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Math Help - slope of tangent

  1. #1
    Member
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    Oct 2009
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    slope of tangent

    the question is

    use implicit differentiation to find the slope of tangent at x = sqrt 2 for x^2 + 4y^2 = 4

    what i did was

    dx^2 / dx = 2x
    d4y^2 / dx = 8yy'
    d4/dx = 0

    2x + 8yy' = 0
    8yy' = -2x
    y'= -2x/8y
    = -x/4y

    since x = sqrt 2

    (sqrt 2)^2 + 4y^2 = 4
    2+4y^2 = 4
    4y^2 = 2
    y^2 = 2/4 = 1/2
    y= +/- sqrt 1/2

    since y' = -x / 4y and x=sqrt 2, and y = +/- sqrt 1/2

    y'= - sqrt2 / 4(sqrt 1/2)

    and

    y'= - sqrt2 / 4(- sqrt 1/2) = sqrt2 / 4(sqrt 1/2)

    so i got slope of tangent = +/- sqrt2 / 4(sqrt 1/2)

    but i drew graph on the calculator and seems like i did something wrong
    but i can't find where i did go wrong...

    does anyone see what i did wrong?
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  2. #2
    MHF Contributor
    Joined
    Dec 2009
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    Your calculations are fine, haebinpark.

    Maybe the problem was with the graphing.
    The slopes of the tangents are 0.5 and -0.5,
    which you will get if you simplify a little more

    \frac{\sqrt{2}}{\left(\frac{4}{\sqrt{2}}\right)}=\  frac{\sqrt{2}\sqrt{2}}{4}
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  3. #3
    Newbie
    Joined
    Mar 2010
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    First find f(sqrt(2)) first which is sqrt(1/2)

    x=sqrt(2)
    y=sqrt(1/2)

    derivative of x^2+4y^2=4

    2x+8y*y'=0
    2(x+4y*y')=0
    x+4y*y'=0
    y'=-x/(4y)

    your slop y'=(-1/2)
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