Differential Geometry: Showing a curve is a sphere curve

[^_^physicist]

Homework Statement

Show that a(x) =( -cos(2x)), -2*cos(x), sin (2x)) is a sphere curve by showing that (-1,0,0) belongs to each normal plane.

Homework Equations

Not quite sure (part of my question)
T= a'(x)
N= T'/norm(T')
B= T x N (T cross N)

The Attempt at a Solution

Ok I found the values of the Tangent field, the Binormal Field, and the Normal field vectors to be certain values and have attempted to demonstrate, by finding a value of x for each of the fields, that the point exists for all three of the planes. However, I don't think this is what should have been done. In fact I am frankly confused as to how exactly to go about showing the asked statement is true. So I guess what I am asking for is did I do this right (see work below), or am I in the wrong frame of mind?

T= (2 sin(2x), 2 sin(x), 2 cos(2x))
N=( (4 cos(2x), 2 cos(x), -4sin(2x))/(sqrt(cos^2(x)+4)

(B is kind of nasty to write up without latex...so I will work on that one, but for the time being if you need to figure it out just go with B = T cross N).

Thanks for any help.

 

[lippyka]

So if I understand the question right, I have to find a value of x for which all the fields equal (-1,0,0). So for T: 2 sin(2x) = -1 2 sin(x) = 0 2 cos (2x) = 0 Solving for x I get x = -pi/4 and pi/4. For N: 4cos(2x) = -1 2 cos (x) = 0 -4sin(2x) = 0 Solving for x I get x = -3pi/4 and 3pi/4. For B: 2cos(2x) = -1 2sin(x) = 0 -4cos(2x) = 0 Solving for x I get x = -pi/2 and pi/2. So since each of the points x = -pi/4, pi/4, -3pi/4, 3pi/4, -pi/2, pi/2 have all the fields equal (-1,0,0), then we can conclude that (-1,0,0) belongs to each normal plane.