Locate the stationary points of x4+y4-2x2+4xy-2y2.
Answers
Answer: The stationary points of x⁴+y⁴-2x²+4xy-2y² are (0,0) (√2, -√2) and (-√2, √2)
Step-by-step explanation:
Let f(x,y) = x⁴+y⁴-2x²+4xy-2y²
We now differentiate f(x,y) with respect to x, treating y as a constant.
⇒ fₓ = 4x³-4x+4y
Similarly, differentiating f(x,y) with respect to y, treating x as a constant, we have,
fᵧ = 4y³+4x-4y
We know solve fₓ and fᵧ = 0.
fₓ = 0 ⇒ 4x³-4x+4y = 0
⇒ x³-x+y = 0.......(i)
⇒ x = x³+y
Again, if fᵧ = 0,
⇒ 4y³+4x-4y = 0
⇒ y³+x-y = 0........(ii)
Putting x = x³+y in the above equation, we have,
y³+x³+y-y = 0
⇒ y³+x³ = 0
⇒ y³ = -x³
Thus, we have, y = -x.
Putting y=-x in (i), we have,
x³-x-x = 0
⇒ x³-2x = 0
⇒ x (x²-2) = 0
⇒ x = 0 or x²-2 = 0
⇒ x = 0 or x² = 2
⇒ x = 0 or x = ±√2
Now, if x=0, y=0, if x=√2, then y= -√2 and if x=-√2, y= √2
Thus, the stationary points are (0,0) (√2, -√2) and (-√2, √2).