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Question #40952

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \mathbb{R}^2 \to \mathbb{R}$ be defined by $f(x,y) = \begin{cases} \frac{x^3 y - x y^3}{x^2+y^2} & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0) \end{cases}$ Let $H(x,y) = \sqrt{|f(x,y)|}$ . Let $D_{\vec{u}} H(0,0) = \lim_{t \to 0^+} \frac{H(t\cos\theta, t\sin\theta) - H(0,0)}{t}$ denote the directional derivative of $H$ at $(0,0)$ in the direction of the unit vector $\vec{u} = (\cos\theta, \sin\theta)$ for $\theta \in [0, 2\pi)$ . Let $M$ be the maximum value of $D_{\vec{u}} H(0,0)$ over all such unit vectors, and let $L$ be the number of distinct unit vectors for which this maximum is attained. Find the value of $L \times M$ .

Options

A.
0
B.
2
C.
4
D.
8

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Chain Rule Complexity|Homogeneous Function Homogeneity|Differentiability vs Continuity

Why This Question Is Tricky

Common Mistakes

This question tests the critical distinction between the existence of partial derivatives and true differentiability at a point, specifically within the context of composite homogeneous functions.

Students often fall into the trap of computing $H_x(0,0) = 0$ and $H_y(0,0) = 0$ , and then incorrectly applying the gradient formula $D_{\vec{u}} H = \nabla H \cdot \vec{u}$ to conclude the directional derivative is $0$ in all directions (leading to Option A).

However, the outer function $g(t) = \sqrt{|t|}$ combined with the degree-2 homogeneous inner function creates a surface that is not locally linear at the origin.

By forcing the use of the fundamental limit definition of the directional derivative, the problem exposes the flaw in blindly chasing algebraic derivative formulas without verifying differentiability.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Limits and Derivatives - Class 11
Concept	Chain Rule Complexity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free