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Question #41621

Difficulty: 8/10 | 🟢 Free

The Question

A conservative vector field is given by F⃗(x,y)=M(x,y)i^+N(x,y)j^\vec{F}(x,y) = M(x,y)\hat{i} + N(x,y)\hat{j}. It is known that M(x,y)=2xex2−y2+3yM(x,y) = 2x e^{x^2 - y^2} + 3y and the scalar potential function ϕ(x,y)\phi(x,y) associated with F⃗\vec{F} satisfies ϕ(0,y)=e−y2\phi(0,y) = e^{-y^2}. Find the unit vector in the direction of F⃗\vec{F} at the point (1,1)(1, 1).

Options

  • A.

    526i^+126j^\frac{5}{\sqrt{26}}\hat{i} + \frac{1}{\sqrt{26}}\hat{j}

  • B.

    12i^+12j^\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}

  • C.

    529i^−229j^\frac{5}{\sqrt{29}}\hat{i} - \frac{2}{\sqrt{29}}\hat{j}

  • D.

    526i^−126j^\frac{5}{\sqrt{26}}\hat{i} - \frac{1}{\sqrt{26}}\hat{j}

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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

Unit Vector Normalization|Exact Differential Recognition|Chain Rule Complexity

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to reconstruct a conservative vector field from its partial components and a boundary condition, specifically targeting the chain rule during partial differentiation.

The trap lies in calculating N(x,y)=∂ϕ∂yN(x,y) = \frac{\partial \phi}{\partial y}.

Students frequently miss the negative sign when applying the chain rule to the exponent −y2-y^2, incorrectly yielding a derivative of 2yex2−y22y e^{x^2 - y^2}.

This error leads to N(1,1)=5N(1,1) = 5, resulting in the highly deceptive distractor vector 5i^+5j^5\hat{i} + 5\hat{j} and its normalized form 12i^+12j^\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}.

The clean magnitude of this incorrect vector creates false confidence.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterVector Algebra - Class 12
ConceptUnit Vector Normalization
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree