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# Solution of problem 1.10(a the unit normal vector is

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Unformatted text preview: Solution of problem 1.10: (a) The unit normal vector is given by −→ N = d −→ T dt vextendsingle vextendsingle vextendsingle d −→ T dt vextendsingle vextendsingle vextendsingle where −→ T is the unit tangent vector. To compute −→ T we compute the velocity vector d −→ r /dt = ( 2 cos(2 t ) , − 2 sin(2 t ) , 1 ) and normalize −→ T = 1 | d −→ r /dt | d −→ r /dt = 1 radicalbig 4 cos 2 (2 t ) + 4 sin 2 (2 t ) + 1 ( 2 cos(2 t ) , − 2 sin(2 t ) , 1 ) = (bigg 2 3 cos(2 t ) , − 2 3 sin(2 t ) , 1 3 )bigg . Hence d −→ T dt = (bigg − 4 3 sin(2 t ) , − 4 3 cos(2 t ) , )bigg , 16 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 ) , )bigg = (− sin(2 t ) , − cos(2 t ) , ) . 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 ) = (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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Solution of problem 1.10(a The unit normal vector is given...

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