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Definujme: $$ a^n := \underbrace{a \cdot a \cdot \ldots \cdot
a}_{\textrm{$n$-kr�t}}, \qquad \textrm{a tak�} \qquad b^m:=
\overbrace{b \cdot b \cdot \ldots \cdot b}^{\textrm{$m$-kr�t}}.$$

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Vyu�ijeme ob� parametrizace, nahrad�me $t$ prom�nnou $\varphi $, a
pou�ijeme konverzi mezi pol�rn�mi a kart�zsk�mi sou�adnicemi:
\begin{eqnarray} \vec r & = &   \big( \,
\rho(\varphi ) \cdot \cos \varphi \, , \, \rho(\varphi ) \cdot
\sin \varphi \, \big) , \\ \vec t & = & \Big( \, \frac{\partial
}{\partial \varphi} \big( \, \rho(\varphi ) \cdot \cos \varphi \,
\big) ,\, \frac{\partial }{\partial \varphi} \big( \, \rho(\varphi
) \cdot \sin \varphi \, \big) \Big) = \nonumber \\ &= & \big( \,
\rho '(\varphi ) \cdot \cos \varphi - \rho (\varphi ) \cdot \sin
\varphi , \, \rho ' (\varphi ) \cdot \sin \varphi + \rho (\varphi
) \cdot \cos \varphi \, \big) .
\end{eqnarray}

Potom \begin{eqnarray}  \vec r \cdot \vec t \ &=& \rho (\varphi )
\cdot \cos \varphi  \cdot \big( \rho '(\varphi ) \cdot \cos
\varphi - \rho (\varphi ) \cdot \sin \varphi  \big) + \nonumber \\
 & +& \rho(\varphi ) \cdot \sin \varphi \cdot \big( \rho
' (\varphi ) \cdot \sin \varphi + \rho (\varphi ) \cdot \cos
\varphi \big) = \nonumber \\ &=& \rho '(\varphi ) \cdot \rho
(\varphi ) \cdot \cos ^2 \varphi - \rho ^2(\varphi ) \cdot \cos
\varphi \cdot \sin \varphi + \nonumber \\ &+ &  \rho '(\varphi )
\cdot \rho (\varphi ) \cdot \sin ^2 \varphi + \rho ^2(\varphi )
\cdot \cos \varphi \cdot \sin \varphi = \nonumber \\ &=&  \rho
'(\varphi ) \cdot \rho (\varphi ) . \label{eq3}
\end{eqnarray}

Proto�e $\beta \neq 90^{\circ }$, tak vektor $\vec t $ nen� kolm�
na vektor $\vec r$. Tedy, $\vec r \cdot \vec t \neq 0$.

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P�edpokl�dejme, �e  \begin{equation} f(x,y) = \left\{
\begin{array}{ll}
                                      \frac{xy}{x^2+y^2}
                                      & \textrm{for } [x,y] \in
                                      \mathbb{R} \times \mathbb{R}
                                      \setminus \{
                                      [0,0]\}, \\
                                      & \\
                                      1 & \textrm{for } [x,y] =
                                      [0,0],
                                       \end{array} \right.
                                       \end{equation}
nebo mo�n� m�te rad�ji z�pis ve tvaru \smallskip $$ f(x,y) =
\left\{
\begin{array}{ll}
                                      \frac{xy}{x^2+y^2}
                                      & \textrm{for } [x,y] \in
                                      \mathbf{R} \times \mathbf{R}
                                      \setminus \{
                                      [0,0]\}, \\
                                      & \\
                                      1 & \textrm{for } [x,y] =
                                      [0,0].
                                      \end{array} \right.
                                      $$


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Matice a jej� determinant: $$ \mathbf{A} = \left(
\begin{array}{ccccc}
                                 1&2&3&\ldots & n\\
                                 2&3&4&\ldots &n+1 \\
                                 3&4&5& \ldots &n+2 \\
                                 \vdots &\vdots & \vdots & \ddots
                                 & \vdots \\
                                 n&n+1&n+2& \ldots& 2n+1
                                 \end{array} \right) ,
\ \textrm{det } \mathbf{A} = \left| \begin{array}{ccccc}
                                 1&2&3&\ldots & n\\
                                 2&3&4&\ldots &n+1 \\
                                 3&4&5& \ldots &n+2 \\
                                 \vdots &\vdots & \vdots & \ddots
                                 & \vdots \\
                                 n&n+1&n+2& \ldots& 2n+1
                                 \end{array} \right| .
                                 $$

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