Thursday January 17 Solutions Parametric Curves Defined Using Vector Arithmetic 1 a Plot of f x x 2 x 2 10 8 1 5 6 4 2 4 3 2 1 1 2 3 2 b f 0 x 2x 1 so the equation for the tangent line to f x at x 2 is T x f 2 f 0 2 x 2 4 5 x 2 5x 6 2 c A vector in the direction of the tangent line has a slope of 5 so the vector 1 5 is a good choice It is shown on the graph above based at 2 4 x t a Plot of for 0 t 4 This is different from the graph above because the y t2 t 2 domain is restricted 15 10 5 1 2 3 4 b The vectors based at 0 0 and ending at x t y t for t 0 1 2 3 are shown on the graph above c x 0 2 y 0 2 1 5 This represents velocity this vector is shown on the curve in the graph below 1 a p p d The speed of the particle is the magnitude of the velocity or 12 52 26 3 a d shown below The red arrows from left to right are the vectors 8 3 5 2 2 1 and 1 0 The black arrows show how these are obtained by adding the multiples v 0 v and 2v of the vector v 3 1 to the vector 5 2 3 0 2 5 2 0 1 5 1 0 0 5 8 4 6 2 e If we allow the scalar t to vary in the parametric equation 5 2 t 3 1 we get a line through the point 5 2 in the direction of the vector 3 1 4 a l t 5 2t 2 3t 1 t 5 2 1 t 2 3 1 so p 5 2 1 and v 2 3 1 b Plot of the line from part a 10 20 5 z 0 5 0 y 20 10 20 x 0 10 c v is called the direction vector because it points in the direction of the line p 5 Let a 3 0 1 0 and b 1 1 0 1 be vectors in R4 q p p p p a The distance between 3 0 1 0 and 1 1 0 1 is 1 3 2 12 12 12 7 2 3 b The angle between a and b is found by p a b 3 arccos arccos p arccos 1 2 2 3 a b 2 3
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