ENGI 5432 Advanced Calculus

    Faculty of Engineering and Applied Science
    2009 Winter

    Problem Set 2   -   Solutions

    1. For the vector field   F = < 1/x, e^y, -1 >
      1. find the equations of the lines of force and

        dx/ds = t F     ==> < dx/ds, dy/ds, dz/ds > = k < 1/x, e^y, -1 >
        Þ     k ds   =   x dx   =   e -y dy   =   -dz
        Integral (x) dx = Integral (e^(-y)) dy = Integral (-1) dz >
        (x^2)/2 + c1 = -z and -e^(-y) + c2 = -z
        x^2 = -2c1 - 2z and e^(-y) = c2 + z
        In order to avoid the problem of multiple square roots, make   x   the parameter
        Then,
        z = -c1 - (1/2)x^2 and -y = ln(c2 + z) = ln(c2 - c1 - (1/2)x^2)
        Let   A = c 2 - c 1   and   B = -c1   then the lines of force are

        y = -ln(A - (1/2)x^2),     z = B - (1/2)x^2     (provided   x2 < 2A).

        Alternate forms for the answer include
        x = +-(A-2z)^.5 , y = -ln(B + z), (provided -B < z =< (1/2)A )
        and
        x = ±(A - 2e^(-y))^.5 , z = B + e^(-y) (provided y >=ln(2/A) )

      2. find the equations of the particular line of force passing through the point   (2, 0, 4).

        The line of force passes through (2, 0, 4)
        Þ     0   =   -ln(A - 4/2)     Þ     A - 2 = 1     Þ     A = 3
        and     4 = B - 4/2     Þ     B = 4 + 2 = 6

        Therefore the line of force is

        y = -ln( 3 - (1/2)x^2 ),     z = 6 - (1/2)x^2     (provided x2 < 6)

        or
        with   z   as the parameter
        x = +-(12 - 2z)^.5 , y = -ln(z - 3) (provided 3 < z <= 6)
        or
        with   y   as the parameter
        x = +-( 6 - 2e^(-y) )^.5 , z = 3 + e^(-y), (provided y >= ln 3)

    1. For the vector field   F = 2*e^z j - cos y k
      1. find the equations of the lines of force and

        dx/ds = k F ==> <dx/ds , dy/ds , dz/ds> = k <0, 2e^z, -cos y>
        ==> dx/ds = 0 and k ds = dy/2e^z = -secy dz
        ==> x = A and (Integ) cosy dy = -2 (Integ) e^z dz
        ==> sin y = c - 2e^z
        Therefore the lines of force are

        x = A,     y = Arcsin (c - 2e^z)     ln( (c - 1)/2 ) <= z <= ln( (c + 1)/2 )

        or
        x = A , z = ln( (c - sin y)/2 ) provided sin y < c

      2. find the equations of the particular line of force passing through the point   (3, p, 0).
        The line of force passes through   (3, p, 0)
        Þ     3 = A     and     sin p   =   c - 2e0
        Þ     0 = c - 2     Þ     c = 2

        Therefore the line of force is
        x = 3 , y = Arcsin(2 - 2e^z) ( ln(1/2) <= z <= ln(3/2) )
        or
        with   y   as the parameter:

        x = 3,   z = ln( (2 - sin y)/2 )             (for all real y)

    1. Find the family of vector fields in real 3 space whose lines of force are straight lines.

      With   s = arc length and r0, v constant vectors, the vector parametric form of
      the equation of a straight line can be written as
      r = r0 + sv ==> dr/ds = v
      But, along a line of force, dr/ds = kF
      Therefore F = (1/k)v, which is parallel to v for all t
      Þ     F     must be a family of parallel vectors:
      (the direction never changes, although the magnitude may vary)

      F = f(s) <a, b, c> ,

      where a, b, c = scalar constants (not all zero) and
      f (s) = some scalar function of s.
      Any vector field of this type is a valid solution.

      Check [not required]:
      dr/ds = tF ==> < (dx/ds), (dy/ds), (dz/ds) > = k f(s) <a, b, c>
      If any of   a, b, c   is zero, then the corresponding coordinate (x, y or z) is constant
      along the line of force.   For the non-zero case,
      kf(s)ds = dx/a = dy/b = dz/c, whose solution is
      (x-x0)/a = (y-y0)/b = (z-z0)/c, (where (xo , yo , zo ) is any point on the line of force),
      which is a straight line.
      When exactly one of   a, b, c   is zero, the line is parallel to the corresponding coordinate plane.
      When two of   a, b, c   are zero, the line is parallel to the remaining coordinate axis.

    1. The plane P   passes through the point (2, -1, 5) and contains the line L : y = -3 ,   z = x - 2 .

      1. Find the Cartesian equation of the plane P.

        [diagram]
        L :   y = -3 ,   z = x - 2
        (x-2)/1 = (z-0)/1 ; y = -3
        r = a + tv   where
        a = <2, -3, 0>, v = <1, 0, 1>

        Let Q be (2, -3, 0)   (which a point known to be on the line)
        Let P be (2, -1, 5) and u = QP = OP - OQ = <2-2, -1-(-3), 5-0> = <0, 2, 5>
        The required plane contains both of the vectors u and v.
        The plane normal is n = u x v = | i j k | | 0 2 5 | | 1 0 1 |
        = <2-0, 5-0, 0-2> = <2, 5, -2>
        ==> a * n = <2, -3, 0> * <2, 5, -2> = 4 - 15 + 0 = -11
        r * n = 2x + 5y - 2z

        2x + 5y - 2z = -11

        or     2x + 5y - 2z + 11   =   0

      2. Find the Cartesian equations of the normal line to   P   that passes through the origin.

        Normal Line: a = <0,0,0> v = n = <2, 5, -2>     (line must pass through origin)
        therefore the line is

        (x-0)/2 = (y-0)/5=(z-0)/-2

    1. Find the [acute] angle q   (to the nearest 0.1°) between the line
              r = < 2, 7, -10 > + t < 3, 1, 2 >
      and the plane
              r dot < 2, 1, -3 > = 1

      [diagram]
      L : r = <2, 7, -10> + t<3, 1, 2>
      Plane : r * <2, 1, -3> = 1
      v = <3, 1, 2>
      and
      n = <2, 1, -3>
      sin(theta) = (|v dot n|)/vn = (|3 x 2 + 1 x 1 + 2(-3)|)/( (9+1+4) x (4x1x9) )^.5 = 1/14

      theta = 4.1 degrees

    1. A temperature distribution for a region within 75 metres of the origin is given by
                  T = (10000 - x^2 - y^2) / (z + 100)

      1. Find the gradient of the temperature function   T.

        T = {partial} <dT/dx, dT/dy, dT/dz> = <(-2x)/(z+100), (-2y)/(z+100), (x^2 + y^2 - 10000)/(x+100)^2>

      2. Find the [instantaneous] rate at which the temperature is changing at the point (50, 50, 0) in the same direction as the vector i - j .

        P is the point (50, 50, 0)
        Direction a = i - j ==> a = |a| = (1+1)^.5 = (2)^.5
        ==> u = a/a = (1/2)*(2)^.5 <1, -1, 0>
        T_p = < (-2)(50)/(0+100), (-2)(50)/(0+100), (50^2 + 50^2 - 10000)/(0+100)^2 > = < -1, -1, -1/2 >
        ==> DuTp = Tp dot u = <-1, -1, -1/2> dot (1/2)*(2)^.5<1, -1, 0> = (1/2)(2)^.5(-1 + 1 + 0) = 0
                                  0

        [The vector i - j is at right angles to the gradient vector (and therefore tangential to a level surface) at the point (50, 50, 0).]

      3. Is the field formed by the gradient vector purely radial?
        [That is, does the gradient vector point directly towards or directly away from the origin at every point?]

        At P (50, 50, 0),   r = <50, 50, 0>   but
        Tp = <-1, -1, -1/2> not= kr   for any scalar k.
        Therefore,  
          No , the gradient vector is not purely radial.  

    1. Find the equations of the tangent plane and the normal line to the sphere
              x2 + y2 + z2 = 9
      at the point (-2, 1, 2).

      f (x, y, z) = x2 + y2 + z2 = 9
      ==> f = <2x, 2y, 2z> = 2 <x, y, z>
      Use   n = <x, y, z>
      P (-2, 1, 2) n = <-2, 1, 2> ==> n = sqrt{4+1+4} = 3
      Tangent plane at P:
      r dot n = a dot n
      ==> <x, y, z> dot <-2, 1, 2> = <-2, 1, 2> dot <-2, 1, 2>

      -2x + y + 2z = 9

      Normal line at P:
      r = a + tv with v = n
      ==> <x, y, z> = <-2, 1, 2> + t<-2, 1, 2>
      [from which is it easy to deduce that this normal line passes through the origin.]
      One form of the equations of the normal line is

      (x+2)/-2 = (y-1)/1 = (z-2)/2

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    Created 2008 01 01 and most recently modified 2008 12 27 by Dr. G.H. George