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Examples from Section 13.1 #28. At what points does the helix r(t) = ⟨ sint, cost, t ⟩ intersect the sphere x2 + y2 + z2 = 5? Well, since x = sint and y = cost, So x2 + y2 = 1, always. x2 + y2 + z2 = 5 1 + z2 = 5 z2 = 4, but since z = t t2 = 4, so t = ± 2, giving us the points (sin2, cos2, 2) and (-sin2, cos2, -2) #38: Find the vector function that represents the curve of intersection of the paraboloid z = 4x2 + y2 and the parabolic cylinder y = x2 What do we have here? We know z = 4x2 + y2 and y = x2, x = t, y = t 2, z = 4t2 + t4 so if we let x = t, we have r(t) = ⟨ t, t2, 4t2 + t4 ⟩ and we are done. OR, we could let x = T5 , then: x = T5 , y = T10 , z = 4T10 + T20 , which gives r(T) = ⟨ T5, T10, 4T10 + T20 ⟩ There are many (∞) possible (correct) answers. #42 Two particles travel along the space curves: r1(t) = ⟨ t, t2, t3 ⟩ and r2(t) = ⟨ 1 + 2t, 1 + 6t, 1 + 14t ⟩ Do the particles collide? Do their paths intersect? Do they collide? Is there a value of t such that x1 = x2, y1 = y2, and z1 = z2? For x1 = x2, we need t = 1 + 2t which yields t = -1, but that value of t does not satisfy the other 2 equations (y1 = y2, and z1 = z2). So the particles do not collide. Do their paths intersect? Rewrite the curves as: r1(t) = ⟨ t, t2, t3 ⟩ and r2() = ⟨ 1 + 2, 1 + 6, 1 + 14 ⟩ Can we satisfy: t = 1 + 2, t 2 = 1 + 6, and t3 = 1 + 14 simultaneously? Yes, t = 1, = 0 works as does t = 2, = ½. So the paths intersect, but the particles do NOT collide. The points of intersection are: (1,1,1) (t=1, = 0) and (2, 4, 8) (t = 2, = ½)