Given, f : R 2 → R f:\mathbb{R}^2\rightarrow \mathbb{R} f : R 2 → R is a smooth map such that f f f is defined as below
f ( x , y ) = x 2 + y 2 f(x,y)=x^2+y^2 f ( x , y ) = x 2 + y 2 Now, when c=0 then
f ( x , y ) = 0 ⟹ x 2 + y 2 = 0 ⟹ x = y = 0 f(x,y)=0\implies x^2+y^2=0\implies x=y=0 f ( x , y ) = 0 ⟹ x 2 + y 2 = 0 ⟹ x = y = 0 Thus, f − 1 ( c = 0 ) = { ( 0 , 0 ) } f^{-1}(c=0)=\{(0,0)\} f − 1 ( c = 0 ) = {( 0 , 0 )} .
Now, for positive c c c ,
f ( x , y ) = c = x 2 + y 2 f(x,y)=c=x^2+y^2 f ( x , y ) = c = x 2 + y 2 Represents locus of radius c \sqrt{c} c and center is (0,0), thus
f − 1 ( c ) = { ( x , y ) ∈ R 2 : f ( x , y ) = c } f^{-1}(c)=\{(x,y)\in \mathbb{R}^2:f(x,y)=c\} f − 1 ( c ) = {( x , y ) ∈ R 2 : f ( x , y ) = c } is the set of points on the circle mentioned above.
Thus, level curves for c=0,1,2,3,4 will be
Where, origin is f − 1 ( 0 ) f^{-1}(0) f − 1 ( 0 ) , Blue, Green, Violet and black circles are respectively f − 1 ( 1 ) , f − 1 ( 2 ) , f − 1 ( 3 ) & f − 1 ( 4 ) f^{-1}(1),f^{-1}(2),f^{-1}(3)\&f^{-1}(4) f − 1 ( 1 ) , f − 1 ( 2 ) , f − 1 ( 3 ) & f − 1 ( 4 ) respectively.
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