Question

how to parameterize x^3 + y^3 = 1?

Answer 1

Can you solve for y in terms of x?

Answer 2

z=0 by the way

Answer 3

x = cos^3(t) y= sin^3(t) z = 0 right?

Answer 4

the answer is given as the same as what you put, but the powers are (2/3), not 3
and i'm unsure how they got that

Answer 5

I was close then :) I dont know how they got ^(2/3) either... id have to go back many years to remember ;)

Answer 6

its a cubic equation that crosses the x and y axis 3 times.... i see that much

Answer 7

i tried to use http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtI.aspx
to help but :( cant seem to figure it out

Answer 8

lets see if I can raeson this..... or muddy the waters :)

x = rcos ; y = rsin ; 1 = sin^2 + cos^2 right?

Answer 9

r^3 (cos^3 + sin^3) = sin^2 + cos^2

Answer 10

yeah

Answer 11

I see where the ^1/3 is coming from.... just how to get it :)

Answer 12

it seems as though thats not too far away

Answer 13

Do you see that if you plug in x = (cos theta)^{2/3) and y=(sin theta)^(2/3) you get the original equation? That's why the answer is correct, but doesn't give much insight into how you were supposed to guess it.

Answer 14

yeah, the solution being correct is the easy part ;)

Answer 15

1 is also the "radius" or vector measure from the origin.... we could try that route.

Answer 16

are there any trig identitied we could utilize for this?

Answer 17

ermm, not that i can think of

Answer 18

r^3 (cos^3 + sin^3) = r^2 ; r^2 = 1 so r^3 = 1 ?

Answer 19

ye, i think r^3 must have to be 1

Answer 20

cos^3 + sin^3 = cos^2 + sin^2

Answer 21

cos^3 - cos^2 = sin^3 - sin^3 ??

Answer 22

typoed it...but you can see that lol

Answer 23

ye, i see lol...what do you have to do to a square power to make it ^2/3?
do you cube root it?

Answer 24

yes; ^2/3 means cbrt(^2)

Answer 25

it's because you want cosx^2+sinx^2=1. So if x=cost^(2/3) when you plug it in, the square will distrbute in to leave cos^2

Answer 26

ok, i think i get it, thanks so much for the help!