convert space curve to cartesian

yanyin

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

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

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

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

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).

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yanyin	convert space curve to cartesian Tweet 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	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 Recognitions: Homework Help Science Advisor	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]

HallsofIvy Recognitions: Gold Member Science Advisor Staff Emeritus	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).