微分法累乗、和、曲線、接線、円、アステロイド

学習環境

微分積分演習〈理工系の数学入門コース/演習新装版〉 (和達三樹(著)、十河清(著)、岩波書店)の第3章(微分法)、3-2(微分法)、問題3の解答を求めてみる。

\begin{array}{l} \frac{d}{dx} (x^{2} + y^{2}) = \frac{d}{dx} 1 \\ 2 x + 2 y \frac{dy}{dx} = 0 \\ \frac{dy}{dx} = - \frac{x}{y} \end{array}

よって、求める接線の方程式は、

\begin{array}{l} y - y_{0} = - \frac{x_{0}}{y_{0}} (x - x_{0}) \\ y_{0} y + x_{0} y = x_{0}^{2} + y_{0}^{2} \\ y_{0} y + x_{0} y = 1 \end{array}

\begin{array}{l} \frac{d}{dx} (x^{3} + y^{3}) = \frac{d}{dx} 1 \\ 3 x^{2} + 3 y^{2} \frac{dy}{dx} = 0 \\ \frac{dy}{dx} = - \frac{x^{2}}{y^{2}} \end{array}

よって、求める接線の方程式は、

\begin{array}{l} y - y_{0} = - \frac{x_{0}^{2}}{y_{0}^{2}} (x - x_{0}) \\ y_{0}^{2} y + x_{0}^{2} x = x_{0}^{3} + y_{0}^{3} \\ y_{0}^{2} y + x_{0}^{2} x = 1 \end{array}

\begin{array}{l} \frac{d}{dx} (x^{\frac{2}{3}} + y^{\frac{2}{3}}) = \frac{d}{dx} 1 \\ \frac{2}{3} x^{- \frac{1}{3}} + \frac{2}{3} y^{- \frac{1}{3}} \frac{dy}{dx} = 0 \\ \frac{dy}{dx} = - \frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}} \end{array}

よって、求める接線の方程式は、

y - y_{0} = - {(\frac{y_{0}}{x_{0}})}^{\frac{1}{3}} (x - x_{0})

{(y_{0})}^{- \frac{1}{3}} y + {(x_{0})}^{- \frac{1}{3}} x = 1

コード(Wolfram Language, Jupyter)

ContourPlot[
    {x^2+y^2 == 1, 1/2y+(1-(1/2)^2)^(1/2)x==1},
    {x, -2, 2},
    {y, -2, 2}
]

ContourPlot[
    {x^3+y^3 == 1, (1/2)^2y+((1-1/2^3)^(1/3))^2x==1},
    {x, -2, 2},
    {y, -2, 2}
]

ContourPlot[
    {x^(2/3)+y^(2/3) == 1, (1/2)^(-1/3)y+(((1-1/2^(2/3)))^(3/2))^(-1/3)x==1},
    {x, -2, 2},
    {y, -2, 2}
]