Theory of functions of a complex variable, 1st edition by Andrew Russell Forsyth

By Andrew Russell Forsyth

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Ka = |k| a . , a vector of length 1) in the direction of a nonzero vector a is given by a/ a . PROOF Part 1 is left as an exercise. ) For part 2, we must check that the length of a/ a is 1: a a 1 1 a = a a = a = 1, ■ by part 1 (since 1/ a is a positive scalar). ” That is, proja b = ± |a · b| a length of proja b × a a =± a b |cos θ| a . a a unit vector in direction of a Note that the angle θ keeps track of the appropriate sign of proja b; that is, when 0 ≤ θ < π/2, cos θ is positive and proja b points in the direction of a, and when π/2 < θ ≤ π , cos θ is negative and proja b points in the direction opposite to that of a.

From our experience with circular geometry and, perhaps, polar coordinates, we see that −→ B P is described by π π −→ i + aθ sin θ − j = aθ sin θ i − aθ cos θ j. B P = aθ cos θ − 2 2 Hence, −→ −→ −→ O P = O B + B P = a(cos θ + θ sin θ) i + a(sin θ − θ cos θ ) j. 31 The involute. 31. 2 Exercises In Exercises 1–5, write the given vector by using the standard basis vectors for R2 and R3 . 1. (2, 4) 2. (9, −6) 4. (−1, 2, 5) 5. (2, 4, 0) 3. (3, π, −7) In Exercises 6–10, write the given vector without using the standard basis notation.

45. (a) Describe the curve given parametrically by x = 2 cos 3t y = 2 sin 3t 33. Find where the line x = 3t − 5, y = 2 − t, z = 6t in- 0≤t < 2π . 3 tersects the plane x + 3y − z = 19. 34. Where does the line x = 1 − 4t, y = t − 3/2, z = 2t + 1 intersect the plane 5x − 2y + z = 1? 35. Find the points of intersection of the line x = 2t − 3, What happens if we allow t to vary between 0 and 2π ? (b) Describe the curve given parametrically by y = 3t + 2, z = 5 − t with each of the coordinate planes x = 0, y = 0, and z = 0.

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