【微分積分】8-2-1 ヤコビアン｜問題集

\(\displaystyle \frac{\partial u}{\partial x}=1\)
\(\displaystyle \frac{\partial u}{\partial y}=-1\)
\(\displaystyle \frac{\partial v}{\partial x}=y\)
\(\displaystyle \frac{\partial v}{\partial y}=x\)
よって、ヤコビアンは
\(\displaystyle \frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}1 & -1 \\ y & x\end{vmatrix}=x+y\)

\(\displaystyle \frac{\partial u}{\partial x}=2x\)
\(\displaystyle \frac{\partial u}{\partial y}=-2y\)
\(\displaystyle \frac{\partial v}{\partial x}=2y\)
\(\displaystyle \frac{\partial v}{\partial y}=2x\)
よって、ヤコビアンは
\(\displaystyle \frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}2x & -2y \\ 2y & 2x\end{vmatrix}=4(x^2+y^2)\)

\(\displaystyle \frac{\partial x}{\partial r}=\cos\theta\)
\(\displaystyle \frac{\partial x}{\partial\theta}=-r\sin\theta\)
\(\displaystyle \frac{\partial y}{\partial r}=\sin\theta\)
\(\displaystyle \frac{\partial y}{\partial\theta}=r\cos\theta\)
よって、ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(r,\theta)}=\begin{vmatrix}\cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta\end{vmatrix}=r(\cos^2\theta+\sin^2\theta)=r\)

\(\displaystyle \frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}\)
\(\displaystyle \frac{\partial r}{\partial y}=\frac{y}{\sqrt{x^2+y^2}}\)
\(\displaystyle \frac{\partial\theta}{\partial x}=-\frac{y}{x^2+y^2}\)
\(\displaystyle \frac{\partial\theta}{\partial y}=\frac{x}{x^2+y^2}\)
よって、ヤコビアンは
\(\displaystyle \frac{\partial(r,\theta)}{\partial(x,y)}=\begin{vmatrix}\displaystyle \frac{x}{\sqrt{x^2+y^2}} & \displaystyle \frac{y}{\sqrt{x^2+y^2}} \\ \displaystyle -\frac{y}{x^2+y^2} & \displaystyle \frac{x}{x^2+y^2}\end{vmatrix}=\frac{1}{\sqrt{x^2+y^2}}\)