Question

Consider the paraboloid z=x^2+y^2. The plane 6x−6y+z−8=0 cuts the paraboloid, its intersection being a curve.

Find "the natural" parametrization of this curve.

Answer 1

I have x^2 + y^2 = 6y - 6x + 8

Answer 2

no clue what to do now

Answer 3

Yes parameterize that circle

Answer 4

x^2 -6x + y^2 -6y = 8 is a circle?

Answer 5

i thought a circle was x^2 + y^2 = R^2

Answer 6

\[(x-h)^2 + (y-k)^2 = r^2\]

is a circle with radius \(r\) and center \((h, k)\)

Answer 7

hmm, ok so i have to complete the square

Answer 8

yes

Answer 9

AH

Answer 10

see if this helps
https://answers.yahoo.com/question/index?qid=20080221162628AAGcxo6

i don't remember "natiral parameterization", il need to review i guess

Answer 11

ah thank you :)

Answer 12

I have (x-3)^2 + (y-3)^2 = 26

Answer 13

that looks good!

Answer 14

do we have

\(x-3 = \sqrt{26}\cos t\)
\(y-3 = \sqrt{26}\sin t\)

?

Answer 15

hmm

Answer 16

@ganeshie8 not working :(

Answer 17

what are you entering?

Answer 18

x(t) = 3 + sqrt(26) cos(t)

Answer 19

what about y and z

Answer 20

y(t) = 3+sqrt(26)sin(t) worked though

Answer 21

nvm got it, sign error, ugh !

Answer 22

@ganeshie8 , ty :)

Answer 23

nice nice