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Math Help - Hypocycloid. Oh so fun!

  1. #1
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    Hypocycloid. Oh so fun!

    At the risk of asking too many questions on my first day.

    I need some more help.

    How do I "Show that every tangent line to the curve x^(2/3) +y^(2/3)=1 in the first quadrant has the property that the portion of the line in the first quadrant has length 1" using implicit differentiation?

    Thanks bunches for any help.
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  2. #2
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    Hello, Columbo!

    Show that every tangent line to the curve x^{\frac{2}{3}} +y^{\frac{2}{3}} \;=\;1
    in the first quadrant has the property that the portion of the line
    in the first quadrant has length 1, using implicit differentiation.
    Find the derivative . . .

    . . \frac{2}{3}x^{-\frac{1}{3}} + \frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx} \;=\;1 \quad\Rightarrow\quad \frac{dy}{dx} \;=\;-\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}


    Let A(p,q) be a point on the astroid in Quadrant 1.
    . . Note that: . p^{\frac{2}{3}} + q^{\frac{2}{3}} \:=\:1

    The slope of the tangent at A is: . m \;=\;-\frac{q^{\frac{1}{3}}}{p^{\frac{1}{3}}}


    The equation of the line through A with slope m is:

    . . y -q \;=\;-\frac{q^{\frac{1}{3}}}{p^{\frac{1}{3}}}(x - p) \quad\Rightarrow\quad y \;=\;-\frac{q^{\frac{1}{3}}}{o^{\frac{1}{3}}}x + p^{\frac{2}{3}}q^{\frac{1}{3}} + q

    . . y \;=\;-\frac{q^{\frac{1}{3}}}{p^{\frac{1}{3}}}x + q^{\frac{1}{3}}\underbrace{\left(p^{\frac{2}{3}} + q^{\frac{2}{3}}\right)}_{\text{This is 1}} . \Rightarrow\quad\boxed{ y \;=\;-\frac{q^{\frac{1}{3}}}{p^{\frac{1}{3}}}x + q^{\frac{1}{3}}}


    This tangent has: .x-intercept \left(p^{\frac{1}{3}},\:0\right) . . . and y-intercept \left(0,\:q^{\frac{1}{3}}\right)


    The length of that segment is: . L \;=\;\sqrt{\left(p^{\frac{1}{3}}-0\right)^2 + \left(0-q^{\frac{1}{3}}\right)^2}

    . . \text{and we have: }\;L \;=\;\sqrt{\underbrace{p^{\frac{2}{3}} + q^{\frac{2}{3}}}_{\text{This is 1}}} \;=\;\sqrt{1} \;=\;1 \quad\hdots\quad \text{ There!}

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