Problem 30 - The Calculus with Analytic Geometry

Question

In Exercises 30 through 33 , we show that a function may be differentiable at a point even though it is not continuously differentiable there. Hence, the conditions of Theorem \(19.5 .5\) are sufficient but not necessary for differentiability. The function \(f\) in these exercises is defined by In Exercises 30 through 33 , we show that a function may be differentiable at a point even though it is not continuously differentiable there. Hence, the conditions of Theorem \(19.5 .5\) are sufficient but not necessary for differentiability. The function \(f\) in these exercises is defined by $$ f(x, y)= \begin{cases}\left(x^{2}+y^{2}\right) \sin \frac{1}{\sqrt{x^{2}+y^{2}}} & \text { if }(x, y) \neq(0,0) \\ 0 & \text { if }(x, y)=(0,0)\end{cases} $$ $$ \text { Find } \Delta f(0,0) \text {. } $$

Step-by-Step Solution

Answer

The function is differentiable at (0, 0) with ∆f(0, 0) = 0.

Step 1 - Understanding the function

The function given is a piecewise function with two definitions based on the point (x, y). For (x, y) = (0, 0), the function is zero. For any other point, the function is given by \[f(x, y)= \left(x^{2}+y^{2}\right) \sin \frac{1}{\sqrt{x^{2}+y^{2}}}\]

Step 2 - Definition of Differentiability

Check if the function is differentiable at (0,0). This means checking if \[\lim_{(x, y) \rightarrow (0, 0)} \frac{f(x, y) - f(0, 0) - (D_x f (0, 0) x + D_y f (0, 0) y)}{\sqrt{x^2 + y^2}} = 0\]

Step 3 - Partial Derivatives

Find the partial derivatives \(D_x f\) and \(D_y f\) at (0, 0). For function \[f(x, y)= \left(x^{2}+y^{2}\right) \sin \frac{1}{\sqrt{x^{2}+y^{2}}}\], the partial derivatives at (0, 0) can be determined by taking limits.Calculate: \[D_x f(0, 0) = \lim_{h \rightarrow 0} \frac{f(h, 0) - f(0, 0)}{h}\] and \[D_y f(0, 0) = \lim_{h \rightarrow 0} \frac{f(0, h) - f(0, 0)}{h}\]

Step 4 - Evaluating Limits

Evaluate \[\lim_{h \rightarrow 0} \frac{f(h, 0)}{h} = \lim_{h \rightarrow 0} \frac{h^2 \sin \frac{1}{h}}{h} = \lim_{h \rightarrow 0} h \sin \frac{1}{h} = 0\]Similarly, \[\lim_{h \rightarrow 0} \frac{f(0, h)}{h} = \lim_{h \rightarrow 0} \frac{h^2 \sin \frac{1}{h}}{h} = \lim_{h \rightarrow 0} h \sin \frac{1}{h} = 0\]Thus, both partial derivatives are zero at (0, 0).

Step 5 - Applying the definition at (0, 0)

Using the differentiability limit from Step 2:\[\lim_{(x, y) \rightarrow (0, 0)} \frac{f(x, y)}{\sqrt{x^2 + y^2}} = \lim_{(x, y) \rightarrow (0, 0)} \frac{\left(x^{2}+y^{2}\right) \sin \frac{1}{\sqrt{x^{2}+y^{2}}}}{\sqrt{x^2 + y^2}}\]Simplify this to \[\lim_{r \rightarrow 0} r \sin \frac{1}{r} = 0\]since \(r = \sqrt{x^2 + y^2}\).

Step 6 - Conclusion

Since the limit equals 0, the function is differentiable at (0, 0). Thus, \(\Delta f(0, 0) = 0\).

Key Concepts

Partial Derivatives

In multivariable calculus, the concept of partial derivatives helps us understand how a function changes as each variable changes individually. For a function \(f(x, y)\), the partial derivative with respect to \(x\) is found by fixing \(y\) and differentiating with respect to \(x\). Similarly, the partial derivative with respect to \(y\) is found by fixing \(x\) and differentiating with respect to \(y\).
For example, given the function:
\[f(x, y)= \left(x^{2}+y^{2}\right) \sin \frac{1}{\sqrt{x^{2}+y^{2}}}\]
we can find the partial derivative with respect to \(x\) at any point other than the origin. But, at the origin \( (0, 0) \), we use the definition and the limit process to find that both partial derivatives are zero:
\[D_x f(0, 0) = \lim_{h \rightarrow 0} \frac{f(h, 0) - f(0, 0)}{h} = 0\]
and
\(D_y f(0, 0) = \lim_{h \rightarrow 0} \frac{f(0, h) - f(0, 0)}{h} = 0\).
This means the initial rates of change in both \(x\) and \(y\) directions are zero at the origin.

Piecewise Functions

Piecewise functions are defined by different expressions based on the values of the input variables. These different parts can describe the behavior of a function in various regions.
In the given exercise, our function \(f(x, y)\) is defined piecewise as:
\[ f(x, y)= \begin{cases}\left(x^{2}+y^{2}\right) \sin \frac{1}{\sqrt{x^{2}+y^{2}}} & \text { if }(x, y) eq(0,0)\ 0 & \text { if }(x, y)=(0,0) \end{cases} \]
This means that the function takes on one form when \((x, y)\) is not equal to \((0, 0)\) and another form at the exact point \((0, 0)\).
For understanding and working with piecewise functions:

• Understand each piece of the function separately in terms of their definitions and behavior.
• Check for continuity and differentiability within each piece and at the boundaries (in this case, the origin).

Limits in Calculus

Limits play a crucial role in calculus, especially when dealing with continuity and differentiability. A limit helps us understand the behavior of a function as the input approaches a particular point.
For the function \(f(x, y)\) given in the exercise, we need to determine if the function is differentiable at \((0,0)\). This involves evaluating the following limit:
\[\lim_{(x, y) \rightarrow (0, 0)} \frac{f(x, y) - f(0, 0) - (D_x f (0, 0) x + D_y f (0, 0) y)}{\sqrt{x^2 + y^2}} = 0\]
To simplify this:
Since the partial derivatives \(D_x f(0, 0)\) and \(D_y f(0, 0)\) are both zero, the limit simplifies to:
\[\lim_{(x, y) \rightarrow (0, 0)} \frac{f(x, y)}{\sqrt{x^2 + y^2}} = \lim_{r \rightarrow 0} r \sin \frac{1}{r} = 0 \]
This is because the expression involves \(r = \sqrt{x^2 + y^2}\) and evaluating the limit as \(r\) approaches zero shows the value approaches zero.
Understanding limits can help us determine important properties such as continuity, and differentiability, which are essential in mathematical analysis of functions.