ENGI 5432/5435 Advanced Calculus

    Faculty of Engineering and Applied Science
    2008 Winter

    Term Test 1   -   Solutions

    1. For the ellipsoid defined by   x^2 / 6 + y^2 / 3 + z^2 / 2 = 1

      1. Show that the equation of the tangent plane to the ellipsoid at the point (1, 1, –1) is   x + 2y – 3z = 6.

        First verify that the point (1, 1, –1) is indeed a point on the ellipsoid:
        1/6 + 1/3 + 1/2 = 1
        gradient = < x/3, 2y/3, z >
        gradient at point = < 1/3, 2/3, -1 >
        We can take as the normal to the tangent plane any convenient non-zero multiple of this gradient vector:   normal = < 1, 2, -3 >
        a dot n = 6
        Therefore the equation of the tangent plane to the ellipsoid at (1, 1, –1) is

        x + 2y - 3z = 6

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

        Any plane   z = constant   is at right angles to the z-axis.
        A normal vector to   z = –1   is therefore   k.
        The angle between the ellipsoid and the plane is the angle between their normal vectors:
        cos theta = 3/sqrt{14}
        Therefore the angle, correct to the nearest degree, is

        theta = 37 degrees

    1. A vector field is defined in cylindrical polar coordinates by rho exp(-rho) phiHat.
      Find   curl vector F   in terms of cylindrical polar components.

      curl F = 1/rho d/drho (rho^2 exp(-rho)) kHat

      curl F = (2 - rho) exp(-rho) kHat

      [This result can also be obtained by transforming the vector into Cartesian form, then converting the curl back into cylindrical polar form, but this method is far less efficient, as shown on this separate page.]

    1. The location vector r, function of time of a particle moving in a fluid at any time   t   is given by

      r = > t sqrt{3}, sin t, cos t <

      1. Show that the tangential component of acceleration is zero at all times.

        velocity = < sqrt{3}, cos t, -sin t > ==> speed = 2
        a_T = dv/dt = 0

      2. Find the radius of curvature   r   at all times.

        acceleration vector = < 0, -sin t, -cos t >
        v × a = < -1, sqrt{3}cos t, -sqrt{3}sin t >
        |v × a| = 2
        | v | = 2
        curvature = 1/4

        radius of curvature = 4 metres

        [The curvature may also be found from
        a_N = kappa v^2 or | dT^/ds |

      3. Find the angle between the tangent vector to the path of the particle and the y-z coordinate plane.

        This is equivalent to the problem of finding the angle between a line and a plane.
        vectors v = < sqrt{3}, cos t, -sin t > , n = < 1, 0, 0 >
        sin theta = sqrt{3}/2

        theta = 60 degrees

      BONUS QUESTION

    1. Find the equations of the line of force for the vector field     that passes through the point (0, –1, 0).

      dx/ds = ky, dy/ds = -kx, dz/ds = kxy
      dx/y = dy/(-x) = dz/(xy)
      -Integ{x dx} = Integ{y dy}
      x^2 + y^2 = A^2
      Each line of force therefore spirals around the z axis at a constant distance from the axis.
      z = x^2 / 2 + B
      Choosing   x   to be the parameter, the family of lines of force is
      r(t) = < x, ±sqrt{A^2 - x^2}, x^2 / 2 + B >
      We require the line of force through (0, –1, 0).
      y < 0     Þ     we must take the negative square root:
      A = 1 and B = 0
      The required line of force is therefore

      r(t) = < x, -sqrt{1 - x^2}, x^2 / 2 >

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    Created 2008 01 26 and most recently modified 2008 01 27 by Dr. G.H. George