UPSC 2016 Maths Optional Paper 2 Q5a — Step-by-Step Solution

10 marks · Section B

Quasilinear first-order PDEs (Lagrange's method) · PDEs · asked 9× in 13 yrs · Read the full method →

Question

Find the general equation of surfaces orthogonal to the family of spheres given by $x^2+y^2+z^2=cz$ .

Technique

Orthogonal-trajectory PDE in 3D — the orthogonal surfaces are integral surfaces of $Pp+Qq=R$ where $(P,Q,R)$ is the sphere normal $\nabla F$ after eliminating the parameter $c$ ; solve by Lagrange’s auxiliary equations, extracting two independent first integrals.

Solution

Setup. The given one-parameter family is $x^2+y^2+z^2=cz$ ( $c$ the parameter). At each point the family has a definite normal direction; the orthogonal surfaces are the integral surfaces of the linear first-order PDE $Pp+Qq=R$ whose direction ratios $(P,Q,R)$ are those of the normal to the spheres.

Step 1 — Normal direction to the sphere family

Eliminate the parameter $c$ . From $x^2+y^2+z^2=cz$ we get $c=\dfrac{x^2+y^2+z^2}{z}$ . Write $F=x^2+y^2+z^2-cz$ . Treating $c$ as fixed on each member, the normal direction is

\nabla F=(2x,\ 2y,\ 2z-c).

Substituting $c=\dfrac{x^2+y^2+z^2}{z}$ :

2z-c=2z-\frac{x^2+y^2+z^2}{z}=\frac{2z^2-x^2-y^2-z^2}{z}=\frac{z^2-x^2-y^2}{z}.

So the normal direction ratios (clearing the common factor $2$ and multiplying through by $z$ ) are

\big(P,Q,R\big)=\Big(2xz,\ 2yz,\ z^2-x^2-y^2\Big).

Step 2 — PDE of the orthogonal surfaces

A surface $z=z(x,y)$ with normal $(p,q,-1)$ is orthogonal to the spheres when its normal is perpendicular to the sphere normal — equivalently the orthogonal surface contains the sphere-normal direction, so it satisfies

Pp+Qq=R,\qquad 2xz\,p+2yz\,q=z^2-x^2-y^2.

Step 3 — Lagrange auxiliary equations

\frac{dx}{2xz}=\frac{dy}{2yz}=\frac{dz}{z^2-x^2-y^2}.

First integral. From the $x,y$ pair, $\dfrac{dx}{x}=\dfrac{dy}{y}\Rightarrow \dfrac{y}{x}=c_1$ (planes through the $z$ -axis).

Second integral. Use the multiplier set $(x,y,z)$ :

\frac{x\,dx+y\,dy+z\,dz}{2x^2z+2y^2z+z(z^2-x^2-y^2)}=\frac{x\,dx+y\,dy+z\,dz}{z(x^2+y^2+z^2)}.

Also $\dfrac{dz}{z^2-x^2-y^2}$ — but cleaner is to pair $\dfrac{x\,dx+y\,dy+z\,dz}{z(x^2+y^2+z^2)}$ with $\dfrac{dy}{2yz}$ :

\frac{x\,dx+y\,dy+z\,dz}{x^2+y^2+z^2}=\frac{dy}{2y}\quad\Big(\text{both denominators carried a factor }z\Big).

The left side is $\tfrac12\,d\!\left[\ln(x^2+y^2+z^2)\right]$ and the right is $\tfrac12\,d(\ln y)$ . Integrating,

\ln(x^2+y^2+z^2)=\ln y+\text{const}\ \Rightarrow\ \frac{x^2+y^2+z^2}{y}=c_2.

Step 4 — General orthogonal surface

The general orthogonal surface is an arbitrary relation between the two independent integrals:

Answer

\boxed{\;\Phi\!\left(\frac{y}{x},\ \frac{x^2+y^2+z^2}{y}\right)=0,\qquad\text{i.e. }x^2+y^2+z^2=y\,\psi\!\left(\frac{y}{x}\right),\;}