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1.Solve the equation $f(x,y,z):= x^{4}+2x\cos(y) + \sin (z) = 0$ in a neighbourhood of $p=(0,0,0)$ for z in the form : $z=g(x,y)$
2. Give the maximal domain for $g$ and the neighbourhood of $p$ and
3.calculate the differential of $g$. Check the connection between the theorem of implicit function for the differential of $g$ and the differential of $f$.

I solve like this:

  1. $z= \arcsin(-x^{4}-2x\cos(y)$ ,
  2. we can see that the maximal domain must be : $Do_{x}=[-1,1]$ and $Do_{y}=[\frac{1}{2}(3\pi) , \frac{1}{2}(7\pi)]$ . How does one calculate the neighbourhood of $p$?

  3. $\displaystyle Dg(x,y)= \begin{pmatrix} \frac{-2(2x^{3}+\cos(y))}{(1-(x^{4}+2x\cos(y))^{2})^{1/2}} & \frac{2x\sin(y)}{(1-(x^{4}+2x\cos(y))^{1/2}} \end{pmatrix}$

how can I show the connection, if put $Df(x,y,g(x,y))$ then i get $0$ and that is not even invertible.

Thanks for any efforts to explain.

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1 Answer 1

up vote 1 down vote accepted
+50

Ad 1: $g(x,y)=\arcsin\bigl(-x^4-2x\cos (y)\bigr)$ is correct.

Ad 2: One has to determine the maximal connected set in the $(x,y)$-plane containing $(0,0)$, on which the auxiliary function

$$h(x,y):=-x^4-2x\cos(y)$$

has an absolute value $<1$. To this end I drew the curves $h(x,y)=-1$, $h(x,y)=0$, and $h(x,y)=1$ with the help of Mathematica. Here is the result:

contour

The domain you are looking for is the infinite wavy strip going vertically upwards, minus the holes.

Ad 3: You computed correctly the gradient $\nabla g(x,y)$. It's coordinates are the coefficients of the differential $dg$, when expressed as a linear combination of the coordinate differentials: $dg=g_x\ dx+g_y\ dy$.

There is no connection between the differential $df=f_x\ dx+f_y\ dy+ f_z\ dz$ and the differential $dg$ in variable terms, because these differentials live in completely different spaces. If, however, you are given a point $p$ on the surface $S$ defined by the equation $f(x,y,z)=0$ then there is a relation between the differentials $df(p)$ and $dg(p')$, where $p'$ denotes the point $p$ projected onto the $(x,y)$-plane. This relation is best viewed in terms of partial derivatives, namely $$g_x(p')=-{f_x(p)\over f_z(p)}\ ,\qquad g_y(p')=-{f_y(p)\over f_z(p)}\ .$$

You can easily check that for $p=(0,0,0)$, $p'=(0,0)$ everything is o.k.

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Thank you! Very good answer! –  VVV Dec 20 '11 at 13:42

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