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Functions of several variables: exercises
(1) Consider
$$f(x,y) = ( x^2 + ( y - 2 )^2 - y )^{\frac{1}{2}} + 1.$$
(a) Domain:
$$x^2 + ( y - 2 )^2 - y \geq 0,$$
$$x^2 + ( y - 2 )^2 \geq 4$$
region with boundary
$$x^2 + ( y - 2 )^2 = 2^2.$$
This circle of radius $2$ with center at $(0, 2)$.
(b) Graph:
$$f( 0, y ) = ( ( y - 2 )^2 - y )^{\frac{1}{2}} + 1 = z,$$
hence
$$z = ( x^2 + ( y - 2 )^2 - y )^{\frac{1}{2}} + 1$$
$$( z - 1 )^2 = x^2 + ( y - 2 )^2.$$
Then with $x = 0$:
$$( z - 1 )^2 = ( y - 2 )^2 {\rightarrow} z - 1 = \pm ( y - 2 ),$$
and with $y = 0$:
$$( z - 1 )^2 = x^2 {\rightarrow} z - 1 = \pm x.$$
(c) This is funnel, a truncated cone.
(2) To prove that $|| x ||$ is continuous, use
$$|| x || = ( x_1^2 + ... + x_n^2 )^{\frac{1}{2}},$$
and show that
$$|| x - a || < \delta \rightarrow | || x || - || a || | < \delta,$$
by the Triangle Inequality.
(3) Suppose $A$ and $B$ are path-connected and $a$ is in the intersection. If $P$ and $Q$ belong to the union, find a path from $P$ to $a$, from $a$ to $Q$. This gives you a path from $P$ to $Q$.
(6) Show that
$$T() = x_2^2 + 2x_2 + 1$$
is the best affine approximation of the function
$$f() = x_1^4 + x_2^2.$$
Consider
$$\begin{array}{} \displaystyle\lim_{x \rightarrow a} \frac{f(x) - T(x)}{|| x - a ||} &= \displaystyle\lim_{x \rightarrow (0,-1)} \frac{x_1^4 + x_2^2 + 2x_2 + 1}{|| ( x_1, x_2) - ( 0, -1)} \\ &= \displaystyle\lim_{x \rightarrow (0,-1)} \frac{x_1^4 + x_2^2 + 2x_2 + 1}{( ( x_1 - 0 )^2 + ( x_2 + 1 )^2}^{\frac{1}{2}} \\ &= \displaystyle\lim_{x_2 \rightarrow (-1)} \frac{x_2^2 + 2x_2 + 1}{| x_2 + 1 |} \\ &= \displaystyle\lim_{x_2 \rightarrow -1} \frac{(x_2 + 1 )^2}{| x_2 + 1 |} {\rm \hspace{3pt} by \hspace{3pt} canceling \hspace{3pt}} ( x_2 + 1 ) \\ &= \displaystyle\lim_{x_2 \rightarrow -1} | x_2 + 1 | \\ &= 0. \end{array}$$
(7) Let
$$h() = ( {\rm sin \hspace{3pt}} e^t, {\rm cos \hspace{3pt}} e^t).$$
Then $h$ is the composition of
$g: {\bf R} {\rightarrow} {\bf R}^2, g(x) = ( {\rm sin \hspace{3pt}} x, {\rm cos \hspace{3pt}} x)$.
Then
$$f'(t) = e^t, {\nabla}g = ( {\rm cos \hspace{3pt}} x, -{\rm sin \hspace{3pt}} x),$$
and by the Chain Rule
$$h' = ( g {\circ}f )' = {\nabla}g \cdot f' = ( {\rm cos \hspace{3pt}} e^t, -{\rm sin \hspace{3pt}} e^t ) e^t.$$
Also consider Find the point of the graph nearest to $0$.