ENGI 5432 Advanced Calculus

    Faculty of Engineering and Applied Science
    2009 Winter

    Problem Set 3   -   Solutions

    1. A curve is defined in vector parametric form by

      r = < t^2, t^3, 0 >

      1. Show that the unit tangent is well-defined everywhere along the curve except at the origin.

        dr/dt = < 2t, 2t^2, 0 > = t < 2, 3t, 0 >         ==> ds/dt = |dr/dt| = |t|(4 + 9t^2)^.5
        dr/st and ds/dt are well-defined everywhere, but
        ds/dt = 0 at t=0 (where vector r = 0)
        Therefore, T = (dr/dt) / |dr/dt| = (t/|t|) * ( (< 2, 3t, 0 >) / ((4 + 9t^2)^.5) )
        is well-defined everywhere except at the origin (0, 0, 0).

      2. Eliminate the parameter t to find the Cartesian equations of the curve.

        x = t2       y = t 3       z = 0
        ==> t = x^.5 ==> y = x^(3/2);
        Therefore the equations of the curve are           y = x^(3/2), z = 0

      3. Sketch the curve.   Display on your sketch the unit tangent T-hat and the unit principal normal N-hat at two points close to the origin, one in the region where t < 0 and the other in the region where t > 0 .

        [graph showing cusp]

      4. From your sketch, state briefly why the unit tangent is not defined at the origin.

        T-hat “bounces” from minus i-hat to plus i-hat abruptly at   t = 0
        (and N-hat also changes abruptly at t = 0, from minus j-hat to plus j-hat )

      5. From your sketch, show that the unit binormal B-hat has a well defined limit at the origin, (even though the unit tangent and unit principal normal do not).

        Everywhere except t = 0, B-hat = T-hat x N-hat = + k-hat
        lim(as t -> 0) B-hat = + k-hat

    1. For the surface whose equation in cartesian coordinates is z2 = x2 - y2,

      1. what type of quadric surface is this surface?

        [Maple plot of cone] z2 = x2 - y2     Þ     - x2 + y2 + z2 = 0
        This quadric surface is of the type     X^2/a^2 + Y^2/b^2 - Z^2/c^2 = 0     with a = b
        It is therefore a

          circular cone  

        whose axis of symmetry is the x-axis

      2. find the cartesian equation of the tangent plane to the surface at the point (1, 1, 0).

        f(x, y, z) = -x^2 + y^2 + z^2 = 0 ==> del f = 2 < -x, y, z >
        At (1,1,0)     At (1, 1, 0) del f = 2 < -1, 1, 0 > leads to n = <-1, 1, 0 >
        (Any non-zero multiple of the gradient vector is a normal vector.)
        A point on the tangent plane is   (1, 1, 0) ==> a = < 1, 1, 0 >
        ==> n dot a = < -1,1,0 > dot < 1,1,0 > = -1 + 1 + 0 = 0
        Þ     the tangent plane is   n dot r = n dot a ==> -1x + 1y = 0   or

        y = x

      3. find the cartesian equations of the normal line to the surface at the point (1, 1, 0).

        Normal line:     v = < -1,1,0 > a = < 1,1,0 >
        ==> r = < 1,1,0 > + t < -1,1,0 >

        (x-1)/-1 = (y-1)/1 ; z = 0

        or     y = 2 - x,   z = 0
        [Note that the line   y = x, z = 0   is one of the generators of the circular cone.]

    1. For the surfaces whose cartesian equations are   z = 3x2 + 2y2   and   -2x + 7y2 - z = 0 :

      1. what types of quadric surface are these surfaces?

        [Maple plot of parabolic cylinder intersecting elliptic paraboid] z = 3x2 + 2y2   is of type
        Z/c = X^2/a^2 + Y^2/b^2     elliptic paraboloid
        with its axis of symmetry along the z-axis.

        -2x + 7y2 - z = 0     Þ     7y2 = +2x + z
        is of type     X^2/a^2 = Y/b     parabolic cylinder
        Its vertex line is   2x + z = 0,   y = 0

      2. find the angle between the surfaces at the point (1, 1, 5).

        f(x, y, z) = 3x^2 + 2y^2 - z = 0 ==> del f = < 6x, 4y, -1 >
        ==> del f|(1,1,5) = < 6,4,-1 > = n1
        g(x, y, z) = -2x + 7y^2 - z = 0 ==> del g = < -2, 14y, -1 >
        ==> del g|(1,1,5) = < -1, 14, -1 > = n2
        [One should check that (1, 1, 5) does indeed lie on both surfaces]
        q   =   Angle between surfaces   =   Angle between gradients of f & g at the point
        ==> cos(theta) = (n1 dot n2)/(n1*n2) = ( <6, 4, -1 > dot < -2, 14, -1 > ) / ( (36+16+1)*(4+196+1) )^.5 =(-12 + 56 + 1) / (53 * 201)^.5
        45/(10653)^.5 ~ .4360     theta = 64.15 degrees       (to 2 d.p.) ( ~ 64 degrees 9' )

    1. Calculate the directional derivative of phi(r) = x*ln(y) - e^(x*z^3) at the point (5, 1, -2) in the direction of the vector a = 3 i - 2 j + k.

      phi(r) = x ln y - e^(xz^3) ==> phi = < ln y - z^3e^(xz^3) , x/y , -3xz^2e^(xz^3) >
      ==gt; phi|(5,1,-2) = < 0-(-8)e^(-40), 5, -60e^(-40) >
      a = < 3, -2, 1 > ==> a = (9 + 4 + 1)^.5 = (14)^.5
      ==> a-hat = ( (14)^.5 )/14 <3, -1, 1 >
      D sub a-hat phi|(5,1,-2) = < 8e^(-40), 5, -60e^(-40) dot ((14)^.5)/14 ^lt; 3,-1,1 >
      = ((14)^.5)/14 (24e^(-40) -10 -60e^(-40) = - ((14)^.5)/14(10 + 36e^(-40))
      -((14)^.5)/7 (5 + 18e^(-40)       ~ (-5(14)^.5)/7 ~~ 2.673

    1. Given that the velocity of a particle at any time   t   is v(t) = < 10, -8t, 6t > and that the particle is at the origin when   t = 0 , find the unit binormal vector B-hat to the curve along which the particle moves and hence find the equation of the plane in which the trajectory of the particle lies.

      v = 10(1 + t^2)^.5
      ==> T-hat = [1/( 10(1+t^2)^.5)] < 10,-8t,6t > = [1 / (1+t^2)] < 1, -4t/5, 3t/5 >
      ==> dT-hat/dt = [ <0,-4/5,3/5 >(1+t^2) - < 1, -4t/5,3t/5 > (t/(1+t^2).5) ]/1+t^2

      =[ <0-t,-45/-4t^2/5+4t^2/5, 3/5+3t^2/5-3t^2/5 > ] / (1+t^2)^(3/2) = [ < -t, -4/5, 3/5 > ] / (1+t^2)^(3/2)
      ==> abs(dT-hat/dt) = ((t^2 + 16/25 + 9/25)^.5) / (1+t^2)^(3/2) = ((1+t^2)^.5) / (1+t^2)^(3/2)

      ==> N-hat = 1 / (1+t^2)^.5 < -t,-4/5, 3/5; >
      ==> B-hat = T-hat cross N-hat = (1 / (1+t^2)) * [ i j k; 1 -4t/5 3t/5; -t -4/5 3/5]
      (1 / (1+t^2)) < 0, (-3/5)(1+t^2), (-4/5)(1+t^2) > = - < 0, 3/5, 4/5 >
      Therefore   B-hat = - < 0, 3/5, 4/5 >   everywhere along the curve
      B-hat = constant     Þ     the curve lies in one plane
      The plane normal n-arrow   =   any non-zero multiple of B-hat.
      Choose   n-arrow = < 0, 3, 4 >
      The curve passes through the origin     ==> a-arrow = zero vector     ==> n-arrow dot a-arrow = 0
      The equation of the plane is   n-arrow dot r-arrow = n-arrow dot a-arrow

      3y + 4z = 0

      [Note:   the curve is a parabola in this plane.]

      Alternative Calculation of Unit Principal Normal   N-hat:

      Use the acceleration vector:
      a-arrow = d(r-arrow)/dt = < 0,-8,6 > ==> a = (0+64+36)^.5 = 10
      a_T = dv/dt = 10d/dt(1+t^2)^.5 = 10t/(1+t^2)^.5
      a_N = ( a^2 - (A sub T)^2).5 = ( 100 - (100t^32)/(1+t^2))^.5 = 10 / (1+t^2))
      But   vector a = (a sub T)T-hat + (a sub N)N-hat ==> N-hat = (vector-a - (a_T)T-hat) / a_N
      therefore, N-hat = (((1+t^2)^.5 )/10)( <0,-8,6 > - ((10t)/(1+t^2))(1/(1+t^2))<1,-4t/5,3t/5>
      = (((1+t^2)^.5)/10)(1/(1+t^2)) < 0-10t, -8-8t^2+8t^2, 6+6t^2-6t^2 >
      = (1/(10(1+t^2)^.5))<-10t,-8,6> = (1/(1+t^2))<-t,-4/5,3/5>

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            Created 2008 01 02 and most recently modified 2008 12 27 by Dr. G.H. George