Question

Let \( f: \mathbb{R}^2 \to \mathbb{R} \) be defined by \[ f(x, y) = \begin{cases} \frac{2x^2y}{x^2 + y^2}, & (x, y) \neq (0,0) \\ 0, & (x, y) = (0,0) \end{cases} \] Then

• A. the directional derivative of \( f \) at \( (0,0) \) in the direction of \( (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \) is \( \frac{1}{\sqrt{2}} \)
• B. the directional derivative of \( f \) at \( (0,0) \) in the direction of \( (0, 1) \) is 1
• C. the directional derivative of \( f \) at \( (0,0) \) in the direction of \( (1, 0) \) is 0
• D. \( f \) is NOT differentiable at \( (0,0) \)

Hint:

A common test for non-differentiability at the origin is to check if \( D_{\mathbf{u}}f(0,0) = f_x(0,0)u_1 + f_y(0,0)u_2 \). If this equality fails for any single direction \( \mathbf{u} \), the function is not differentiable. This is often faster than computing the differentiability limit.

Answer 1

The Correct Option is A, C

Step 1: Understanding the Concept:
This question tests the concepts of directional derivatives, partial derivatives, and differentiability for a function of two variables at the origin. We need to apply the definitions to check each statement.
Step 2: Key Formula or Approach:
The directional derivative of \( f \) at \( (0,0) \) in the direction of a unit vector \( \mathbf{u} = (u_1, u_2) \) is given by: \[ D_{\mathbf{u}}f(0,0) = \lim_{h \to 0} \frac{f(0 + hu_1, 0 + hu_2) - f(0,0)}{h} = \lim_{h \to 0} \frac{f(hu_1, hu_2)}{h} \] For differentiability at \( (0,0) \), if the partial derivatives \( f_x(0,0) \) and \( f_y(0,0) \) exist, we must check if: \[ \lim_{(h,k) \to (0,0)} \frac{f(h,k) - f(0,0) - h f_x(0,0) - k f_y(0,0)}{\sqrt{h^2 + k^2}} = 0 \] Step 3: Detailed Explanation or Calculation:
First, let's find a general formula for the directional derivative at \( (0,0) \) for any unit vector \( \mathbf{u} = (u_1, u_2) \). \[ f(hu_1, hu_2) = \frac{2(hu_1)^2(hu_2)}{(hu_1)^2 + (hu_2)^2} = \frac{2h^3 u_1^2 u_2}{h^2(u_1^2 + u_2^2)} = \frac{2h^3 u_1^2 u_2}{h^2(1)} = 2h u_1^2 u_2 \] So, the directional derivative is: \[ D_{\mathbf{u}}f(0,0) = \lim_{h \to 0} \frac{2h u_1^2 u_2}{h} = 2u_1^2 u_2 \] Analysis of Statement (A):
The direction is \( (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \). Here, \( u_1 = \frac{1}{\sqrt{2}} \) and \( u_2 = \frac{1}{\sqrt{2}} \). \[ D_{\mathbf{u}}f(0,0) = 2 \left(\frac{1}{\sqrt{2}}\right)^2 \left(\frac{1}{\sqrt{2}}\right) = 2 \left(\frac{1}{2}\right) \left(\frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}} \] Therefore, statement (A) is correct.
Analysis of Statement (B):
The direction is \( (0, 1) \). Here, \( u_1 = 0 \) and \( u_2 = 1 \). \[ D_{\mathbf{u}}f(0,0) = 2(0)^2(1) = 0 \] The statement says the derivative is 1. Therefore, statement (B) is incorrect.
Analysis of Statement (C):
The direction is \( (1, 0) \). Here, \( u_1 = 1 \) and \( u_2 = 0 \). \[ D_{\mathbf{u}}f(0,0) = 2(1)^2(0) = 0 \] Therefore, statement (C) is correct. Note that this is also the partial derivative \( f_x(0,0) \). Similarly, from (B), \( f_y(0,0)=0 \).
Analysis of Statement (D):
For \( f \) to be differentiable at \( (0,0) \), the directional derivative must be a linear function of \( u_1, u_2 \). Specifically, \( D_{\mathbf{u}}f(0,0) \) should equal \( f_x(0,0)u_1 + f_y(0,0)u_2 \). From our calculations, \( f_x(0,0) = 0 \) and \( f_y(0,0) = 0 \). So, if \( f \) were differentiable, we would need \( D_{\mathbf{u}}f(0,0) = 0 . u_1 + 0 . u_2 = 0 \) for all unit vectors \( \mathbf{u} \). However, from (A), we found a direction \( (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \) for which the directional derivative is \( \frac{1}{\sqrt{2}} \neq 0 \). This contradiction shows that \( f \) is not differentiable at \( (0,0) \).
Alternatively, using the limit definition for differentiability: \[ \lim_{(h,k) \to (0,0)} \frac{f(h,k) - 0 - h . 0 - k . 0}{\sqrt{h^2 + k^2}} = \lim_{(h,k) \to (0,0)} \frac{2h^2k/(h^2+k^2)}{\sqrt{h^2+k^2}} = \lim_{(h,k) \to (0,0)} \frac{2h^2k}{(h^2+k^2)^{3/2}} \] Let's check the limit along the path \( k = h \): \[ \lim_{h \to 0} \frac{2h^2(h)}{(h^2+h^2)^{3/2}} = \lim_{h \to 0} \frac{2h^3}{(2h^2)^{3/2}} = \lim_{h \to 0} \frac{2h^3}{2\sqrt{2} |h|^3} = \frac{1}{\sqrt{2}} \neq 0 \] Since the limit is not 0, the function is not differentiable at \( (0,0) \).
Therefore, statement (D) is correct.
Step 4: Final Answer:
The correct statements are (A), (C), and (D).
Step 5: Why This is Correct:
The function exhibits different behaviors along different paths to the origin. Although all directional derivatives exist, they are not related in the linear way required for differentiability. The formula \( D_{\mathbf{u}}f = 2u_1^2 u_2 \) is not linear in \( u_1 \) and \( u_2 \), which is a key indicator that the function is not differentiable.