解答例 1

\begin{alignat*}{2} \frac{\partial x}{\partial r} &= \cos\theta,\quad &\frac{\partial x}{\partial \theta} &= -r\sin\theta, \\ \frac{\partial y}{\partial r} &= \sin\theta,\quad &\frac{\partial y}{\partial \theta} &= r\cos\theta \end{alignat*} より, \begin{align*} \frac{\partial z}{\partial r} &= \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r} = \frac{\partial z}{\partial x}\cos\theta + \frac{\partial z}{\partial y}\sin\theta, \\ \frac{\partial z}{\partial \theta} &= \frac{\partial z}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta} = \frac{\partial z}{\partial x}(-r\sin\theta) + \frac{\partial z}{\partial y}(r\cos\theta). \end{align*} よって, \begin{align*} \left( \frac{\partial z}{\partial r} \right)^2 + \frac{1}{r^2}\left( \frac{\partial z}{\partial \theta} \right)^2 &= \left( \frac{\partial z}{\partial x}\cos\theta + \frac{\partial z}{\partial y}\sin\theta \right)^2 + \left(-\frac{\partial z}{\partial x}\sin\theta + \frac{\partial z}{\partial y}\cos\theta \right)^2 \\ &= \left( \frac{\partial z}{\partial x} \right)^2(\cos^2\theta+\sin^2\theta) + \left( \frac{\partial z}{\partial y} \right)^2(\sin^2\theta+\cos^2\theta) \\ &= \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2. \end{align*}

最終更新日：2011年11月02日