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Math Help - slope of tangent

  1. #1
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    slope of tangent

    Use "implicit differentiation" to find the slope of tangent when x= sqrt 3 for x^2 + y^2 = 4

    i did

    dx^2/dx + dy^2/dx = d4/dx

    dx^2/dx = 2x
    dy^2/dx = 2yy'
    d4/dx = 0

    2x + 2yy' = 0
    y' = - x/y

    and then what do i have to do to find the slope of the tangent ?
    Last edited by haebinpark; October 28th 2009 at 04:40 PM.
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  2. #2
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    Quote Originally Posted by haebinpark View Post
    Use "implicit differentiation" to find the slope of tangent when x= sqrt 3 for x^2 + y^2 = 4

    i did

    dx^2/dx + dy^2/dx = d4/dx

    dx^2/dx = 2x
    dy^2/dx = 2yy'
    d4/dx = 0

    2x + 2yy' = 0
    y' = x/y

    and then what do i have to do to find the slope of the tangent ?
    All you have to do is solve the original equation x^2+y^2=4 for y and the substitute into the equation for y'.

    3+y^2=4

    y=1

    y'=\frac{x}{y}=\sqrt{3}
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  3. #3
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    isn't y should be plus minus 1 ?
    how can you tell if it is +1 ?

    and i got y' = - x / y ?

    hm... did i do something wrong?
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  4. #4
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    Quote Originally Posted by haebinpark View Post
    isn't y should be plus minus 1 ?
    how can you tell if it is +1 ?

    and i got y' = - x / y ?

    hm... did i do something wrong?
    Actually, I you're right.

    3+y^2=4
    y=\pm1
    y'=\frac{x}{y}=\pm\sqrt{3}

    Notice that if you graph the circle, that there are two values of y that correspond to the point x=\sqrt{3}. Both of these points (\sqrt{3},\pm 1) will have tangents that have opposite slopes.
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  5. #5
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    thank you !!!
    helped a lot !!

    so slope of tangent will be +/- sqrt 3. right?
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