← 2015 Paper 1

UPSC 2015 Maths Optional Paper 1 Q5e — Step-by-Step Solution

10 marks · Section B

Gradient: definition, geometric meaning, computation · Vector Analysis · asked 6× in 13 yrs · Read the full method →

Question

Find the angle between the surfaces $x^2+y^2+z^2-9=0$ and $z=x^2+y^2-3$ at $(2,-1,2)$.

Technique

Angle between surfaces at a common point = angle between gradient vectors (normals); standard $\cos\theta=\vec n_1\cdot\vec n_2/(|\vec n_1||\vec n_2|)$.

Solution

Setup. Verify the point is on both surfaces:

• $S_1:\;x^2+y^2+z^2-9=0$. At $(2,-1,2)$: $4+1+4-9=0$ ✓.
• $S_2:\;z-x^2-y^2+3=0$. At $(2,-1,2)$: $2-4-1+3=0$ ✓.

The angle between two surfaces at a common point is the angle between their normals.

Step 1 — Normal to $S_1$

$\nabla(x^2+y^2+z^2-9)=(2x,2y,2z)$. At $(2,-1,2)$: $(4,-2,4)$, or simplified $\vec n_1=(2,-1,2)$.

$|\vec n_1|=\sqrt{4+1+4}=3$.

Step 2 — Normal to $S_2$

Write $S_2:\;F=z-x^2-y^2+3=0$. $\nabla F=(-2x,-2y,1)$. At $(2,-1,2)$: $(-4,2,1)$.

$|\vec n_2|=\sqrt{16+4+1}=\sqrt{21}$.

Step 3 — Angle from dot product

The angle between surfaces is usually taken as the acute angle, so take absolute value:

Answer