← 2013 Paper 2

UPSC 2013 Maths Optional Paper 2 Q6b — Step-by-Step Solution

15 marks · Section B

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

Question

Find the surface which intersects the surfaces of the system $z(x+y)=C(3z+1)$, $(C$ being a constant$)$ orthogonally and which passes through the circle $x^{2}+y^{2}=1,\;z=1$.

Technique

Orthogonal trajectories of a surface family = integral curves of $\nabla\phi$; two first integrals; specific surface from initial-curve condition.

Solution

Strategy. For a one-parameter family of surfaces $\phi=C$, orthogonal trajectories are integral curves of $\nabla\phi$, satisfying $\dfrac{dx}{\phi_x}=\dfrac{dy}{\phi_y}=\dfrac{dz}{\phi_z}$. Two functionally independent first integrals $u(x,y,z)=$ const and $v(x,y,z)=$ const describe the curves; a surface in the orthogonal family is then any relation $F(u,v)=0$.

Step 1 — Compute $\nabla\phi$

The family: $\phi(x,y,z)=\dfrac{z(x+y)}{3z+1}=C$.

$\phi_x=\phi_y=\dfrac{z}{3z+1}$.

$\phi_z=\dfrac{(x+y)(3z+1)-z(x+y)(3)}{(3z+1)^{2}}=\dfrac{(x+y)}{(3z+1)^{2}}$.

Step 2 — Orthogonal-trajectory ODEs

Pair 1 ($dx,dy$): Equal denominators give $dx=dy$, so $x-y=c_1$ is one integral.

Pair 2 ($dx,dz$): Cross-multiply:

Using $x-y=c_1$, $y=x-c_1$, so $x+y=2x-c_1$. The ODE becomes

Integrate: $x^{2}-c_1 x=z^{3}+\dfrac{z^{2}}{2}+c_2$.

Substitute $c_1=x-y$: $x^{2}-x(x-y)=z^{3}+z^{2}/2+c_2$, i.e., $xy=z^{3}+z^{2}/2+c_2$.

Second integral: $v=xy-z^{3}-\dfrac{z^{2}}{2}=c_2$.

Step 3 — Surface through the given circle

On the circle $x^{2}+y^{2}=1,\;z=1$: compute $u$ and $v$.

$u=x-y$.

$v=xy-1-\tfrac{1}{2}=xy-\tfrac{3}{2}$.

From $x^{2}+y^{2}=1$: $(x-y)^{2}=x^{2}-2xy+y^{2}=1-2xy$, so $xy=\dfrac{1-u^{2}}{2}$.

Substituting: $v=\dfrac{1-u^{2}}{2}-\dfrac{3}{2}=-1-\dfrac{u^{2}}{2}$.

Rearranging: $u^{2}+2v+2=0$.

Step 4 — Surface equation

$u^{2}+2v+2=0$ becomes

Expand $(x-y)^{2}+2xy=x^{2}-2xy+y^{2}+2xy=x^{2}+y^{2}$. So

i.e.

Answer