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Unformatted text preview: r = < 5 4sin , 4cos , 4sin >, 2 . To get the equation for the tangent line to the curve at the point ( 1 , , 4 ) , we first find the parameter for that points. Set up the equation ( 1 , , 4 ) = ( 5 4sin , 4cos , 4sin ) , we get = / 2. Now we have r = < 4cos , 4sin , 4cos >, r ( / 2 ) = < 4cos 2 , 4sin 2 , 4cos 2 > = < , 4 , > . 1 Hence < , 4 , > is a tangent vector to the curve at ( 1 , , 4 ) . So an equation of the tangent line is given by x = 1 , y = 4 t , z = 4 . Extra Credit. The moves to identify the two knots are given as follows 2...
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