← 2015 Paper 1

UPSC 2015 Maths Optional Paper 1 Q6c — Step-by-Step Solution

12 marks · Section B

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

Question

Find the values of $\lambda$ and $\mu$ so that the surfaces $\lambda x^2-\mu yz=(\lambda+2)x$ and $4x^2 y+z^3=4$ may intersect orthogonally at $(1,-1,2)$.

Technique

Two conditions (point-on-surface for Surface 1; perpendicularity of gradients); two unknowns $\lambda,\mu$.

Solution

Two surfaces intersect orthogonally at a common point iff (i) the point lies on both surfaces and (ii) the gradients (normals) are perpendicular at that point.

Step 1 — Condition (i): $(1,-1,2)$ on both surfaces

Surface 2: $4x^2 y+z^3=4$. At $(1,-1,2)$: $4(1)(-1)+8=-4+8=4$ ✓.

Surface 1: $\lambda x^2-\mu yz-(\lambda+2)x=0$. At $(1,-1,2)$: $\lambda(1)-\mu(-1)(2)-(\lambda+2)(1)=\lambda+2\mu-\lambda-2=2\mu-2=0$.

Step 2 — Condition (ii): gradients perpendicular at $(1,-1,2)$

$F_1=\lambda x^2-\mu yz-(\lambda+2)x$, $\nabla F_1=(2\lambda x-(\lambda+2),\,-\mu z,\,-\mu y)$.

At $(1,-1,2)$ with $\mu=1$: $\nabla F_1=(2\lambda-(\lambda+2),\,-2,\,1)=(\lambda-2,\,-2,\,1)$.

$F_2=4x^2 y+z^3-4$, $\nabla F_2=(8xy,\,4x^2,\,3z^2)$.

At $(1,-1,2)$: $\nabla F_2=(-8,4,12)$.

Orthogonality: $\nabla F_1\cdot\nabla F_2=0$:

Step 3 — Verify

With $\lambda=5/2$, $\mu=1$, recheck Step 1 for Surface 1: at $(1,-1,2)$, $2\mu-2=2(1)-2=0$ ✓.

Recheck Step 2: $\nabla F_1=(5/2-2,-2,1)=(1/2,-2,1)$. $\nabla F_1\cdot\nabla F_2=(1/2)(-8)+(-2)(4)+(1)(12)=-4-8+12=0$ ✓.

Answer