13 CALCULUSOF
                                                                                       VECTOR-VALUED
                                                                                       FUNCTIONS
                                                                     13.1 Vector-Valued Functions                                        (LT Section 14.1)
                                                                     Preliminary Questions
                                                                      1. Which one of the following does not parametrize a line?
                                                                                                         
                                                                     (a) r (t) = 8−t,2t,3t
                                                                            1
                                                                     (b) r (t) = t3i − 7t3j + t3k
                                                                            2                                        
                                                                     (c) r (t) = 8 −4t3,2+5t2,9t3
                                                                            3
                                                                     solution
                                                                                                                                                                                                                                              
                                                                     (a) Thisisaparametrizationofthelinepassingthroughthepoint(8,0,0)inthedirectionparalleltothevector −1,2,3 ,
                                                                     since:
                                                                                                                                                                                 
                                                                                                                                8−t,2t,3t = 8,0,0 +t −1,2,3
                                                                     (b) Using the parameter s = t3 we get:
                                                                                                                              3        3    3                                      
                                                                                                                             t   , −7t ,t        = s,−7s,s =s 1,−7,1
                                                                                                                                                                                                            
                                                                    This is a parametrization of the line through the origin, with the direction vector v = −1,7,1 .
                                                                                                                  3,2+5t2,9t3 does not parametrize a line. In particular, the points (8,2,0) (at t = 0),
                                                                     (c) The parametrization 8 − 4t
                                                                     (4,7,9) (at t = 1), and (−24,22,72) (at t = 2) are not collinear.
                                                                      2. What is the projection of r(t) = ti + t4j + etk onto the xz-plane?
                                                                                                                                                                                                   t                t 
                                                                     solution The projection of the path onto the xz-plane is the curve traced by ti + e k = t,0,e . This is the curve
                                                                     z = ex in the xz-plane.
                                                                      3. Which projection of cost,cos2t,sint is a circle?
                                                                     solution Theparametricequations are
                                                                                                                                 x = cost,          y=cos2t, z=sint
                                                                                                                                                        2       2          2            2
                                                                    Theprojection onto the xz-plane is cost,0,sint . Since x + z = cos t + sin t = 1, the projection is a circle in the
                                                                                                                                                                                        
                                                                     xz-plane. The projection onto the xy-plane is traced by the curve cost,cos2t,0 . Therefore, x = cost and y = cos2t.
                                                                    Weexpressy in terms of x:
                                                                                                                                y = cos2t = 2cos2t −1 = 2x2 −1
                                                                    The projection onto the xy-plane is a parabola. The projection onto the yz-plane is the curve 0,cos2t,sint. Hence
                                                                     y = cos2t and z = sint. We find y as a function of z:
                                                                                                                                 y = cos2t = 1−2sin2t = 1−2z2
                                                                    Theprojection onto the yz-plane is again a parabola.
                                                                      4. What is the center of the circle with parametrization
                                                                                                                             r(t) = (−2+cost)i+2j+(3−sint)k?
                                                                     solution Theparametricequations are
                                                                                                                             x =−2+cost, y=2,z=3−sint
                                                                    Therefore, the curve is contained in the plane y = 2, and the following holds:
                                                                                                                                       2                2          2            2
                                                                                                                            (x +2) +(z−3) =cos t +sin t =1
                                                                    Weconcludethat the curve r(t) is the circle of radius 1 in the plane y = 2 centered at the point (−2,2,3).
                              250
                                                                                        May16,2011
                                                                                                        SECTION 13.1           Vector-Valued Functions  (LT SECTION 14.1)      251
                                                                                                                        
                                                      5. Howdothepathsr (t) = cost,sint and r (t) = sint,cost around the unit circle differ?
                                                                              1                         2
                                                     solution Thetwopathsdescribetheunitcircle.However,ast increasesfrom0to2π,thepointonthepathsinti+costj
                                                     movesinaclockwisedirection, whereas the point on the path costi + sintj moves in a counterclockwise direction.
                                                      6. Which three of the following vector-valued functions parametrize the same space curve?
                                                     (a) (−2+cost)i+9j+(3−sint)k                                     (b) (2 +cost)i −9j+(−3−sint)k
                                                     (c) (−2+cos3t)i+9j+(3−sin3t)k                                   (d) (−2−cost)i+9j+(3+sint)k
                                                     (e) (2 + cost)i + 9j + (3 + sint)k
                                                     solution Allthe curves except for (b) lie in the vertical plane y = 9. We identify each one of the curves (a), (c), (d)
                                                     and (e).
                                                     (a) The parametric equations are:
                                                                                              x =−2+cost, y=9,z=3−sint
                                                     Hence,
                                                                                                  2           2           2            2
                                                                                          (x +2) +(z−3) =(cost) +(−sint) =1
                                                     This is the circle of radius 1 in the plane y = 9, centered at (−2,9,3).
                                                     (c) The parametric equations are:
                                                                                            x =−2+cos3t, y=9,z=3−sin3t
                                                     Hence,
                                                                                                 2           2            2             2
                                                                                         (x +2) +(z−3) =(cos3t) +(−sin3t) =1
                                                     This is the circle of radius 1 in the plane y = 9, centered at (−2,9,3).
                                                     (d) In this curve we have:
                                                                                              x =−2−cost, y=9,z=3+sint
                                                     Hence,
                                                                                                  2           2             2          2
                                                                                          (x +2) +(z−3) =(−cost) +(sint) =1
                                                     Again, the circle of radius 1 in the plane y = 9, centered at (−2,9,3).
                                                     (e) In this parametrization we have:
                                                                                               x = 2+cost,       y=9,z=3+sint
                                                     Hence,
                                                                                                   2           2           2          2
                                                                                           (x −2) +(z−3) =(cost) +(sint) =1
                                                     This is the circle of radius 1 in the plane y = 9, centered at (2,9,3).
                                                     Weconcludethat (a), (c) and (d) parametrize the same circle whereas (b) and (e) are different curves.
                                                     Exercises
                                                                                          t    1            −3
                                                      1. What is the domain of r(t) = e i + t j + (t + 1)      k?
                                                     solution r(t)isdefinedfort = 0 and t =−1, hence the domain of r(t) is:
                                                                                                          {                       }
                                                                                                    D= t ∈R:t =0,t=−1
                                                                                                   π    2    2     −1
                                                                                                   √
                                                      3. Evaluate r(2) and r(−1) for r(t) = sin       t,t ,(t +1)       .
                                                                                              s     2
                                                             Whatisthedomainofr(s) = e i+            sj +cossk?
                                                                                     π    2   2      −1
                                                     solution Sincer(t) = sin 2t,t ,(t +1)                , then
                                                                                                                    −1          1
                                                                                                  r(2) = sinπ,4,5        = 0,4, 5
                                                     and
                                                                                                            −π       −1            1
                                                                                              r(−1) = sin 2 ,1,2           = −1,1, 2
                                                                   May16,2011
                     252   CHAPTER 13           CALCULUSOFVECTOR-VALUEDFUNCTIONS (LTCHAPTER14)
                                                                                                                                                  
                                                  5. Find a vector parametrization of the line through P = (3,−5,7) in the direction v = 3,0,1 .
                                                                                                                                               2  4
                                                        Doeseither of P = (4,11,20) or Q = (−1,6,16) lie on the path r(t) = 1 + t,2 + t ,t ?
                                                 solution Weusethevectorparametrization of the line to obtain:
                                                                                    −→                                                   
                                                                            r(t) = OP +tv = 3,−5,7 +t 3,0,1 = 3+3t,−5,7+t
                                                 or in the form:
                                                                                   r(t) = (3 + 3t)i − 5j + (7 + t)k,     −∞