View Full Version : convert space curve to cartesian
if R = sinti+sqrt(2)costj+sintk, 0<=t<=Pi/2
please eliminate t to determine the cartesian equation of R(t). Put limits on the variables and verbally describe the curve
himanshu121
Feb15-04, 02:48 AM
x= sint, y=sqrt(2)cost, z=sint
u can clearly see that
x2 + y2 +z2=2{sin2t +cos2t}
=2
x2 + y2 +z2=2
matt grime
Feb15-04, 05:06 AM
And eqaully clearly, surely you can see there is more to it than that? You've just replaced a locally 1-d structure (a curve) with a locally 2-d structure, a sphere.
yes, the x, y, and z coordinates necessarily satisfy that, but that isn't sufficient. You need to intersect with the plane x=z (or similar) at the very least.
generally the equation is x=z=(1-y^2)^{1/2}/\sqrt 2
Originally posted by matt grime
And eqaully clearly, surely you can see there is more to it than that? You've just replaced a locally 1-d structure (a curve) with a locally 2-d structure, a sphere.
yes, the x, y, and z coordinates necessarily satisfy that, but that isn't sufficient. You need to intersect with the plane x=z (or similar) at the very least.
generally the equation is x=z=(1-y^2)^{1/2}/\sqrt 2
Thanks matt grime, i've checked yours is correct.
but can you show me how the above equation is reached.
HallsofIvy
Feb15-04, 10:08 PM
If matt grime will forgive me for sticking in my oar:
x= sint, y=sqrt(2)cost, z=sint so obviously x= z.
x2= sin2t= (1-cos2t). But
y2= 2 cos2t so cos2t= y2/2. That is x2= 1- y2/2 and
x= z= √(1- y2/2).
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.