\begin{eqnarray*}\vec f(x_1,x_2)&=&\vect{x_1+x_2^2\\x_1^2}\\\rightarrow \tilde A(x_1,x_2)&=&\left(\begin{array}{cc} 1 & 2 x_2a \\ 2 x_1 & 0 \end{array} \right)\;\;\mbox{as before!}\end{eqnarray*}
\begin{eqnarray*} \vec f(r,\vartheta,\varphi)&=&\vect{r\sin\vartheta\cos\varphi\\r\sin\vartheta\sin\varphi\\r\cos\vartheta}=\vect{f_1(r,\vartheta,\varphi)\\f_2(r,\vartheta,\varphi)\\f_3(r,\vartheta,\varphi)}=\vect{x\\y\\z}\hat=\;\mbox{spherical coordinates!}\\ \tilde A&=&\left(\begin{array}{ccc}\sin\vartheta\cos\varphi&r\cos\vartheta\cos\varphi&-r\sin\vartheta\sin\varphi\\ \sin\vartheta\sin\varphi&r\cos\vartheta\sin\varphi&r\sin\vartheta\cos\varphi\\ \cos\vartheta&-r\sin\vartheta&0\end{array}\right)\\ \det \tilde A&=& r^2 \sin\vartheta\;\;\;\;\mbox{(volume!)} \end{eqnarray*}
\begin{eqnarray*} \vec f(r,\varphi,z)&=&\vect{r\cos\varphi\\r\sin\varphi\\z}\hat=\;\mbox{cylindrical coordinates}\\ \tilde A&=&\left(\begin{array}{ccc}\cos\varphi&-r\sin\varphi&0\\\sin\varphi&r\cos\varphi&0\\0&0&1\end{array}\right)\\ \det \tilde A &=& r \;\;\;\;\mbox{(volume!)} \end{eqnarray*}
\begin{eqnarray*} f:\mathbb{R}^2&\to&\mathbb{R}\qquad f(x_1,x_2)=x_1^2+x_2^3\\ \vec A&=&\vec\nabla f=\vect{2x_1\\3x_2^2}\qquad\vec\nabla f:\mathbb{R}^2\to\mathbb{R}^2\\ \mbox{second derivative is now a matrix:}\\ \tilde D_2f&=&\left(\begin{array}{cc} 2 & 0 \\ 0 & 6 x_2 \end{array} \right)\leftarrow\mbox{$M=1$, Hesse$(\vec\nabla f)^\top$} \end{eqnarray*}
\begin{eqnarray*} f&:&\mathbb{R}^N\to\mathbb{R}\\ f(\vec x)&=&|\vec x|^2=x_1^2+x_2^2+\ldots+x_N^2\\ \vec A&=&\vec\nabla f=\vect{2x_1\\2x_2\\\vdots\\2x_N}=2\vect{x_1\\\vdots\\x_N}=2\vec x\\ \mbox{second derivative:}\\ \tilde D_2 f&=&\left(\begin{array}{ccccc}2&0&0&\cdots&0\\0&2&0&\cdots&0\\0&0&\ddots& &\vdots\\\vdots&&&\ddots&\vdots\\0&0&0&\cdots&2\end{array}\right)=2\tilde I \end{eqnarray*}
physics example: \(g\)-measurement with pendulum
| \[\mbox{period } T=2\pi\sqrt{\frac{l}{g}}\to g=\frac{4\pi^2l}{T^2}=g(l,T)\] |
Measurement: \(l\pm\Delta
l,\;T\pm\Delta T\); \(\Delta l,\Delta T\) errors of measurement. (\(l=1\)m, \(\Delta l=0.001\), \(T=2\)s, \(\Delta T=\)0.1s) What is the error \(\Delta g\) of g? Total differential: \begin{eqnarray*}\Delta g&=&|\frac{\partial g}{\partial l}|\Delta l+|\frac{\partial g}{\partial T}|\Delta T\\ &=&\left|\frac{4\pi^2}{T^2}\right| \Delta l+\left|\frac{4\pi^2l}{T^3}(-2)\right| \Delta T\\ &=&0.01\frac{m}{s^2}+1\frac{m}{s^2}=1.01\frac{m}{s^2}\end{eqnarray*}
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With frame
Total Derivatives