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Convert space curve to cartesian

by yanyin
Tags: cartesian, convert, curve, space
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yanyin
#1
Feb14-04, 08:58 PM
P: 21
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
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himanshu121
#2
Feb15-04, 02:48 AM
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P: 658
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
#3
Feb15-04, 05:06 AM
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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 [tex]x=z=(1-y^2)^{1/2}/\sqrt 2[/tex]

yanyin
#4
Feb15-04, 06:05 PM
P: 21
Convert space curve to cartesian

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 [tex]x=z=(1-y^2)^{1/2}/\sqrt 2[/tex]
Thanks matt grime, i've checked yours is correct.
but can you show me how the above equation is reached.
HallsofIvy
#5
Feb15-04, 10:08 PM
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Thanks
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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= &radic;(1- y2/2).


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