Problem 24 - Calculus On Manifolds: A Modern Approach To Classical Theorems Of Advanced Calculus

Question

2-24. Define \(f: \mathbf{R}^{2} \rightarrow \mathbf{R}\) by $$ f(x, y)= \begin{cases}x y \frac{x^{2}-y^{2}}{x^{2}+y^{2}} & (x, y) \neq 0 \\\ 0 & (x, y)=0\end{cases} $$ (a) Show that \(D_{2} f(x, 0)=x\) for all \(x\) and \(D_{1} f(0, y)=-y\) for all \(y\). (b) Show that \(D_{1,2} f(0,0) \neq D_{2,1} f(0,0)\)

Step-by-Step Solution

Answer

Partial derivatives: \( D_2 f(x, 0) = x \) and \( D_1 f(0, y) = -y \). Mixed partial derivatives at \((0,0)\) are not equal.

Step 1 - Define Partial Derivatives

The partial derivative of a function with respect to a variable is the derivative of the function while keeping the other variables constant. For function \(f(x, y)\), we need to find \(D_2 f(x, 0)\) and \(D_1 f(0, y)\).

Step 2 - Compute \(D_2 f(x, 0)\)

To find \(D_2 f(x, 0)\), differentiate \(f(x, y)\) with respect to \(y\): \(f(x, y) = x y \frac{x^2 - y^2}{x^2 + y^2}\) Fix \(y = 0\): \(f(x, 0) = 0\) The partial derivative \(D_2 f(x, y) = \frac{\partial f}{\partial y}\) evaluated at \(y = 0\) is \[ D_2 f(x, 0) = x \]

Step 3 - Compute \(D_1 f(0, y)\)

To find \(D_1 f(0, y)\), differentiate \(f(x, y)\) with respect to \(x\): \(f(x, y) = x y \frac{x^2 - y^2}{x^2 + y^2}\) Fix \(x = 0\): \(f(0, y) = 0\) The partial derivative \(D_1 f(x, y) = \frac{\partial f}{\partial x}\) evaluated at \(x = 0\) is \[ D_1 f(0, y) = - y \]

Step 4 - Calculate Mixed Partial Derivatives at (0,0)

Mixed partial derivatives refer to partial derivatives of a function with respect to multiple variables. For \(f(x, y)\), we are interested in computing \(D_{1,2} f(0,0)\) and \(D_{2,1} f(0,0)\): 1. Find \(D_{1,2} f(0,0) = \frac{\partial^2 f}{\partial x \partial y}\). To get \(D_{1,2} f(0,0)\), differentiate \(D_2 f(x,y)\) with respect to \(x\): \(D_2 f(x, y) = x(\frac{x^2-y^2}{x^2+y^2}) + xy(\frac{-2xy}{(x^2+y^2)^2})\). Substitute \((0,0)\), yields 0 due to fractions terms.2. Find \(D_{2,1} f(0,0) = \frac{\partial^2 f}{\partial y \partial x}\). To get \(D_{2,1} f(0,0)\), differentiate \(D_1 f(x,y)\) with respect to \(y\): Likewise the term \(D_2f(0, 0) =0\).

Step 5 - Conclude Inequality of Mixed Derivatives

Given the results: \[D_{1,2} f(0,0)= 0 eq0 = D_{2,1} f(0,0)\]. Therefore, the mixed partial derivatives are not equal.

Key Concepts

Mixed Partial Derivatives

Mixed partial derivatives involve taking the derivative of a multivariable function with respect to one variable, and then taking the derivative of the result with respect to another variable. For a function \(f(x, y)\), these derivatives would be \(D_{1,2} f(x, y) = \frac{\partial^2 f}{\partial x \partial y}\) and \(D_{2,1} f(x, y) = \frac{\partial^2 f}{\partial y \partial x}\).

In the exercise, we are asked to compute the mixed partial derivatives at the point \( (0,0) \). These derivatives are crucial for understanding the behavior of functions in multiple dimensions.

Here's a step-by-step to compute \(D_{1,2} f(0,0)\) and \(D_{2,1} f(0,0):\)

• First, calculate \(D_{2} f(x,y)\) and then differentiate it with respect to \(x\) to find \(D_{1,2} f(x,y)\).
• Next, calculate \(D_{1} f(x,y)\) and differentiate it with respect to \(y\) to find \(D_{2,1} f(x,y)\).
• In our exercise, the values were determined to be unequal, showing that \(D_{1,2} f(0,0) eq D_{2,1} f(0,0)\).

Mixed partial derivatives can tell us a lot about the curvature and behavior of surfaces defined by multivariable functions.

Continuous Partial Derivatives

For partial derivatives to be continuous, they must be well-defined and smooth across the domain of the function. Continuity of partial derivatives means that as you approach a point from different directions, the derivative must converge to the same value.

In our given function \(f(x, y)\), the partial derivatives need to be examined for continuity at all points, especially at the critical point \( (0,0) \). If the mixed partial derivatives are not equal at a point, this implies that the partial derivatives may not be continuous at that point.

• To check continuity of partial derivatives: Calculate the partial derivatives at the critical point.
• Verify if approaching from different paths gives consistent results.
• Note that discontinuous partial derivatives can indicate sharp corners or abrupt changes in the surface defined by the function.

Ensuring continuity can simplify many problems in multivariable calculus and is essential in various applications such as physics and engineering.

Multivariable Functions

Multivariable functions are functions with more than one input variable and often map these inputs to a single output. In this case, we consider a function \(f(x, y)\) which takes two inputs and produces an output. Understanding the behavior of multivariable functions involves examining how changes in one variable affect the function while keeping other variables constant.

Key aspects of multivariable functions include:

• **Partial Derivatives:** These derivatives describe the rate of change of the function concerning one variable, holding the others constant.
• **Critical Points:** Points where the first derivatives (gradients) are zero, which can indicate maxima, minima, or saddle points.
• **Mixed Derivatives:** As seen in the exercise, mixed derivatives can provide insights into the curvature and specific behavior around critical points.
• **Graphical Interpretation:** Visualizing multivariable functions can help understand their behavior. Functions can be represented as surfaces in three-dimensional space.

By analyzing partial and mixed derivatives, as well as understanding the continuity and critical points, students can develop a deeper comprehension of multivariable functions.