Math Problem Statement
STDY THE DIFFERENTIABLITY AT R abd tell me what is a vectorial function
Solution
To study the differentiability of the function at , let’s analyze the function's form and determine if it meets the conditions for differentiability.
The function is defined as:
\frac{xy}{\sqrt{x^2 + y^2}} & \text{if } (x, y) \neq (0, 0), \\ 0 & \text{if } (x, y) = (0, 0). \end{cases}$$ ### Step-by-Step Analysis of Differentiability at $$(0, 0)$$ 1. **Check Continuity at $$(0, 0)$$**: For $$f(x, y)$$ to be differentiable at $$(0,0)$$, it must first be continuous there. Let’s evaluate the limit of $$f(x, y)$$ as $$(x, y) \to (0, 0)$$: $$\lim_{(x, y) \to (0, 0)} f(x, y) = \lim_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2 + y^2}}.$$ If we approach along the $$x$$-axis (where $$y = 0$$), $$f(x, 0) = 0$$, and similarly, if we approach along the $$y$$-axis (where $$x = 0$$), $$f(0, y) = 0$$. Thus, the limit is zero along these paths. Since $$f(0,0) = 0$$, the function is continuous at $$(0,0)$$. 2. **Partial Derivatives at $$(0, 0)$$**: To check differentiability, we calculate the partial derivatives $$f_x$$ and $$f_y$$ at $$(0, 0)$$. However, directly evaluating partial derivatives at the origin is challenging here due to the square root term. Instead, we can examine the differentiability by checking the directional derivative or employing the definition of differentiability, involving a limit. 3. **Limit for Differentiability**: A function is differentiable at a point if there exists a linear approximation at that point, such that: $$\lim_{(x, y) \to (0, 0)} \frac{f(x, y) - f(0,0) - (f_x(0, 0)x + f_y(0, 0)y)}{\sqrt{x^2 + y^2}} = 0.$$ Substituting $$f(0,0) = 0$$ and noting that $$f_x(0,0)$$ and $$f_y(0,0)$$ (partial derivatives) are both 0, the expression simplifies to: $$\lim_{(x, y) \to (0, 0)} \frac{\frac{xy}{\sqrt{x^2 + y^2}}}{\sqrt{x^2 + y^2}} = \lim_{(x, y) \to (0, 0)} \frac{xy}{x^2 + y^2}.$$ This limit depends on the path taken toward the origin. For example: - Along $$y = x$$: $$\frac{xy}{x^2 + y^2} = \frac{x^2}{2x^2} = \frac{1}{2}$$. - Along $$y = -x$$: $$\frac{xy}{x^2 + y^2} = \frac{-x^2}{2x^2} = -\frac{1}{2}$$. Since the limit depends on the path, it does not exist, indicating that the function $$f(x, y)$$ is **not differentiable at $$(0,0)$$**. ### Understanding a Vectorial Function A **vectorial function** (or vector-valued function) is a function that takes one or more variables and returns a vector. Instead of producing a scalar output, a vectorial function gives an output in the form of a vector with multiple components. For example: $$\vec{F}(x, y) = \begin{pmatrix} f_1(x, y) \\ f_2(x, y) \end{pmatrix},$$ where $$f_1(x, y)$$ and $$f_2(x, y)$$ are scalar functions of $$x$$ and $$y$$. Would you like further details or have any questions? --- Here are five questions that delve deeper into this topic: 1. What is the difference between continuity and differentiability in multivariable functions? 2. How can we determine the differentiability of a function using partial derivatives? 3. What are common techniques to approach limits in two variables for differentiability analysis? 4. How do vectorial functions differ from scalar functions in terms of applications? 5. What is the significance of path-dependence in evaluating multivariable limits? **Tip:** When evaluating differentiability, checking multiple paths can help confirm if a limit exists or not, as path-dependence often indicates non-differentiability.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Differentiability in multivariable calculus
Limits
Continuity
Path dependence
Vector-valued functions
Formulas
f(x, y) = \begin{cases} \frac{xy}{\sqrt{x^2 + y^2}} & \text{if } (x, y) \neq (0, 0), \\ 0 & \text{if } (x, y) = (0, 0).\end{cases}
\lim_{(x, y) \to (0, 0)} \frac{f(x, y) - f(0,0)}{\sqrt{x^2 + y^2}} = 0
Theorems
Definition of differentiability
Definition of vector-valued functions
Suitable Grade Level
College level
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