Question

How do i parametrize this?

x^2 + y^2 + z^2 = 4(x+y)

x+y=4

Answer 1

allow x= t solve your bottom and sub i'm assuming

Answer 2

i did horrible on this test so ... lol

Answer 3

\[x ^{2}+ y ^{2} + z ^{2} = 4(x+y) \rightarrow (x-2)^{2} +(y-2)^{2}+z^{2} = 8 \]

Then i do x = t in the second equation and get y = 4-t, replacing in the fisrt... do you think thats correct? z ends a bit ugly

Answer 4

not sure how you got x-2?

Answer 5

parametrize wont alter the equation i believe and you just made your center offset

Answer 6

is x+y=4 you parameter?

Answer 7

i was saying
\[x^2+y^2+z^2=4(4)=16\]

let t=x
y=4-x
y=4-t

Answer 8

\[z^2=16-t^2-(4-t)^2\]

Answer 9

\[z^2=16-16-t^2-t^2+8t=-2t^2+8t\]

Answer 10

\[z=\sqrt{-2t^2+8t}\]

Answer 11

so parameter would be

\[t^2,(t-4)^2,\sqrt{-2t^2+8t}\]

Answer 12

now i'm like around 50 on this one cause i failed this test lol

Answer 13

Exactly that is what i ended with, but that square root seemed a little ugly to me, but now that you confirm it, im more confident, however, i think i can get some other things from this... let me check... i think there are 2 curves, one with -square root and the other with +squereroot....

Answer 14

well the parameter they picked was ugly to begin with... if you were to pick it you could find a much more simple one

Answer 15

both positive and negative i believe you have what looks like a plane intersecting a sphere

Answer 16

or line

Answer 17

Well i think i got it, the final answer would be:

\[\Gamma _{1}: x= t // y = 4-t // z = \sqrt{-2t^{2}+8t}\] 0 <= t <= 4

\[\Gamma _{2}: x= t // y = 4-t // z = -\sqrt{-2t^{2}+8t}\] 0 <= t <= 4


The only thing that changes is the minus sign in t.

Answer 18

in z i mean

Answer 19

Thanks for your help.