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File: Vector Calculus Pdf 170954 | Tema1 Apostol Secc14 14
tom m apostol calculus volume 1 one variable calculus with an introduction to linear algebra second edition john wiley sons inc new york l santa barbara l london l sydney ...

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             Tom M. Apostol
       CALCULUS
                    VOLUME 1
    One-Variable Calculus, with an
    Introduction to Linear Algebra
                 SECOND EDITION
                   John Wiley  & Sons, Inc.
           New York l Santa Barbara l London l Sydney l Toronto
                      536                                                      Calculus          of vector-valuedfunctions
                      18. If 0 < 6 < 4a,  let r(f) = a(t - sin t)i + a(1 - COS t)j + b sin ht k. Show that the length of
                              the path traced  out  from  t = 0 to t = 2~ is 8&(k),  where E(k) has the meaning given in
                              Exercise  17 and k2 = 1  - (b/4~)~.
                      19. A particle moves with position vector
                                                                                    r(t)  =  tA  +  t2B +  2(5t)3/3   A x  B ,
                              where A and B are two fixed unit vectors making an angle of ~13  radians with each  other.
                              Compute the speed of the particle at time  t and find how long it takes for  it  to move a distance
                              of 12 units of arc length from the initial position r(0).
                      20. (a) When a circle rolls (without slipping) along  a straight line, a point on the circumference
                              traces  out  a curve called a  cycloid.                           If the fixed line is the x-axis and if the tracing point (x, y)
                              is originally at  the origin, show that  when the circle rolls through an angle 0 we have
                                                                               x = a(0  - sin 0) ,                       y = a(1 - COS e)  >
                              where a is the radius of the circlle.                             These serve as parametric equations for the cycloid.
                              (b) Referring to part (a), show that dy/dx  = cet  $0 and deduce that the tangent line of the
                              cycloid at (x, y) makes an angle + (7~ - 0) with the x-axis. Make a sketch and show that the
                              tangent line passes through the highest point on the circle.
                      21. Let C be a curve described by two equivalent functions                                                          X and Y, where Y(t) = X[u(t)]  for
                              c 5 t 5 d. If the function                      u which defines  the change of parameter has a continuous                                                       deriv-
                              ative  in [c,  d] prove that
                                                                                           ‘u(d)    IIX’(u)Ij     du =j””  11 Y’(t)i/  dt ,
                                                                                         iu(c)                                 c
                              and deduce that the arc length of C is invariant under such  a change of parameter.
                      22. Consider the plane curve whose vector equation is r(t) = ti +f(t)j,  where
                                                                                                      rr
                                                                          f(t) = f COS z                            if t#O,                    f(0)  = 0 .
                                                                                                   0
                              Consider the following partition of the interval                                          [0, 11:
                                                                                                                1                    1     1
                                                                                                  2n >2n _  1 9*. . >3 > 2 >1                          .
                                                                                   p=i 0,’                 ~ -- 1
                               Show that the corresponding inscribed polygon  T(P)   has length
                               and deduce that this curve is nonrectifiable.
                       14.14 Curvature of a curve
                            For a straight line the unit tangent vector  T  does  not change its direction, and hence
                       T’  = 0. If the curve is not a straight line, the derivative  T’ measures the tendency of the
                       tangent to change its direction. The rate of change of the unit tangent with respect to arc
                                                       Curvature of a curve                                                537
         length   is called the  curvature vector  of the curve.           We denote  this by dT/ds,  where s repre-
         sents arc length. The chain rule, used in conjunction with the relation s’(t)  = v(t),  tells us
         that the curvature vector dT/ds  is related to the “time” derivative T’ by the equation
                                            dT       dt dT
                                            -c--c-
                                            ds       ds dt       &> T’(t)  = & T’(t).
         Since   T’(t)  = II T’(t)  //  N(t), we obtain
                                                        dT  IITYOII   Nctj
         (14.19)                                        -=-
                                                        ds         40            ’
         which shows that the curvature vector has the same direction as the principal normal N(t).
         The scalar  factor which multiplies  N(t)  in (14.19) is a nonnegative number called the
          curvature of the curve at t and it is denoted by K(t) (K is the Greek letter kappa). Thus the
          curvature K(t), defined to be the length of the curvature vector, is given by the following
          formula :
          (14.20)                                          K(t) - 11 T’(t)11  .
                                                                      40
             EXAMPLE 1. Curcature of a circle. For a circle of radius a, given by r(t) = a  COS t i +
          a sin tj,  we have  v(t)  =  -a  sin t i +  a  COS  t  j,  v(t)  =  a, T(t)  =  -sin   t i +  COS  tj,  and
          T’(t)  =  -COS  t i - sin t  j. Hence we have  11  T’(t)11   =  1  SO  K(t)   =  I/a.   This shows that a
          circle has constant curvature. The reciprocal of the curvature is the radius of the circle.
             When K(t) # 0, its reciprocal is called the radius of curvature and is denoted by p(t)
          (p is the Greek letter rho). That circle in the osculating plane with radius p(t)  and tenter
          at the tip of the curvature vector is called the osculating circle. It cari  be shown that the
          osculating circle is the limiting position of circles passing through three nearby points on
          the curve as two of the points approach the third.                 Because of this property, the osculating
          circle is often called the circle that “best fits the curve” at each  of its points.
             EXAMPLE   2. Curvature  of  a plane curve.           For a plane curve, we have seen  that 11 T’(t)  Ij =
          Iu’(t)l,   where u(t) is the angle the tangent vector makes with the positive x-axis, as shown
          in Figure 14.11. From the chain rule,  we  have  cr’(t)  =  dcc/dt =  (dcc/ds)(ds/dt)                 = v(t)dcc/ds,
          SO Equation (14.20) implies
          In other words, the curvature of a plane curve is the absolute value of the rate of change of
          cx  with respect to arc length. It measures the change of direction per unit distance along the
          curve.
             EXAMPLE  3. Plane  curves   of  constant curvature.  If  dulds   is a nonzero constant,  say
          da/ds   = a, then tc = as + b, where b  is a constant.                 Hence, if we use the arc length s as
                                                                                                             Calcur’us   of  vector-valuedfunctions
                             538
                             a parameter, we have  T  =  COS  (as  +  b)i + sin (as  +  ZI)~.  Integrating, we find that  r =
                             (lia)  sin (as + b)i - (l/a)   COS  (as  +  6)j + A, where A is a constant vector. Therefore
                              11~  -  A  11 =  I/lal,   SO the curve is a circle (or an arc of a circle) with tenter at A and radius
                             l/lal.  This proves that a plane curve of constant curvature K # 0 is a circle (or an arc of a
                             circle) with radius l/~.
                                    Now we prove a theorem which relates the curvature to the velocity and acceleration.
                                    THEOREM  14.14. For any  motion with  velocity v(t), speed v(t), acceleration u(t), and
                              curvature  K(t),   we  have
                             (14.21)                                                                            u(t)  -= v’(t)T(t)                           + K(t)v’(t)N(t).
                              This formula, in  turn,   implies
                              (14.22)                                                                                          K(t)  =  Ila  x V(al
                                                                                                                                                               u3(t)
                                    Proof.                 TO prove (14.21), we rewrite (14.20) in the form I/ T’(t)  11 = K(t)V(t),  which gives us
                              T’(t)  =  K(t)V(f)N(t).                                    Substituting this expression for  T’(t)  in Equation  (14.8), we obtain
                              (14.21).
                                    TO prove (14.22), we form the cross product  u(t)  x  v(t),  using (14.21) for  u(t)  and the
                              formula  v(t)  = v(t)T(t)  for the velocity. This gives us
                              (14.23)                                                        a x v = v’vT  x T + KV~N  x  T = KV~N   x T
                              since   T  x  T  = 0. If we take the length of  each  member of (14.23) and note that
                                                                                                             Il N x Tlj = lINl/   II TII  sin &T  =  1 ,
                              we obtain Ila  x  vlj =  KV~,  which proves (14.22).
                                     In practice it is fairly easy to compute the vectors u and a (by differentiating the position
                              vector  Y); hence Equation (14.22)  provides a useful method for computing the curvature.
                              This method is usually simpler than determining the curvature from its definition.
                                     For a straight line we have  u x v = 0,  SO the curvature is everywhere zero. A curve
                              with a small curvature at a point has a large radius of curvature there and hence does  not
                              differ  much from a straight line in the immediate vicinity of the point. Thus the curvature
                              is a measure of the tendency of a curve to deviate from a straight line.
                               14.15 Exercises
                                  1. Refer to the  curves  described in  Exercises 1 through 6 of Section 14.9 and in  each  case determine
                                         the curvature K(I) for the value of t indicated.
                                 2. A helix is described by the position function                                                                             r(t) = a COS ot i + a sin otj  + bwtk.  Prove that
                                         it has constant curvature K = a/(~?  + @).
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...Tom m apostol calculus volume one variable with an introduction to linear algebra second edition john wiley sons inc new york l santa barbara london sydney toronto of vector valuedfunctions if a let r f t sin i cos j b ht k show that the length path traced out from is where e has meaning given in exercise and particle moves position ta tb x are two fixed unit vectors making angle radians each other compute speed at time find how long it takes for move distance units arc initial when circle rolls without slipping along straight line point on circumference traces curve called cycloid axis tracing y originally origin through we have radius circlle these serve as parametric equations referring part dy dx cet deduce tangent makes make sketch passes highest c be described by equivalent functions d function u which defines change parameter continuous deriv ative prove iix ij du dt iu invariant under such consider plane whose equation ti rr z o following partition interval n p corresponding in...

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