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Curves in euclidean 3-space

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To find the tangent vector you just differentiate the components of the vector w.r.t. the parameter and then normalize the result. Let's denote the position vector by V. Then if:

V = [lambda, (lambda-1)^2, - lambda]

dV/dlambda = [1, 2(lambda-1), - 1]

at point P the parameter lambda = 1, so the derivative there is [1, 0, - 1]. The tangent vector is obtained by normalizing this to 1: 1/sqrt(2) [1, 0, - 1]

Next case:

V = [cos(mu), sin(mu), mu-1]

dV/d mu = [-sin(mu), cos(mu), 1]

At P the parameter mu = 0, so there the derivative is [0,1,1] and the ...

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