UPSC 2015 Maths Optional Paper 1 Q7c — Step-by-Step Solution

12 marks · Section B

Curl: definition, physical meaning, computation · Vector Analysis · asked 4× in 13 yrs · Read the full method →

Question

A vector field is given by $\vec F=(x^2+xy^2)\hat\imath+(y^2+x^2 y)\hat\jmath$ . Verify that the field $\vec F$ is irrotational or not. Find the scalar potential.

Technique

Irrotational test via $\partial Q/\partial x=\partial P/\partial y$ ; potential by partial integration $\phi_x=P$ , then differentiate w.r.t. $y$ and match to $Q$ .

Solution

Let $P=x^2+xy^2$ , $Q=y^2+x^2 y$ . (No $z$ -component, so treat as 2D field in the $xy$ -plane.)

Step 1 — Test for irrotational

\nabla\times\vec F=\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)\hat k.

$\dfrac{\partial Q}{\partial x}=2xy$ . $\dfrac{\partial P}{\partial y}=2xy$ .

$\nabla\times\vec F=0$ ✓. The field is irrotational.

Step 2 — Find scalar potential $\phi$ with $\nabla\phi=\vec F$

$\phi_x=P=x^2+xy^2$ . Integrate w.r.t. $x$ :

\phi=\dfrac{x^3}{3}+\dfrac{x^2 y^2}{2}+g(y).

Differentiate w.r.t. $y$ and equate to $Q$ :

\phi_y=x^2 y+g'(y)=y^2+x^2 y\;\;\Longrightarrow\;\;g'(y)=y^2\;\;\Longrightarrow\;\;g(y)=\dfrac{y^3}{3}+\text{const}.

Answer

\boxed{\;\phi(x,y)=\dfrac{x^3}{3}+\dfrac{x^2 y^2}{2}+\dfrac{y^3}{3}+C.\;}

UPSC 2015 Maths Optional Paper 1 Q7c — Step-by-Step Solution

Question

Technique

Solution

Step 1 — Test for irrotational

Step 2 — Find scalar potential ϕ\phiϕ with ∇ϕ=F⃗\nabla\phi=\vec F∇ϕ=F

Answer

This solution is part of the Maths Coverage Map — 13 years, mapped. Get the take-away PDF free.

Step 2 — Find scalar potential $\phi$ with $\nabla\phi=\vec F$