PLEASE EXPLAIN ! (Strange Saddle) Let f (x,y) = (y - x2)(y - x4). Verify that (0

Question

PLEASE EXPLAIN !

(Strange Saddle) Let f (x,y) = (y - x2)(y - x4). Verify that (0,0) is a critical point for f and that the discriminant at (0,0) is zero. Verify that 0 = f(0,0) is the minimum value of f(x, y) along every line through (0,0). (These lines are of the form y = kx and x = 0.) Show f does not have a local minimum value at (0,0) by showing that every circle centered at (0,0) contains points where f is negative. (Hint: do one of the following: (i) describe the solution set of the inequality f(x.y)

Explanation / Answer

f(x,y) = (y - x^2)(y - x^4)

a) critical points are:

(f,y) = 0

--> (y - x^2)(y - x^4) = 0;

--> y - x^2 = 0 or y -x^4 = 0

--> y = x^2 or y = x^4

---> x^2 = x^4

--> (x^2)(x^2 -1) =0

--> x = 0,0,+i,-i

but x,y are real.

hence, x = 0 --> y = 0

So, critical points is (0,0)

f(0,0) = 0

2) lines passing through (0,0) are y = kx or x = 0;

(i) y = kx

f(x,y) = (kx - x^2) (kx - x^4)

--> f(x) = (kx - x^2) (kx - x^4)

which is function in one variable. it's points extremum are at f'(x,y) = 0.

f'(x) = (kx - x^2) (k - 4x^3) + (k - x) (kx - x^4) = (x)(k - x)(2k - 5x^3) ---- (1)

f'(x) = 0

--->  (x)(k - x)(2k - 5x^3) = 0

--> x = 0, k, (2k/5)^(1/3)

f(x) is minimum at x = 0 or x = k.

since it has to satisfied for all k.

x = 0

--> y = kx = 0, and

min value is f(0,0) = 0.

3) we can represent all circles, centered at (0,0) having radius r, in polar form as:

x = rcos(theta), y = r sin(theta)

f(r,theta) = {rcos(theta) - [rsin(theta)]^2}{rcos(theta) - [rsin(theta)]^4}

-->f(r,theta) = (r^2){cos(theta) - rsin^2(theta)}{cos(theta) - (r^3)sin^4(theta)}

Now,since r^2 > = 0, so,

f(r,theta) can be negative for:

{cos(theta) - rsin^2(theta)} < 0 & {cos(theta) - (r^3)sin^4(theta} > 0

or, {cos(theta) - rsin^2(theta)} > 0 & {cos(theta) - (r^3)sin^4(theta} < 0

--> sin(theta)tan(theta) < r & sin^3(theta)tan(theta) > 1/r^3

or, sin(theta)tan(theta) > r & sin^3(theta)tan(theta) < 1/r^3

Hence f(r,theta) can have negative values at some points according to r.

So, f(0,0) is not the local minimum.

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