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Calculus

4. Given ln(x/y) + y^3 - 2x = -1.

A. Find the equation of the normal line to the curve ln(x/y) + y^3 - 2x = -1 at the point (1,1). (I got -2x+3)

B. Find the equation of a tangent line to the curve y=e^(x^2) that "also" passes through the point (1,0). You can approximate your final answer with your calculator after you've shown your work. (idk how to do this)

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  1. Let's rewrite things a bit to get
    lnx - lny + y^3 - 2x = -1
    Now,
    A.
    1/x - 1/y y' + 3y^2 y' - 2 = 0
    y' (3y^2 - 1/y) = 2 - 1/x
    y' = (2 - 1/x) / (3y^2 - 1/y)
    so at (1,1), y' = (2-1)/(3-1) = 1/2
    The normal there has slope -2, so
    y-1 = -2(x-1)
    y = -2x + 3
    You are correct

    B.
    At any point on the curve, say, (m,e^(m^2)) the slope is 2m e^(m^2)
    So now we have a point and a slope, so the equation is
    y-0 = 2m e^(m^2) (x-1)
    Now, we know that at x=m,
    2m e^(m^2) (m-1) = e^m^2
    2m(m-1) = 1
    2m^2-2m-1 = 0
    m = (1±√3)/2
    So there are two lines:
    y = (1+√3)e^(1+√3/2) (x-1)
    y = (1-√3)e^(1-√3/2) (x-1)

    You can see the graph of the first line and its tangency at

    https://www.wolframalpha.com/input/?i=plot+y%3De%5Ex%5E2%2C+y+%3D+%281%2B%E2%88%9A3%29e%5E%281%2B%E2%88%9A3%2F2%29+%28x-1%29+for+0%3Cx%3C2

    The other line is pretty hard to see, and just clutters things up if you plot it too.

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    oobleck

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