← 2022 Paper 2

UPSC 2022 Maths Optional Paper 2 Q5a — Step-by-Step Solution

10 marks · Section B

Family of surfaces · PDEs · asked 7× in 13 yrs · Read the full method →

Question

The equation of any cone with vertex at $(a,b,c)$ is $f\!\left(\dfrac{x-a}{z-c},\dfrac{y-b}{z-c}\right)=0$. Find the differential equation of the cone.

Technique

Standard “PDE of family” derivation: differentiate implicit $f=0$ w.r.t. $x$ and $y$; eliminate the unknown derivatives $f_u,f_v$ to get a relation in $x,y,z,p,q$.

Solution

Strategy. Let $u=\dfrac{x-a}{z-c}$, $v=\dfrac{y-b}{z-c}$. Then $f(u,v)=0$ implicitly defines $z$ as a function of $x,y$. Differentiate w.r.t. $x$ and $y$, then eliminate $f_u,f_v$.

Step 1 — Differentiate $f(u,v)=0$ w.r.t. $x$

$\partial_x f=f_u u_x+f_v v_x=0$.

$u_x=\dfrac{\partial}{\partial x}\!\left(\dfrac{x-a}{z-c}\right)=\dfrac{(z-c)-(x-a)z_x}{(z-c)^2}=\dfrac{1-up}{z-c}$ where $p=z_x$.

$v_x=\dfrac{\partial}{\partial x}\!\left(\dfrac{y-b}{z-c}\right)=\dfrac{-(y-b)z_x}{(z-c)^2}=\dfrac{-vp}{z-c}$.

So

Step 2 — Differentiate w.r.t. $y$

$u_y=\dfrac{-(x-a)z_y}{(z-c)^2}=\dfrac{-uq}{z-c}$ where $q=z_y$.

$v_y=\dfrac{(z-c)-(y-b)z_y}{(z-c)^2}=\dfrac{1-vq}{z-c}$.

So

Step 3 — Eliminate $f_u,f_v$

From $(1)$: $f_u/f_v=\dfrac{vp}{1-up}$.

From $(2)$: $f_u/f_v=\dfrac{1-vq}{uq}$.

Equate: $\dfrac{vp}{1-up}=\dfrac{1-vq}{uq}$.

Cross-multiply: $vp\cdot uq=(1-up)(1-vq)$, i.e. $uvpq=1-vq-up+uvpq$, $0=1-vq-up$, $up+vq=1$.

Substitute back $u=(x-a)/(z-c)$, $v=(y-b)/(z-c)$:

$\dfrac{(x-a)p}{z-c}+\dfrac{(y-b)q}{z-c}=1$,

$(x-a)p+(y-b)q=z-c$.

Answer