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

The "largest neighborhood" is (-1, 1) which is the "largest interval" and the graph of the function is the "largest neighborhood in the xy-plane". So this is just the "Implicit Function Theorem". We have a function $y= \pm\sqrt{1- x^2}$ and we can differentiate that to get $dy/dx= \pm\frac{1}{2}(1- x^2)^{-1/2}(-2x)= \pm\frac{x}{\sqrt{1- x^2}}$ which is continuous for $x\ne 0$ but not at 0. We can't find a "largest neighborhood"
  • #1
chris222
1
0
$f: \mathbb{R^2} \rightarrow \mathbb{R}$, $f(x,y) = x^2+y^2-1$

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

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 $g(a)=b$.----------
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?
 
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  • #2
(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.
 

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

What is the Implicit Function Theorem for f(x,y) = x^2+y^2-1?

The Implicit Function Theorem for f(x,y) = x^2+y^2-1 is a mathematical theorem that states that under certain conditions, a multivariable function can be represented as an explicit function of one of its variables.

What are the conditions for the Implicit Function Theorem to hold?

The conditions for the Implicit Function Theorem to hold are that the function must be continuously differentiable and that the partial derivative of the function with respect to the dependent variable must not be equal to zero at the point in question.

How is the Implicit Function Theorem used in practical applications?

The Implicit Function Theorem is used in many practical applications, such as optimization problems, engineering, and physics. It allows for the determination of relationships between variables in a system, even when it is not possible to solve for one variable explicitly in terms of the others.

What is the difference between the Implicit Function Theorem and the Inverse Function Theorem?

The Implicit Function Theorem deals with multivariable functions, while the Inverse Function Theorem deals with single-variable functions. Additionally, the Implicit Function Theorem finds relationships between variables, while the Inverse Function Theorem finds the inverse of a function.

Can the Implicit Function Theorem be extended to functions with more than two variables?

Yes, the Implicit Function Theorem can be extended to functions with more than two variables. In this case, the function would be represented as an explicit function of multiple variables, rather than just one variable as in the case of two variables.

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