Becky T.

asked • 02/07/18

curve in a plane

Consider the curve given parametrically by x = acost,y = β sint,z = csint, where a,b,c are positive real numbers. Show that it lies in a plane. Find the equation of the plane.
 
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Bobosharif S. answered • 02/07/18

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x = a cos(t),y = b sin(t), z = c sin(t)
x/a = cos(t), y/b = sin(t), z/c=sin(t)
(x/a)2+(y/b)2+(z/c)2=1+sin2(t)
(x/a)2+(y/b)2=1.
 
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Becky T.

Hi,
 
Thank you for answering. I am still a little confused however. What you have written is the equation of the plane, correct? Is the fact that such an equation can be found proof of the curve being planar? Any problems I've seen have had truly complex ways of finding it (using binormal vectors and torsion equations that need to equate to 0). please elaborate a little more.
 
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02/08/18

Bobosharif S.

Alright.You have an equation given in a parametric form. I think it is clear how the last equation is derived. 
Now, your question asks to show that it lies on a plane but doesn't specify which plane. Indeed given equation lies on a (x, y) plane. This is (basically) equation of ellipse.  If you imagine it in 3D, its an equation of elliptic cylinder. 
I hope it is clear now.
 
 
 
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02/08/18

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