ENGI 5432 Advanced Calculus

    Faculty of Engineering and Applied Science
    2009 Winter

    Term Test 1   -   Solutions

    1. For the surface defined by   f = y^3 - x^2 y + z^2 = 1,

      1. Find the equations of the tangent plane and the normal line to the surface at the point (–1, 1, 1).

        f = y^3 - x^2 y + z^2 = 1 ; P(-1, 1, 1)
        gradient of f at P = 2 < 1, 1, 1, >
        The vectors defining the tangent plane are
        n = < 1, 1, 1, > ; a = < -1, 1, 1, > ==> n dot a = 1
        The equation of the tangent plane is n dot r = n dot a

        x + y + z = 1

        The vectors defining the normal line are   v = n ; a = < -1, 1, 1, >
        The equation of the line is
        (x - x0) / v1 = (y - y0) / v2 = (z - z0) / v3

        x+1 = y-1 = z-1     [or   z = y = x + 2]

      2. Find, to the nearest degree, the angle that the surface makes with the plane   x + z = 0   at the point (–1, 1, 1).

        The angle between the surface and the plane is also the angle between the normal of the tangent plane to the surface at that point and the normal to the plane   x + z = 0 .
        n1 = < 1, 1, 1, > ; n2 = < 1, 0, 1, > ;
        cos theta = sqrt(2/3)
        theta = 35.26... degrees     Correct to the nearest degree,

          q = 35°  

    1. The location of a particle is defined in spherical polar coordinates by vector r = scalar r . unit vector r^.
      At any instant,   r = 3, theta = 2t, phi = pi/2.
      Find the velocity   v   in spherical polar coordinates.

      d(theta)/dt = 2, dr.dt = d(phi)/dt = 0
      velocity = d(vector r)/dt = ...
      [only one non-zero term for the velocity]

      v = 6 theta^

    1. A thin wire is in the shape of the unit circle in the xy-plane, C: x^2 + y^2 = 1 , z = 0, which can be parameterized by . r = < cos t, sin t, 0 > , (-pi < t <= pi).
      The line density of the wire is   rho = y + 2 - z.

      1. Find the mass and the Cartesian coordinates of the centre of mass of the wire.

        [circular wire]
        rho = y + 2 - z
        On C, rho = 2 + sin t
        The element of mass for the wire is
        Delta m = rho Ds/Dt Dt
        m = Int_C rho |dr/dt| dt
        vector dr/dt = < -sin t, cos t, 0 >
        |dr/dt| = 1
        m = 4 pi

        The wire lies entirely in the xy-plane zBar = 0
        Both the wire and the density are symmetric about   x = 0 xBar = 0
        The wire is symmetric about   y = 0, but the density is not.
        Taking moments about the xz plane,
        M = Integral rho y |dr/dt| dt
        Delta M = y Delta m
        M = pi
        yBar = M/m = 1/4
        Therefore the mass and centre of mass of the wire are

        m = 4 pi ; (x,y,z)Bar = (0, 1/4, 0)

        Note that, without consideration of symmetry, the evaluation of the vector moment would have been
        M = Integral rho r |dr/dt| dt

      2. Find the work done by a force F = < -y, x, 0 > in travelling once around the wire.

        F = < -sint, cost, 0 >
        F dot dr/dt = 1
        W = Integral F dot dr/dt

        W = 2 pi

    1. For the velocity field   v = < x - y^2, x^2 - y, 0 >

      1. Find the divergence   div v.

        [divergence]

        div v = 0

        BONUS QUESTION

      1. Find the stream function psi(x, y) and the family of streamlines for   v.

        z   never changes and   div v = 0 ==> stream function exists
        [match partial derivatives of psi]

        psi = x^3 / 3 - xy + y^3 / 3

        and the family of streamlines is   psi = C, z = B
        or, re-writing the family as   3 psi = A, z = B
        the family of streamlines is

        x^3 - 3xy + y^3 = A, z = B

        The family of streamlines is plotted on this separate page.

      2. Find the equation of the streamline that passes through the point (2, 0, 0).

        x^3 - 3xy + y^3 = A; psi(2,0) = 8 = A
        Also   z = 0   leads obviously to   B = 0.
        Therefore the required streamline is

        x^3 - 3xy + y^3 = 8 , z = 0

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    Created 2009 01 31 and most recently modified 2009 02 01 by Dr. G.H. George