating both sides of the given equation and using the chain rule, d(xy^3)/dt= (8/5)d(1+y^2)/dt, so

    y^3 dx/dt+ 3xy^2 dy/dt= (8/5)(2y)dy/dt.

    Now, you are told "x-coordinate is increasing at a rate of 6 units/s when the particle is at the point (1, 2)." so x= 1, y= 2 and dx/dt= 6. Put those into the equation and solve for dy/dt.

    2. Is the particle rising or falling at that instant?
    Well, you know, I hope, that the particle is "rising" if dy/dt> 0 and "falling" if dy/dt< 0.

    Could you explain me with answers plzz

    Thank you..
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  4. #4
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    differentiate both side with respect to 't'
    => x.(y^2).dy/dt + (y^3).dx/dt = 8/5.2y.dy/dt

    Use the given value dx/dt(1,2) = 6 in the above equation

    1.4.dy/dt + 8.6 = (32/5)dy/dt
    dy/dt(32/5-4) = 48
    dy/dt(12/5) = 48
    dy/dt = 20

    The particle is rising as the dy/dt at that point is +ve.
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  5. #5
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    Does dy/dt definitely equal 20?
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  6. #6
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    I just missed one coefficient.. thanks for pointing that out...

    Please recalculate by replacing
    x.(y^2).dy/dt
    with
    3.x.(y^2).dy/dt

    And if dy/dt > 0 ... it is rising ...
    if dy/dt < 0 ... it is decending...
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