Consider $F(x,y,z,w)=0$ and $G(x,y,z,w)=0$ . Why $(\frac{\partial x}{\partial z})_w$= -$\frac{\frac{\partial (F,G)}{\partial (z,y)}}{\frac{\partial (F,G)}{\partial (x,y)}}$.Noted that the $(\frac{\partial x}{\partial z})_w$= -$\frac{\frac{\partial (F,G)}{\partial (z,y)}}{\frac{\partial (F,G)}{\partial (x,y)}}$=-$\frac{F_3G_2-F_2G_3}{F_1G_2-F_2G_1}$
Why $(\frac{\partial x}{\partial z})_w$ can be expressed as-$\frac{\frac{\partial (F,G)}{\partial (z,y)}}{\frac{\partial (F,G)}{\partial (x,y)}}$
-
0What do you mean by $\left ( \frac {\partial x}{\partial z} \right )_w$ and $\frac{(\partial F,G)}{(\partial x,y)}$? – 2012-01-11
-
0Why do you think it is true? – 2012-01-11
-
0F and G are 2 function with variable xy z w and $(\frac{\partial x}{\partial Z})_w$ means we may find $(\frac{\partial x}{\partial Z})$ while setting w as a constant in the 2 function – 2012-01-11
-
0That's shown in the textbook but i dont quite understand – 2012-01-11
1 Answers
The two equations $F(x,y,z,w_0)=0$, $\ G(x,y,z,w_0)=0$ define a curve $\gamma$ in the three-dimensional $(x,y,z)$-plane $w=w_0$ of ${\mathbb R}^4$. This curve should have a parametrization $$\gamma:\quad z\mapsto \bigl(x(z),y(z),z, w_0\bigr)\qquad \qquad (z_0-h
-
0the question is i dont think that it is just coincident to get $\frac{\frac{\partial (F,G)}{\partial (z,y)}}{\frac{\partial (F,G)}{\partial (x,y)}}$ as the answer and i think it applies to all similar cases but why? – 2012-01-11
-
0Is it true to apply the same step(i.e.$-{\det\left[{\partial(F,G)\over\partial(z,y)}\right] \over\det\left[{\partial(F,G)\over\partial(x,y)}\right]}$) to calculate for the case like n variables – 2012-01-11