B) If the
lim_{(x,y)->(0,0)} x y^3/(x^2+y^4)
exists its value is independent of how (x,y) approaches (0,0).
If we approach (0,0) along the x axis (x=0)
lim_{(0,y)->(0,0)} 0*y^3/(y^4) = 0
If we approach (0,0) along x=y
lim_{(x,x)->(0,0)} x^4/(x^2+4^4) = 1.
Therefore the limit does not exist.
RonL
Hello, m777!
Convert .x^4 + 2x²y² + y^4 - 6x²y + 2y³ .= .0 . to polar coordinates.
We have: .(x² + y²)² - 6x²y + 2y³ .= .0
Then: .(r²)² - 6(r²·cos²θ)(r·sinθ) + 2(r³·sin³θ) .= .0
. . . . . r^4 .= .6r³·sinθ·cos²θ - 2r³·sin³θ
Divide by r³: . r .= .6·sinθ·cos²θ - 2·sin³θ
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Basically, we're done . . . it can be simplified a bit more . . .
. . r .= .6·sinθ(1 - sin²θ) - 2·sin³θ
. . r .= .6·sinθ - 6·sin³θ - 2·sin³θ
. . r .= .6·sinθ - 8·sin³θ
. . r .= .2·sinθ·(3 - 4·sin²θ)