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Question #71547

Difficulty: 8/10 | 🟢 Free

The Question

Let SS be the surface defined by the position vector R⃗(u,v)=ui^+vj^+(u2+v2)k^\vec{R}(u,v) = u\hat{i} + v\hat{j} + (u^2+v^2)\hat{k}. A family of curves CC is generated by projecting the intersections of SS with the family of planes Pc:r⃗⋅(i^−12ck^)=0P_c: \vec{r} \cdot (\hat{i} - \frac{1}{2c}\hat{k}) = 0 onto the xyxy-plane, where c>0c > 0 is a parameter.

Let Γ\Gamma be the family of orthogonal trajectories of CC in the xyxy-plane. Let Γ0\Gamma_0 be the specific curve in Γ\Gamma that passes through the point (7,7)(7, 7).

Evaluate the line integral I=14π∮Γ0(−y2dx+2xydy)I = \frac{1}{4\pi} \oint_{\Gamma_0} \left( -y^2 dx + 2xy dy \right) where Γ0\Gamma_0 is traversed in the counter-clockwise direction.

Options

  • A.

    0

  • B.

    341

  • C.

    343

  • D.

    686

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

Orthogonal Trajectories|Geometric Interpretation of Cross Product|Line lying in a plane|Parametric form pitfalls

Why This Question Is Tricky

Examiner Levers

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

This question tests the geometric interpretation of vector-defined surfaces and the derivation of orthogonal trajectories.

The primary trap lies in the 'Parametric form pitfall' and loss of degrees of freedom: students might attempt to parameterize the 3D intersection curve r⃗(t)\vec{r}(t) and find orthogonal vectors in 3D space before projecting, which conflates the parameter of the curve with the parameter of the family, leading to an incorrect differential equation.

Furthermore, a secondary trap is embedded in the line integral: if a student fails to find the orthogonal trajectory and mistakenly uses the original curve C0C_0 (centered at (7,0)(7,0)), Green's theorem yields ∬4ydA=0\iint 4y dA = 0.

The value 341 is included as a distractor for arithmetic errors near the 737^3 calculation.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterDifferential Equations - Class 12
ConceptOrthogonal Trajectories
Difficulty8
Cognitive Load7
Target Solve Time170 sec
AccessFree