Implicit function theorem for f(x,y) = x^2+y^2-1

[chris222]

$f: \mathbb{R^2} \rightarrow \mathbb{R}$, $f(x,y) = x^2+y^2-1$

$X:= f^{-1} (\{0\})=\{(x,y) \in \mathbb{R^2} | f(x,y)=0\}$

1. Show that $f$ is continuous differentiable.
2. For which $(x,y) \in \mathbb{R^2}$ is the implicit function theorem usable to express $y$ under the condition $f(x,y)=0$ as a function of $x$?
3. Let $(a,b) \in X$ with $b>0$. Find the largest possible neighbourhood $V$ of $a$ in $\mathbb{R}$ and a continuous differentiable function $g:V \rightarrow \mathbb{R}$ such that $f(x,g(x))=0$ and $g(a)=b$.


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1. $[df(x,y)]=(\frac{\delta f}{\delta x}, \frac{\delta f}{\delta y})=(2x, 2y)$
2. $0=x^2+y^2-1$ $\Rightarrow$ $y=\sqrt{-x^2+1}$, it follows that $x \in [-1,1]$ and $y \in [0,1]$.

Is 1 and 2 correct? How do I do 3?

 

[HOI]

(1) is correct though I would have written "$\nabla f$" rather than "|df(x. y)|" and "$\left(\frac{\partial f}{\partial x} \frac{\partial f}{\partial y}\right)$".

(2) Almost. The question asks " For which $(x, y)\in R^2$ is the implicit function theorem usable to express y under the condition f(x,y)=0 as a function of x?"
From $x^2+ y^2- 1= 0$ it follow that $y= \pm\sqrt{1- x^2}$. Yes. in order for that to be a real number x must be in [-1, 1]. y does NOT have to be in [0, 1]. There are actually two functions "implied" by this equation, $y= \sqrt{1- x^2}$ and $y= -\sqrt{1- x^2}$. y can also be in [-1, 1].

(3) X is the circle in the xy-plane with center (0, 0) and radius 1. The condition that "$(a, b)\in X$ with b> 0" means that (a, b) is in the half circle above the x-axis. "a" is the point on the x-axis directly below (a, b). Frankly, "Find the largest possible neighbourhood V of a in R"
and "a continuous differentiable function g:V→R such that f(x,g(x))=0 and g(a)=b." makes no sense to me! The "largest possible neighborhood V of a in R" is all of R!

What they must mean is "find the largest possible neighborhood of R such that a function g exits such that f(x, g(x))= 0 and g(a)= b" and then ask you to find such a g.

The "largest possible neighborhood of R" must be a neighborhood of a such that we can solve for b. Now, what is the definition of "neighborhood" you are using? Some texts just use it to mean "some set containing the point", in which case the "largest neighborhood" is [-1, 1]. Some texts require that a "neighborhood" be an open set, in which case the "largest neighborhood" is (-1, 1). Finally, some texts require that a "neighborhood of a point" be open and symmetric about that point, in which case the "largest neighborhood" is the open set that extends on one side to the nearest point on the circle and on the other side an equal distance. If a= 0 that neighborhood is (-1, 1), if a> 0 the distance from a to 1 is 1- a while the distance from a to -1 is |a- (-1)|= a+ 1. 1- a is smaller so the largest neighborhood, in this sense, is (a- (1- a). 1)= (2a- 1, 1). If a is negative, -1 is the closer boundary. The distance from -1 to a is again |a- (-1)|= a+ 1 which, since a is negative, is now smaller than 1- a. The largest neighborhood is (-1, a+ (a+1))= (-1, 2a+ 1). Notice that taking a= 0 in both of those gives (-1, 1).

The function "g" is just $b= \pm\sqrt{1- a^2}$ using the minus if a< 0 and the plus if a> 0.