← 2013 Paper 1
UPSC Maths 2013 Paper 1 Q3b — Solution 15 marks · Section A
Question
Compute f x y ( 0 , 0 ) f_{xy}(0,0) f x y ( 0 , 0 ) f y x ( 0 , 0 ) f_{yx}(0,0) f y x ( 0 , 0 )
f ( x , y ) = { x y 3 x + y 2 , ( x , y ) ≠ ( 0 , 0 ) 0 , ( x , y ) = ( 0 , 0 ) . f(x,y)=\begin{cases}\dfrac{xy^{3}}{x+y^{2}},&(x,y)\ne(0,0)\\ 0,&(x,y)=(0,0).\end{cases} f ( x , y ) = ⎩ ⎨ ⎧ x + y 2 x y 3 , 0 , ( x , y ) = ( 0 , 0 ) ( x , y ) = ( 0 , 0 ) . Also, discuss the continuity of f x y f_{xy} f x y f y x f_{yx} f y x ( 0 , 0 ) (0,0) ( 0 , 0 )
Technique
Compute partials away from singular point by quotient rule; use limit definition at the singular point; assess continuity by approach along different paths.
Solution
Strategy. Compute first partials away from the origin; use limit definition at the origin; iterate to get the mixed second partials at ( 0 , 0 ) (0,0) ( 0 , 0 )
Step 1 — First partials away from origin
For ( x , y ) ≠ ( 0 , 0 ) (x,y)\ne(0,0) ( x , y ) = ( 0 , 0 ) x + y 2 ≠ 0 x+y^{2}\ne 0 x + y 2 = 0
f x = ∂ ∂ x [ x y 3 x + y 2 ] = y 3 ( x + y 2 ) − x y 3 ⋅ 1 ( x + y 2 ) 2 = y 5 ( x + y 2 ) 2 . f_x=\frac{\partial}{\partial x}\!\left[\frac{xy^{3}}{x+y^{2}}\right]=\frac{y^{3}(x+y^{2})-xy^{3}\cdot 1}{(x+y^{2})^{2}}=\frac{y^{5}}{(x+y^{2})^{2}}. f x = ∂ x ∂ [ x + y 2 x y 3 ] = ( x + y 2 ) 2 y 3 ( x + y 2 ) − x y 3 ⋅ 1 = ( x + y 2 ) 2 y 5 . f y = ∂ ∂ y [ x y 3 x + y 2 ] = 3 x y 2 ( x + y 2 ) − x y 3 ( 2 y ) ( x + y 2 ) 2 = x y 2 ( 3 x + y 2 ) ( x + y 2 ) 2 . f_y=\frac{\partial}{\partial y}\!\left[\frac{xy^{3}}{x+y^{2}}\right]=\frac{3xy^{2}(x+y^{2})-xy^{3}(2y)}{(x+y^{2})^{2}}=\frac{xy^{2}(3x+y^{2})}{(x+y^{2})^{2}}. f y = ∂ y ∂ [ x + y 2 x y 3 ] = ( x + y 2 ) 2 3 x y 2 ( x + y 2 ) − x y 3 ( 2 y ) = ( x + y 2 ) 2 x y 2 ( 3 x + y 2 ) . Step 2 — First partials at ( 0 , 0 ) (0,0) ( 0 , 0 )
By definition,
f x ( 0 , 0 ) = lim h → 0 f ( h , 0 ) − f ( 0 , 0 ) h = lim h → 0 h ⋅ 0 / ( h + 0 ) − 0 h = 0. f_x(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}\frac{h\cdot 0/(h+0)-0}{h}=0. f x ( 0 , 0 ) = h → 0 lim h f ( h , 0 ) − f ( 0 , 0 ) = h → 0 lim h h ⋅ 0/ ( h + 0 ) − 0 = 0. f y ( 0 , 0 ) = lim k → 0 f ( 0 , k ) − f ( 0 , 0 ) k = lim k → 0 0 ⋅ k 3 / ( 0 + k 2 ) − 0 k = 0. f_y(0,0)=\lim_{k\to 0}\frac{f(0,k)-f(0,0)}{k}=\lim_{k\to 0}\frac{0\cdot k^{3}/(0+k^{2})-0}{k}=0. f y ( 0 , 0 ) = k → 0 lim k f ( 0 , k ) − f ( 0 , 0 ) = k → 0 lim k 0 ⋅ k 3 / ( 0 + k 2 ) − 0 = 0. Step 3 — Mixed partial f x y ( 0 , 0 ) f_{xy}(0,0) f x y ( 0 , 0 )
By definition f x y ( 0 , 0 ) = ∂ f x ∂ y ∣ ( 0 , 0 ) = lim k → 0 f x ( 0 , k ) − f x ( 0 , 0 ) k f_{xy}(0,0)=\frac{\partial f_x}{\partial y}\Big|_{(0,0)}=\lim_{k\to 0}\dfrac{f_x(0,k)-f_x(0,0)}{k} f x y ( 0 , 0 ) = ∂ y ∂ f x ( 0 , 0 ) = lim k → 0 k f x ( 0 , k ) − f x ( 0 , 0 )
From Step 1, for x = 0 , y = k ≠ 0 x=0,y=k\ne 0 x = 0 , y = k = 0 f x ( 0 , k ) = k 5 ( 0 + k 2 ) 2 = k 5 k 4 = k f_x(0,k)=\dfrac{k^{5}}{(0+k^{2})^{2}}=\dfrac{k^{5}}{k^{4}}=k f x ( 0 , k ) = ( 0 + k 2 ) 2 k 5 = k 4 k 5 = k
So
f x y ( 0 , 0 ) = lim k → 0 k − 0 k = 1. f_{xy}(0,0)=\lim_{k\to 0}\frac{k-0}{k}=1. f x y ( 0 , 0 ) = k → 0 lim k k − 0 = 1. Step 4 — Mixed partial f y x ( 0 , 0 ) f_{yx}(0,0) f y x ( 0 , 0 )
f y x ( 0 , 0 ) = ∂ f y ∂ x ∣ ( 0 , 0 ) = lim h → 0 f y ( h , 0 ) − f y ( 0 , 0 ) h f_{yx}(0,0)=\frac{\partial f_y}{\partial x}\Big|_{(0,0)}=\lim_{h\to 0}\dfrac{f_y(h,0)-f_y(0,0)}{h} f y x ( 0 , 0 ) = ∂ x ∂ f y ( 0 , 0 ) = lim h → 0 h f y ( h , 0 ) − f y ( 0 , 0 )
From Step 1, for x = h , y = 0 x=h,y=0 x = h , y = 0 f y ( h , 0 ) = h ⋅ 0 ⋅ ( 3 h + 0 ) ( h + 0 ) 2 = 0 f_y(h,0)=\dfrac{h\cdot 0\cdot(3h+0)}{(h+0)^{2}}=0 f y ( h , 0 ) = ( h + 0 ) 2 h ⋅ 0 ⋅ ( 3 h + 0 ) = 0
So
f y x ( 0 , 0 ) = lim h → 0 0 − 0 h = 0. f_{yx}(0,0)=\lim_{h\to 0}\frac{0-0}{h}=0. f y x ( 0 , 0 ) = h → 0 lim h 0 − 0 = 0. f x y ( 0 , 0 ) = 1 , f y x ( 0 , 0 ) = 0. \boxed{\;f_{xy}(0,0)=1,\qquad f_{yx}(0,0)=0.\;} f x y ( 0 , 0 ) = 1 , f y x ( 0 , 0 ) = 0. The two mixed partials differ at the origin, so Clairaut’s theorem (which would force them equal under continuity of the second partials) cannot apply — meaning at least one of f x y , f y x f_{xy},f_{yx} f x y , f y x ( 0 , 0 ) (0,0) ( 0 , 0 )
Step 5 — Continuity of f x y f_{xy} f x y ( 0 , 0 ) (0,0) ( 0 , 0 )
Compute f x y f_{xy} f x y ( x , y ) ≠ ( 0 , 0 ) (x,y)\ne(0,0) ( x , y ) = ( 0 , 0 )
f x y = ∂ ∂ y [ y 5 ( x + y 2 ) 2 ] = 5 y 4 ( x + y 2 ) 2 − y 5 ⋅ 2 ( x + y 2 ) ( 2 y ) ( x + y 2 ) 4 = y 4 ( 5 x + y 2 ) ( x + y 2 ) 3 . f_{xy}=\frac{\partial}{\partial y}\!\left[\frac{y^{5}}{(x+y^{2})^{2}}\right]=\frac{5y^{4}(x+y^{2})^{2}-y^{5}\cdot 2(x+y^{2})(2y)}{(x+y^{2})^{4}}=\frac{y^{4}(5x+y^{2})}{(x+y^{2})^{3}}. f x y = ∂ y ∂ [ ( x + y 2 ) 2 y 5 ] = ( x + y 2 ) 4 5 y 4 ( x + y 2 ) 2 − y 5 ⋅ 2 ( x + y 2 ) ( 2 y ) = ( x + y 2 ) 3 y 4 ( 5 x + y 2 ) . Approach along the x x x (y = 0 y=0 y = 0 f x y = 0 / x 3 = 0 f_{xy}=0/x^{3}=0 f x y = 0/ x 3 = 0 x ≠ 0 x\ne 0 x = 0 lim x → 0 + f x y ( x , 0 ) = 0 \lim_{x\to 0^{+}}f_{xy}(x,0)=0 lim x → 0 + f x y ( x , 0 ) = 0
But f x y ( 0 , 0 ) = 1 f_{xy}(0,0)=1 f x y ( 0 , 0 ) = 1 f x y f_{xy} f x y discontinuous at ( 0 , 0 ) (0,0) ( 0 , 0 )
Step 6 — Continuity of f y x f_{yx} f y x ( 0 , 0 ) (0,0) ( 0 , 0 )
Compute f y x f_{yx} f y x ( x , y ) ≠ ( 0 , 0 ) (x,y)\ne(0,0) ( x , y ) = ( 0 , 0 )
f y x = ∂ ∂ x [ x y 2 ( 3 x + y 2 ) ( x + y 2 ) 2 ] . f_{yx}=\frac{\partial}{\partial x}\!\left[\frac{xy^{2}(3x+y^{2})}{(x+y^{2})^{2}}\right]. f y x = ∂ x ∂ [ ( x + y 2 ) 2 x y 2 ( 3 x + y 2 ) ] . The algebra (quotient rule, then simplify) yields
f y x = y 4 ( 5 x + y 2 ) ( x + y 2 ) 3 . f_{yx}=\frac{y^{4}(5x+y^{2})}{(x+y^{2})^{3}}. f y x = ( x + y 2 ) 3 y 4 ( 5 x + y 2 ) . (Same formula as f x y f_{xy} f x y
Approach along the y y y (x = 0 x=0 x = 0 f y x ( 0 , y ) = y 4 ⋅ y 2 ( y 2 ) 3 = y 6 y 6 = 1 f_{yx}(0,y)=\dfrac{y^{4}\cdot y^{2}}{(y^{2})^{3}}=\dfrac{y^{6}}{y^{6}}=1 f y x ( 0 , y ) = ( y 2 ) 3 y 4 ⋅ y 2 = y 6 y 6 = 1 y ≠ 0 y\ne 0 y = 0
But f y x ( 0 , 0 ) = 0 f_{yx}(0,0)=0 f y x ( 0 , 0 ) = 0 f y x f_{yx} f y x also discontinuous at ( 0 , 0 ) (0,0) ( 0 , 0 )
Conclusion
Answer
Both f x y and f y x are discontinuous at ( 0 , 0 ) . \boxed{\;\text{Both }f_{xy}\text{ and }f_{yx}\text{ are discontinuous at }(0,0).\;} Both f x y and f y x are discontinuous at ( 0 , 0 ) .