Hello,

please try to do these questions.Thanks

Printable View

- March 18th 2007, 11:47 PMm777convert it into polar coordinates,check limits
Hello,

please try to do these questions.Thanks - March 19th 2007, 12:13 AMtopsquark
- March 19th 2007, 04:00 AMm777
Now i think so there is no adware in this file.i will attach it again.

- March 19th 2007, 06:26 AMCaptainBlack
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 - March 19th 2007, 09:32 AMSoroban
Hello, m777!

Quote:

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)