GATE 2024 EC Q.54: solution and answer

GATE 2024 · ECQ.54 · 2 marks · MSQ

Let $F_{1}$ , $F_{2}$ , and $F_{3}$ be functions of $(x, y, z)$ . Suppose that for every given pair of points $A$ and $B$ in space, the line integral $\int_{C} (F_{1} d x + F_{2} d y + F_{3} d z)$ evaluates to the same value along any path $C$ that starts at $A$ and ends at $B$ . Then which of the following is/are true?

(A)
For every closed path $Γ$ , we have $\oint_{Γ} (F_{1} d x + F_{2} d y + F_{3} d z) = 0$ .
(B)
There exists a differentiable scalar function $f (x, y, z)$ such that $F_{1} = \frac{\partial f}{\partial x}, F_{2} = \frac{\partial f}{\partial y}, F_{3} = \frac{\partial f}{\partial z}$ .
(C)
$\frac{\partial F _{1}}{\partial x} + \frac{\partial F _{2}}{\partial y} + \frac{\partial F _{3}}{\partial z} = 0$ .
(D)
$\frac{\partial F _{3}}{\partial y} = \frac{\partial F _{2}}{\partial z}, \frac{\partial F _{1}}{\partial z} = \frac{\partial F _{3}}{\partial x}, \frac{\partial F _{2}}{\partial x} = \frac{\partial F _{1}}{\partial y}$ .

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Answer and solution

ANSWER

(A), (B) and (D) are true. A path-independent line integral means $F = F_{1} \hat{i} + F_{2} \hat{j} + F_{3} \hat{k}$ is conservative. So it has a potential ( $F = \nabla f$ ), its integral around every closed path is zero, and it is irrotational ( $\nabla \times F = 0$ , which is option D written out component by component). Its divergence does not have to be zero, so (C) is false.

Official answer: A, B, DSource: Official answer key, GATE 2024 EC (IISc Bengaluru)

Write $F = F_{1} \hat{i} + F_{2} \hat{j} + F_{3} \hat{k}$ , so the integral is $\int_{C} F \cdot d r$ . The question says this integral depends only on the end points $A$ and $B$ , not on the path. That makes $F$ a conservative field. As usual for such questions, take $F_{1}, F_{2}, F_{3}$ to be continuously differentiable.
(A) Pick two points $A$ and $B$ on any closed path $Γ$ and split $Γ$ into $C_{1}$ (from $A$ to $B$ ) and $C_{2}$ (from $B$ back to $A$ ). Reversing a path flips the sign of its integral, so $\oint_{Γ} F \cdot d r = \int_{C_{1}} F \cdot d r + \int_{C_{2}} F \cdot d r = \int_{C_{1}} F \cdot d r - \int_{- C_{2}} F \cdot d r = 0,$ because $C_{1}$ and $- C_{2}$ both run from $A$ to $B$ and so give the same value. True.
(B) Fix a point $A$ and define $f (P) = \int_{A}^{P} F \cdot d r$ . This is well defined because the value does not depend on the path taken to $P$ . To find $\frac{\partial f}{\partial x}$ at $P$ , reach $P + h \hat{i}$ by going to $P$ first and then along the straight segment parallel to the $x$ -axis. On that segment $d r = \hat{i} d x$ , so $f (P + h \hat{i}) - f (P) = \int_{x}^{x + h} F_{1} d x$ , and dividing by $h$ and letting $h \to 0$ gives $\frac{\partial f}{\partial x} = F_{1}$ . In the same way $\frac{\partial f}{\partial y} = F_{2}$ and $\frac{\partial f}{\partial z} = F_{3}$ . These partial derivatives are continuous, so $f$ is differentiable and $F = \nabla f$ . True.
(D) Because $F = \nabla f$ , we get $\nabla \times F = \nabla \times \nabla f = 0$ , since mixed partial derivatives are equal (Schwarz/Clairaut), e.g. $\frac{\partial F _{3}}{\partial y} = \frac{\partial ^{2} f}{\partial y \partial z} = \frac{\partial ^{2} f}{\partial z \partial y} = \frac{\partial F _{2}}{\partial z}$ . Writing out the three components of $\nabla \times F = 0$ : $\frac{\partial F _{3}}{\partial y} = \frac{\partial F _{2}}{\partial z}, \frac{\partial F _{1}}{\partial z} = \frac{\partial F _{3}}{\partial x}, \frac{\partial F _{2}}{\partial x} = \frac{\partial F _{1}}{\partial y}$ These are exactly the three equations in option (D). True.
(C) The left side is $\nabla \cdot F = \nabla^{2} f$ , and a conservative field does not have to make this zero. Counterexample: $f = x^{2}$ gives $F = 2 x \hat{i}$ . Every line integral of this field equals $f (B) - f (A)$ , so it is path independent, but $\nabla \cdot F = 2 \neq = 0$ . False.
The correct options are (A), (B) and (D).

KEY IDEA

For a smooth field, these three statements say the same thing: the line integral is path independent, the integral around every closed path is zero, and $F = \nabla f$ for some scalar potential $f$ . Each of them gives $\nabla \times F = 0$ , whose components are the cross-derivative equalities in (D). Path independence says nothing about the divergence, $\nabla \cdot F = \nabla^{2} f$ .

WATCH OUT FOR

Ticking (C) by mixing up "conservative" (curl-free) with "solenoidal" (divergence-free). A conservative field has zero divergence only if its potential also satisfies Laplace's equation, which is an extra condition. For example, $F = 2 x \hat{i}$ is conservative but has divergence 2. A second slip is to pair the wrong derivatives in (D). Check each equation against a component of $\nabla \times F$ , e.g. the $\hat{i}$ -component gives $\partial F_{3} / \partial y = \partial F_{2} / \partial z$ .

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GATE 2024 EC Q.54: solution and answer

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