Problem 10 - The Calculus with Analytic Geometry

Question

Given \(f(x, y)= \begin{cases}\frac{3 x^{2} y^{2}}{x^{4}+y^{4}} & \text { if }(x, y) \neq(0,0) \\ 0 & \text { if }(x, y)=(0,0)\end{cases}\) Prove that \(D_{1} f(0,0)\) and \(D_{2} f(0,0)\) exist, but \(f\) is not differentiable at \((0,0)\).

Step-by-Step Solution

Answer

The partial derivatives \(D_{1} f(0,0)\) and \(D_{2} f(0,0)\) are both zero, but the function \(f\) is not differentiable at \((0,0)\) because the limit does not equal zero.

Step 1 - Understanding the function

Given the function: \[f(x, y)= \begin{cases} \frac{3 x^{2} y^{2}}{x^{4}+y^{4}} & \text { if }(x, y) eq(0,0) \ 0 & \text { if }(x, y)=(0,0) \end{cases}\] the goal is to prove that the partial derivatives \(D_{1} f(0,0)\) and \(D_{2} f(0,0)\) exist, but the function \(f\) is not differentiable at \(0,0\).

Step 2 - Compute \(D_{1} f(0,0)\)

Calculate the partial derivative with respect to \(x\) at (0,0). The definition of the partial derivative is: \[D_{1} f(0,0) = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h}.\] Evaluate \(f(h,0)\): \[f(h,0) = \frac{3h^2 \cdot 0^2}{h^4 + 0^4}=0.\] Thus, \[D_{1} f(0,0) = \lim_{h \to 0} \frac{0 - 0}{h} = 0.\].

Step 3 - Compute \(D_{2} f(0,0)\)

Calculate the partial derivative with respect to \(y\) at (0,0). The definition of the partial derivative is: \[D_{2} f(0,0) = \lim_{k \to 0} \frac{f(0,k) - f(0,0)}{k}.\] Evaluate \(f(0,k)\): \[f(0,k) = \frac{3 \cdot 0^2 \cdot k^2}{0^4 + k^4}=0.\] Thus, \[D_{2} f(0,0) = \lim_{k \to 0} \frac{0 - 0}{k} = 0.\].

Step 4 - Check differentiability at (0,0)

For \(f\) to be differentiable at \((0,0)\), the limit \[\lim_{(x,y)\to(0,0)} \frac{f(x,y) - f(0,0) - D_{1}f(0,0)x - D_{2}f(0,0)y}{\sqrt{x^2+y^2}} \] must exist and be equal to zero. Here, since \(f(0,0) = 0\), \(D_{1}f(0,0) = 0\), and \(D_{2}f(0,0) = 0\), we have: \[\lim_{(x,y)\to(0,0)} \frac{\frac{3x^2y^2}{x^4+y^4}}{\sqrt{x^2+y^2}}.\] Evaluating this along \(y = x\), gives \[\lim_{x \to 0} \frac{\frac{3x^4}{2x^4}}{x \sqrt{2}} = \frac{3}{2\sqrt{2}},\] which is not zero. Therefore, \(f\) is not differentiable at \((0,0)\).

Key Concepts

Differentiability

To understand differentiability in the context of multivariable calculus, let's start with a basic definition. A function is said to be differentiable at a point if it can be locally approximated by a linear function. In simpler terms, the function should behave like a flat plane near that point.

For a function of two variables, say, \(f(x, y)\), the function is considered differentiable at \((0,0)\) if the following limit exists and equals zero:
\[\frac{f(x,y) - f(0,0) - D_{1}f(0,0)x - D_{2}f(0,0)y}{\sqrt{x^2+y^2}}\]

Now, in the given exercise, we were able to show that the partial derivatives \(D_{1}f(0,0)\) and \(D_{2}f(0,0)\) exist and are both zero. However, when we checked the differentiability criteria, we found that the limit along the path \(y = x\) did not equal zero, indicating that the function is not differentiable at \((0,0)\).

This example demonstrates how partial derivatives existing and being zero does not necessarily mean that the function is differentiable at that point.

Limits and Continuity

Limits and continuity are fundamental concepts in calculus and are essential in understanding and proving differentiability.

A function of two variables, \(f(x, y)\), is continuous at a point \((a, b)\) if the limit of \(f(x, y)\) as \((x, y)\) approaches \((a, b)\) equals the function's value at that point. Mathematically, this is expressed as:
\[\text{lim}_{(x, y) \to (a, b)} f(x, y) = f(a, b)\]

In our exercise, the function \(f(x, y)\) had divided behavior depending on whether \((x, y)\) is \((0,0)\) or not. Evaluating the limit at these points is crucial to understanding the behavior of the function near this point, which then helps in determining the continuity.

Although we showed that the partial derivatives exist at \((0,0)\), checking the differentiated limit brought essential insight into the function's behavior, ultimately showing that \(f\) is not differentiable at that point. Understanding the behavior of limits and continuity thus forms a critical building block for grasping differentiability.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions of several variables. In dealing with functions like \(f(x, y)\), we need to consider partial derivatives, directional derivatives, and the more complex, overall differentiability.

Partial derivatives of a function indicate the rate of change along the axes of each variable individually. For instance, \(D_{1}f(0,0)\) denotes the rate of change of \(f\) in the \(x\) direction at the point \((0,0)\). Calculating such derivatives involves fixing one variable and differentiating with respect to the other.

To determine differentiability, we must go beyond partial derivatives. Overall differentiability involves examining if the function can be closely approximated by a linear function near the point. This requires that the function behaves consistently in all directions around the point.

In summary, multivariable calculus provides powerful tools to analyze functions of several variables. By understanding partial derivatives, limits, continuity, and differentiability, we gain a comprehensive view of how such functions behave globally and locally, leading to a deeper grasp of their properties and applications.