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Math Help - Compute tangent line to curve... help :/

  1. #1
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    Compute tangent line to curve... help :/

    A) Computer the tangent line to the curve x^4-yx^2+y^4 =1 at the point (-1,1)

    Ok so I did the derivative =
    4x^3-2x y' + 4x^3 y' = 0
    Do I just take the point x (-1) put it in the formula for X to find y' and that's it? :S

    B) Iterate Newton's method 2 times to produce a rational number approximating 4^-3. Use 2 as your initial guess.

    Xn+1 = Xn - f(Xn)/(f'(Xn))

    so... I'm lost about there :/
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    A mistake:

    Ok so I did the derivative =
    4x^3-2x y' + 4x^3 y' = 0
     x^4-yx^2+y^4 =1

    The derivative is:

    4x^3-2xy-y'x^2+4y^3(y')=0
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  3. #3
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    Thanks, can't believe I missed that.
    But what do I do with it now?
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  4. #4
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    Quote Originally Posted by roushrsh View Post
    Thanks, can't believe I missed that.
    But what do I do with it now?
    sub -1 for x , 1 for y and calculate the value of the derivative (slope) at (-1,1)

    use the point-slope form of a linear equation to get the tangent line equation.
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  5. #5
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    Ah ok, makes sense

    So 3y' = 2
    y'=2/3

    y-y1 = m(x-x1)
    y-1 = 2/3(x+1)
    y=2/3x + 5/3

    For the next one do I start with 3 = 2 - (4^-3)/(4^-4) ? Then repeat? :S
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  6. #6
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    Quote Originally Posted by roushrsh View Post
    B) Iterate Newton's method 2 times to produce a rational number approximating 4^-3. Use 2 as your initial guess.

    Xn+1 = Xn - f(Xn)/(f'(Xn))
    please check this problem ...

    why would one wish to use Newton's method to produce a rational number approximating 4^{-3} , when 4^{-3} = \frac{1}{4^3} = \frac{1}{64} is a rational number to start with?

    did you mean to type 4^{\frac{1}{3}} = \sqrt[3]{4} ?
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  7. #7
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    Yes, sorry, I meant the latter with the sqrt
    the
    ^3sqrt4
    Last edited by roushrsh; October 23rd 2010 at 04:33 PM.
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  8. #8
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    Xn+1 = Xn - f(Xn)/(f'(Xn))
    use f(x) = x^3 - 4 with x_1 = 2
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  9. #9
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    Why would I use x^3 -4 ? :S (=3x^2 for the f'(x))
    Thanks
    EDIT: nvm, I found that out, thanks. I'll try it
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  10. #10
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    So would it be
    xn+1 = xn-f(xn)/f'(xn)
    f(x) = x^3-4 x0 = 2
    f'(x) = 3x^2
    x1 = 2-4/12 = 2-1/3 = (1+2/3)
    x2 = (1+2/3) - ((1+2/3)^3-4)/(3(1+2/3)^2) = 1.561111...
    Makes sense
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