Problem 87 - Calculus Early Transcendentals

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 5 and 6 are consistent with the results in parts \((a)-(d).\) $$f(x, y)=\sqrt{|x y|}$$

Step-by-Step Solution

Answer

Based on the given function \(f(x, y) = \sqrt{|xy|}\), we found that it is continuous at the point \((0,0)\), but its partial derivatives \(f_x(0,0)\) and \(f_y(0,0)\) do not exist. Therefore, the function is not differentiable at \((0,0)\). This is consistent with Theorems 5 and 6, as the converse of Theorem 5 is not true (i.e., continuity does not necessarily imply differentiability), and Theorem 6 requires the existence and continuity of the partial derivatives for differentiability, which is not the case for our function at \((0,0)\).

Step 1: Check for continuity at (0,0)

First, we will check whether the function is continuous at \((0,0)\). A function is continuous at a point if the limit exists and is equal to the function's value at that point. So we need to find: $$\lim_{(x,y) \to (0,0)} f(x,y) = \lim_{(x,y) \to (0,0)} \sqrt{|xy|}$$ As \(x\) and \(y\) approach \(0\), the product \(|xy|\) also approaches \(0\). Therefore, the square root of that product approaches \(\sqrt{0} = 0\). Hence, the function is continuous at \((0,0)\).

Step 2: Calculate partial derivatives

Next, we'll find the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\). $$f_x(x, y) = \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \sqrt{|xy|}$$ $$f_y(x, y) = \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \sqrt{|xy|}$$ To avoid issues with the absolute value, we'll handle each case (either \(xy \geq 0\) or \(x \geq 0\), \(y \leq 0\) or \(xy \leq 0\) or \(x \leq 0\), \(y \geq 0\)) separately: - If \(xy \geq 0\), then \(|xy| = xy\), and: $$f_x(x, y) = \frac{1}{2\sqrt{xy}}y = \frac{y}{2\sqrt{xy}}$$ $$f_y(x, y) = \frac{1}{2\sqrt{xy}}x = \frac{x}{2\sqrt{xy}}$$ - If \(xy \leq 0\), then \(|xy| = -xy\), and: $$f_x(x, y) = -\frac{1}{2\sqrt{-xy}}y = \frac{-y}{2\sqrt{-xy}}$$ $$f_y(x, y) = -\frac{1}{2\sqrt{-xy}}x = \frac{-x}{2\sqrt{-xy}}$$

Step 3: Evaluate partial derivatives at (0,0)

We need to find \(f_x(0,0)\) and \(f_y(0,0)\) if possible. However, when we plug \(x=0\) or \(y=0\) into the expressions for \(f_x(x, y)\) and \(f_y(x, y)\), we get a \(0\) in the denominator, which is not defined. Hence, \(f_x(0,0)\) and \(f_y(0,0)\) do not exist.

Step 4: Check for continuity of partial derivatives at (0,0)

Since the partial derivatives \(f_x(0,0)\) and \(f_y(0,0)\) do not exist, we cannot determine their continuity at \((0,0)\).

Step 5: Explain consistency with Theorems 5 and 6

The results in parts \((a)-(d)\) are consistent with Theorems 5 and 6 because: - Theorem 5 states that if a function is differentiable at a point, then it is continuous at that point. In our case, the function is continuous at \((0,0)\). However, the converse of Theorem 5 is not true: if a function is continuous at a point, it does not necessarily mean it is differentiable at that point. Our function is an example of this. - Theorem 6 states that if all partial derivatives of a function exist and are continuous in a region, then the function is differentiable in that region. In our case, the partial derivatives do not exist at \((0,0)\), so the function is not differentiable at \((0,0)\), which is consistent with Theorem 6.

Key Concepts

Continuity

In mathematics, a function is said to be continuous at a point if the limit of the function as it approaches the point is equal to the function's value at that point. To determine continuity at a point (0,0) for the function \(f(x, y) = \sqrt{|xy|}\), we need to check:

• If the limit \(\lim_{(x,y) \to (0,0)} \sqrt{|xy|}\) exists.
• If the limit equals the function value \(f(0,0)\).

In this particular function, as both \(x\) and \(y\) approach 0, the product \(|xy|\) approaches 0. That means \(\sqrt{|xy|} \approaches \sqrt{0} = 0\). Since \(f(0,0) = 0\), these values match, confirming continuity at (0,0).

Continuity is crucial as it is the first step towards analyzing further properties like differentiability, but continuity itself does not guarantee differentiability.

Differentiability

Differentiability is a concept in calculus that indicates if a function can be locally approximated by a linear function at a given point. In simple terms, differentiability tells us if a function has a clear slope or tangent at a point.
To check differentiability of \(f(x, y) = \sqrt{|xy|}\) at (0,0), we consider:

• Existence of partial derivatives \(f_x\) and \(f_y\).
• Continuity of these partial derivatives.

The partial derivatives at (0,0), given the formula, present an issue as they lead to division by zero, making them undefined.

Without defined partial derivatives at (0,0), the function cannot be differentiable at that point. This exemplifies that even continuous functions might not be differentiable, highlighting a critical nuance in calculus.

Multivariable Functions

Multivariable functions involve more than one variable in their expression, such as \(f(x, y) = \sqrt{|xy|}\). These functions relate to surfaces or curves in space, offering a complex view compared to single-variable functions.
Key concepts about these functions include:

• Partial differentiation: breaking down functions to study the change concerning one variable at a time.
• Understanding limits from multiple directions due to more dimensions involved.

With more variables, these functions introduce challenges such as handling absolute values like \(|xy|\) and determining how operations like square roots impact continuity and differentiability.

Working with multivariable functions requires visualization skills and a solid understanding of partial derivatives, which deepen one's grasp of calculus.

Theorems in Calculus

Theorems in calculus provide foundational truths and help us draw conclusions about functions’ behavior around specific points. Two key theorems relevant here are:

• Theorem 5: If a function is differentiable at a point, it is also continuous at that point. However, the reverse is not always true, i.e., continuity does not imply differentiability.
• Theorem 6: If all partial derivatives of a function exist and are continuous in a region, the function is differentiable in that region.

The function \(f(x,y) = \sqrt{|xy|}\) aligns with these theorems:- It is continuous at (0,0) but not differentiable, illustrating the non-reciprocal nature of Theorem 5.- Since the partial derivatives \(f_x\) and \(f_y\) do not exist at (0,0), Theorem 6 holds true as differentiability fails.

Understanding these theorems aids students in linking the mathematical properties and applying them accurately across different contexts and functions.