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Math Help - Gradient and parametric equations help!

  1. #1
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    Gradient and parametric equations help!

    Could anyone help me with the following problems?:

    a)Find the gradient of the curve:

    8x^2/4x^2-3y^3 = 3y at (3,2)

    b) The curve C has parametric equations x=4cos2t and y=3sint for t being between -pi/2 and pi/2.
    A is the point (2,3/2) and lies on C.
    The normal to C at A is 6y-16x+23. The normal at A cuts C again at point B. Find the y coordinate of point B.

    Thank you so much if you can help me
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  2. #2
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    Hello, Confuzzled!

    Can't read the first one . . .


    b) The curve C has parametric equations: .x = 4cos2t and y = 3sin t
    for -π/2 < t < π/2.

    A is the point (2,3/2) and lies on C.

    The normal to C at A is: .6y - 16x + 23 .= .0.

    The normal at A cuts C again at point B.
    Find the y-coordinate of point B.
    Eliminate the parameter:

    . . x .= .4cos2t .= .4(1 - 2sint) .[1]

    . . y .= .3sin t . . sin t .= .y/3 [2]


    Substitute [2] into [1]: .x .= .4(1 - 2[y/9])

    . . and we get: .x .= .4 - 8y/9 . [3] . . . a parabola

    The equation of the normal is: .x .= .3y/8 + 23/16 . [4]


    Equate [3] and [4]: .4 - 8y/9 .= .3y/8 + 23/16

    . . which simplifies to: .128y + 54y - 369 .= .0

    . . which factors: .(2y - 3)(64y + 123) .= .0 . **

    . . and has roots: .y .= .3/2, -123/64


    So the normal at A cuts the parabola at: .y = -123/64


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    Okay, okay! . . . I didn't really factor it.

    I used the Quadratic Formula and was pleasantly surprised
    . . to find that the roots were rational.

    From the roots I constructed the factors,
    . . then created the illusion that I'm a genius.

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  3. #3
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    Quote Originally Posted by Confuzzled? View Post
    Could anyone help me with the following problems?:

    a)Find the gradient of the curve:

    8x^2/4x^2-3y^3 = 3y at (3,2)

    ...
    Hi,

    I hope I read your problem correctly:
    Attached Thumbnails Attached Thumbnails Gradient and parametric equations help!-gradient_tangt.gif  
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  4. #4
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    Quote Originally Posted by Confuzzled? View Post
    Could anyone help me with the following problems?:

    a)Find the gradient of the curve:

    8x^2/4x^2-3y^3 = 3y at (3,2)

    ...
    Ooops,

    sorry, I forgot to attach the diagram
    Attached Thumbnails Attached Thumbnails Gradient and parametric equations help!-gradienttangt_graph.gif  
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  5. #5
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    Hello, Confuzzled?

    Pay no attention to the man behind the curtain . . .


    a) Find the gradient of the curve: .8x/(4x - 3y) = 3y .at (3,2)

    We have: 8x .= .12xy - 9y^4

    Differentiate implicitly: .16x .= .12xy' + 24xy - 36yy'

    Then: . 36yy' - 12xy' .= .24xy - 16x

    Factor: .12(3y - x)y' .= .8x(3y - 2)

    . . . . . . . . . . . . .2x(3y - 2)
    Finally: . y' . = . --------------
    . . . . . . . . . . . . 3(3y - x)


    . . . . . . . . . . . . .2(3)(32 - 2) . . . . .8
    At (3,2): .y' .= . ----------------- .= .- ---
    . . . . . . . . . . . . .3(32 - 3) . . . . .15

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