← 2021 Paper 2

UPSC 2021 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

Obtain the PDE by eliminating arbitrary function $f$ from $f(x+y+z,x^2+y^2+z^2)=0$.

Technique

Differentiate implicit $f(u,v)=0$ w.r.t. $x$ and $y$; eliminate $f_u,f_v$ ratio; collect.

Solution

Setup. Let $u=x+y+z$, $v=x^2+y^2+z^2$. The relation $f(u,v)=0$ defines $z$ implicitly as a function of $x,y$.

Step 1 — Differentiate w.r.t. $x$

$f_u u_x+f_v v_x=0$.

$u_x=1+z_x=1+p$ where $p=z_x$.

$v_x=2x+2z z_x=2x+2zp$.

So $f_u(1+p)+f_v\cdot 2(x+zp)=0$.

Step 2 — Differentiate w.r.t. $y$

$u_y=1+q$ where $q=z_y$.

$v_y=2y+2zq$.

So $f_u(1+q)+f_v\cdot 2(y+zq)=0$.

Step 3 — Eliminate $f_u,f_v$

From Step 1: $\dfrac{f_u}{f_v}=-\dfrac{2(x+zp)}{1+p}$.

From Step 2: $\dfrac{f_u}{f_v}=-\dfrac{2(y+zq)}{1+q}$.

Equate: $\dfrac{x+zp}{1+p}=\dfrac{y+zq}{1+q}$.

Cross-multiply: $(x+zp)(1+q)=(y+zq)(1+p)$.

Expand: $x+xq+zp+zpq=y+yp+zq+zpq$.

$zpq$ cancels: $x+xq+zp=y+yp+zq$, $x-y+xq-yp+zp-zq=0$, $x-y+(xq-yp)+z(p-q)=0$.

Rearrange: $(x-y)+xq-yp+z(p-q)=0$.

Group: $(x-y)=-xq+yp+z(q-p)=p(y-z)+q(z-x)+\ldots$

Let me redo: $(x-y)+(xq-yp)+(zp-zq)=0$.

$(x-y)+xq-yp+zp-zq=0$.

$(x-y)+p(z-y)-q(z-x)\cdot$ wait $p(z-y)$? Let me check: $-yp+zp=p(z-y)$. And $xq-zq=q(x-z)=-q(z-x)$.

So $(x-y)+p(z-y)+q(x-z)=0$, i.e.,

Wait, let me re-sign. From $(x-y)+p(z-y)+q(x-z)=0$: $p(z-y)+q(x-z)=y-x$, or equivalently $p(y-z)+q(z-x)=x-y$.

Answer