Question

Consider the following functions \(f\). a. Is \(f\) continuous at (0,0)\(?\) b. Is \(f\) differentiable at (0,0)\(?\) c. If possible, evaluate \(f_{x}(0,0)\) and \(f_{y}(0,0)\). d. Determine whether \(f_{x}\) and \(f_{y}\) are continuous at (0,0). e. Explain why Theorems 15.5 and 15.6 are consistent with the results in parts \((a)-(d)\). $$f(x, y)=\sqrt{|x y|}$$

Short answer

In summary, the function \(f(x, y) = \sqrt{|x y|}\) is continuous at the point (0,0), but it is not differentiable at (0,0) because its partial derivatives \(f_x\) and \(f_y\) are not defined at that point. The results of this analysis are consistent with Theorems 15.5 and 15.6, as the given function demonstrates that continuity does not necessarily imply differentiability, and the continuity of partial derivatives is not guaranteed in all cases.

Step-by-step solution

Check the continuity of \(f\) at (0,0)

To determine if the function \(f(x, y) = \sqrt{|x y|}\) is continuous at the point (0,0), we need to show that the limit of \(f(x, y)\) exists as \((x, y) \rightarrow (0, 0)\). Let's find \(\lim_{(x, y) \to (0,0)} \sqrt{|x y|}\): $$\lim_{(x, y) \to (0,0)} \sqrt{|x y|} = \lim_{(x, y) \to (0,0)} \sqrt{|0 * 0|} = 0$$ Since the limit exists and equals zero, the function is continuous at (0,0).

Check differentiability of \(f\) at (0,0)

To determine if \(f(x, y) = \sqrt{|x y|}\) is differentiable at (0,0), we need to check if its partial derivatives, \(f_x\) and \(f_y\), exist at (0,0). We will compute these partial derivatives in the next step. If they exist, then we can proceed to Step 4 and consider their continuity at (0, 0).

Evaluate \(f_{x}(0,0)\) and \(f_{y}(0,0)\) if possible

Let's find the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\). \(1.\) Partial derivative with respect to \(x:\) $$f_x(x, y) = \frac{\partial}{\partial x} (\sqrt{|x y|})$$ \(2.\) Partial derivative with respect to \(y:\) $$f_y(x, y) = \frac{\partial}{\partial y} (\sqrt{|x y|})$$ To compute the partial derivatives, we will use the fact that \(\sqrt{|x y|} = (|x y|)^{\frac{1}{2}}\) For \(f_x\), we have: $$f_x(x, y) = \frac{1}{2} (|x y|)^{-\frac{1}{2}} \frac{\partial}{\partial x} (x y)$$ Using the chain rule, we get: $$f_x(x, y) = \frac{y}{2\sqrt{|x y|}}$$ For \(f_y\), we have: $$f_y(x, y) = \frac{1}{2} (|x y|)^{-\frac{1}{2}} \frac{\partial}{\partial y} (x y)$$ Using the chain rule, we get: $$f_y(x, y) = \frac{x}{2\sqrt{|x y|}}$$ Now, let's evaluate these partial derivatives at (0,0). $$f_x(0,0) = \frac{0}{2\sqrt{|0 * 0|}} = \frac{0}{0}$$ $$f_y(0,0) = \frac{0}{2\sqrt{|0 * 0|}} = \frac{0}{0}$$ The partial derivatives are undefined at (0,0), which means \(f(x, y) = \sqrt{|x y|}\) is not differentiable at (0,0).

Check continuity of \(f_x\) and \(f_y\) at (0,0)

Since the partial derivatives \(f_x\) and \(f_y\) do not exist at (0,0), they cannot be continuous at that point.

Consistency with Theorems 15.5 and 15.6

Theorem 15.5 states that if a function is differentiable at a point (0,0), then it is also continuous at that point. In our case, \(f(x, y) = \sqrt{|x y|}\) is continuous at (0,0), but it is not differentiable at (0,0). This does not contradict Theorem 15.5 since the theorem implies a one-way implication (differentiability implies continuity). Theorem 15.6 states that if the partial derivatives \(f_x\) and \(f_y\) are continuous at a point, then the function is differentiable at that point. In our case, \(f_x\) and \(f_y\) are not continuous at (0,0), which does not contradict Theorem 15.6 as it gives a one-way implication (continuity of partial derivatives implies differentiability). The results obtained in parts (a) through (d) are consistent with Theorems 15.5 and 15.6 as they do not contradict these theorems' statements.

Key concepts

Partial Derivatives

Partial derivatives are the foundation of multivariable calculus. They are a way to understand how a function changes as each of its variables change, while keeping the others constant. Let's break this concept down with a simple explanation.

Consider a function of two variables, \(f(x, y) = \sqrt{|x y|}\). When we look at the partial derivative with respect to \(x\), denoted as \(f_x(x, y)\), we are essentially seeing how the output of the function changes as \(x\) changes, while keeping \(y\) constant.

Similarly, the partial derivative with respect to \(y\), \(f_y(x, y)\), examines the rate of change of the function as \(y\) changes with \(x\) held still. In simpler terms, partial derivatives give us the slope of the function's hypersurface along the respective axes.

In the example given, these derivatives help analyze the behavior of the function around the point (0,0). Calculating them, you realize that at this specific point, the derivatives become undefined due to division by zero. This indicates that despite the function being continuous at (0,0), it is not differentiable since the derivatives do not exist there.

Differentiability

Differentiability is a crucial concept when analyzing functions in multivariable calculus. A function is said to be differentiable at a point if it can be locally approximated by a linear function. In essence, this means the function behaves like a flat plane across small areas around that point.

For the function \(f(x, y) = \sqrt{|x y|}\), determining differentiability at (0,0) involves checking if partial derivatives exist and are continuous near that point. However, as we calculated earlier, the partial derivatives are undefined at (0,0). Consequently, the function cannot be differentiated into a linear approximation there.

Notably, differentiability implies continuity, but the reverse is not true. A function can be continuous without being differentiable, as seen in this exercise. The continuous yet jagged or sharp turning nature of \(f\) around (0,0) disrupts its differentiability.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions with two or more variables. It allows the exploration of spaces and rates of change in higher dimensions.

In this realm, you often encounter functions like \(f(x, y) = \sqrt{|x y|}\), where understanding how changes in two variables affect the output is vital.

Key operations in multivariable calculus include:

• Partial Derivatives: Examining changes relative to one variable while other variables are held constant, as described above.


• Differentiability: Understanding when a function can be linearly approximated at a point.


• Gradients and Directional Derivatives: Providing the direction and rate of maximum increase of a function.

These concepts are fundamental in fields ranging from physics to engineering to economics, helping model complex systems in the real world.