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# Hence d t dt bigg 4 3 sin2 t 4 3 cos2 t bigg 16 and

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Hence d −→ T dt = (bigg 4 3 sin(2 t ) , 4 3 cos(2 t ) , 0 )bigg , 16

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and so −→ N = d −→ T dt vextendsingle vextendsingle vextendsingle d −→ T dt vextendsingle vextendsingle vextendsingle = 3 4 (bigg 4 3 sin(2 t ) , 4 3 cos(2 t ) , 0 )bigg = (− sin(2 t ) , cos(2 t ) , 0 ) . This is a horizontal vector pointing radially towards the z -axis. So (a) is true . (b) From the formula for −→ T above we see that the z component of −→ T is constant and equal to 1 / 3. So (b) is true . (c) The unit binormal vector is given by −→ B = −→ T × −→ N = det hatwide ı hatwide hatwide k 2 3 cos(2 t ) 2 3 sin(2 t ) 1 3 sin(2 t ) cos(2 t ) 0 = (bigg 1 3 cos(2 t ) , 1 3 sin(2 t ) , 2 3 )bigg . This vector has non-trivial x and y components so (c) is false . 17
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