Problem 32 - Calculus On Manifolds: A Modern Approach To Classical Theorems Of Advanced Calculus

Question

2-32. (a) Let \(f: \mathbf{R} \rightarrow \mathbf{R}\) be defined by $$ f(x)= \begin{cases}x^{2} \sin \frac{1}{x} & x \neq 0 \\ 0 & x=0\end{cases} $$ Show that \(f\) is differentiable at 0 but \(f^{\prime}\) is not continuous at 0 . (b) Let \(f: \mathbf{R}^{2} \rightarrow \mathbf{R}\) be defined by $$ f(x, y)= \begin{cases}\left(x^{2}+y^{2}\right) \sin \frac{1}{\sqrt{x^{2}+y^{2}}} & (x, y) \neq 0 \\ 0 & (x, y)=0\end{cases} $$ Show that \(f\) is differentiable at \((0,0)\) but \(D_{i} f\) is not continuous at \((0,0)\)

Step-by-Step Solution

Answer

For part (a), \( f \) is differentiable at 0, but \( f' \) is not continuous at 0. For part (b), \( f \) is differentiable at \((0,0)\), but partial derivatives \( D_i f \) are not continuous at \((0,0)\).

Step 1: Define the Function for Part (a)

Given the function \[ f(x)= \begin{cases}x^{2} \sin \frac{1}{x} & x eq 0 \ 0 & x=0 \end{cases} \]We need to show that this function is differentiable at 0 and that the derivative is not continuous at 0.

Step 2: Show Differentiability at Zero for Part (a)

To show that \( f \) is differentiable at \( x=0 \), we need to check if \[ \lim_{{h \to 0}} \frac{f(h) - f(0)}{h} \] exists. Let’s calculate \[ \frac{f(h) - f(0)}{h} = \frac{h^2 \sin \frac{1}{h}}{h} = h \sin \frac{1}{h} \]Since \( -1 \leq \sin \frac{1}{h} \leq 1 \), we get\( -h \leq h \sin \frac{1}{h} \leq h \). As \( h \to 0 \), \( h \sin \frac{1}{h} \to 0 \).Thus,\[ \lim_{{h \to 0}} h \sin \frac{1}{h} = 0 \]Therefore, \( f \) is differentiable at \( x=0 \) and \( f'(0) = 0 \).

Step 3: Check Continuity of the Derivative at Zero for Part (a)

We need to compute the derivative for \( x eq 0 \). Using the quotient rule on \( x^2 \sin \frac{1}{x} \):\[ f'(x) = 2x \sin \frac{1}{x} - \cos \frac{1}{x} \]We analyze the limit:\[ \lim_{{x \to 0}} f'(x) = \lim_{{x \to 0}} \left(2x \sin \frac{1}{x} - \cos \frac{1}{x}\right) \]Since \( \sin \frac{1}{x} \) and \( \cos \frac{1}{x} \) oscillate between -1 and 1, \( \lim_{{x \to 0}} \left(2x \sin \frac{1}{x} - \cos \frac{1}{x}\right) \) does not exist. Thus, \( f' \) is not continuous at \( x=0 \).

Step 4: Define the Function for Part (b)

Given the function\[ f(x, y) = \begin{cases}(x^{2} + y^{2}) \sin \frac{1}{\sqrt{x^{2} + y^{2}}} & (x, y) eq 0 \ 0 & (x, y) = 0 \end{cases} \]We need to show that this function is differentiable at \((0,0)\) and that the partial derivatives \( D_i f \) are not continuous at \((0,0)\).

Step 5: Show Differentiability at (0,0) for Part (b)

To show differentiability, we need to check if\[ \lim_{{(h, k) \to (0,0)}} \frac{f(h, k) - f(0, 0) - (h D_1 f(0, 0) + k D_2 f(0, 0))}{\sqrt{h^2 + k^2}} = 0 \]Since \( f(0,0) = 0 \), we consider\[ \frac{f(h, k)}{\sqrt{h^2 + k^2}} = \frac{(h^2 + k^2) \sin \frac{1}{\sqrt{h^2 + k^2}}}{\sqrt{h^2 + k^2}} = \sqrt{h^2 + k^2} \sin \frac{1}{\sqrt{h^2 + k^2}} \]Since \( -1 \leq \sin \frac{1}{\sqrt{h^2 + k^2}} \leq 1 \),\( -\sqrt{h^2 + k^2} \leq \sqrt{h^2 + k^2} \sin \frac{1}{\sqrt{h^2 + k^2}} \leq \sqrt{h^2 + k^2} \). As \( (h, k) \to (0,0) \),\[ \sqrt{h^2 + k^2} \sin \frac{1}{\sqrt{h^2 + k^2}} \to 0 \]Therefore, \( f \) is differentiable at \((0,0)\).

Step 6: Check Continuity of Partial Derivatives at (0,0) for Part (b)

We need to compute the partial derivatives for \((x, y) eq 0\). Using partial differentiation, we focus on the component forms:\[ f_x(x, y) = (2x \sin \frac{1}{\sqrt{x^2 + y^2}}) - x(x^2 + y^2)^{\frac{-3}{2}} \cos \frac{1}{\sqrt{x^2 + y^2}} \]To check the limit:\[ \lim_{{(x,y) \to (0,0)}} f_x(x, y) \]The oscillatory nature of \( \sin \frac{1}{\sqrt{x^2 + y^2}} \) and \( \cos \frac{1}{\sqrt{x^2 + y^2}} \) implies that the partial derivatives do not tend to a limit. Thus, \( D_i f \) are not continuous at \((0,0)\).

Key Concepts

headline of the respective core concept

Multivariable calculus is a branch of calculus that involves several independent variables. Unlike single-variable calculus, you need to consider how each variable affects the function simultaneously. For instance, a function can depend on both x and y, leading to a more complex analysis of limits, differentiability, and continuity.

headline of the respective core concept

Partial derivatives help us understand how a function changes with respect to one variable while keeping the other variables constant. To find a partial derivative, we differentiate the function with respect to one variable at a time. For example, the partial derivative of a function f(x, y) with respect to x is denoted as \( \frac{abla f}{abla x} \) and is found by differentiating f while treating y as a constant.

headline of the respective core concept

Discontinuous derivatives describe a situation where the derivative of a function exists but is not continuous across its domain. This is important in advanced calculus because continuity is often assumed for the behavior of derivatives. When derivatives are discontinuous, it means that their values do not smoothly transition, leading to abrupt changes in the slope of the function. This can happen even if the function itself is differentiable at a point.