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Thread: Stuck... beginner

  1. #1
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    Stuck... beginner

    Find the normal to the curve x^2 + y^2 = 2x + 2y that is parallel to the line x+y = 0
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  2. #2
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    Hello, yakkow!

    Find the normal to the curve $\displaystyle x^2 + y^2 = 2x + 2y$ that is parallel to the line $\displaystyle x+y = 0$
    Differentiate implicitly: .$\displaystyle 2x + 2y\cdot y' \:= \:2 + 2y' \quad\Rightarrow\quad 2y\cdot y' - 2y'\;=\;2 - 2x$

    Factor: .$\displaystyle 2(y - 1)y'\:=\:2(1 - x)\quad\Rightarrow\quad y'\:=\:\frac{1-x}{y-1}$ . . . slope of the tangent

    The slope of the normal is: .$\displaystyle m\:=\:\frac{y-1}{x-1}$

    The line $\displaystyle x + y\:=\;0\quad\Rightarrow\quad y \,=\,-x$ has slope $\displaystyle -1.$

    So we have: .$\displaystyle \frac{y - 1}{x - 1}\:=\:-1\quad\Rightarrow\quad y\,=\,2 - x$

    Substitute into the original equation:
    . . . $\displaystyle x^2 + (2 - x)^2\:=\:2x + 2(2 - x)\quad\Rightarrow\quad2x^2 - 4x\:=$ $\displaystyle 0\quad\Rightarrow\quad x \,=\,0,\;2$

    The corresponding y-values are: $\displaystyle y\,=\,2,\;0$

    We seem to have two normals: one at (2,0), another at (0,2)


    At (2,0) with slope -1, the equation of the normal is:
    . . . $\displaystyle y - 0\:=\:-1(x - 2)\quad\Rightarrow\quad y \:=\:-x + 2$

    At (0,2) with slope -1, the equation of the normal is:
    . . . $\displaystyle y - 2\:=\:-1(x - 0)\quad\Rightarrow\quad y\:=\:-x + 2$

    Ha! . . . It's the same line!
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