ENGI 5432 Advanced Calculus

    Faculty of Engineering and Applied Science
    2009 Winter

    Problem Set 1   -   Solutions

    1. Let   F = x(t)i + y(t)j + z(t)k   be the position vector of a particle at time   t   as it moves along a curve   C, where the functions   x(t), y(t) and z(t)   are all differentiable for all values of   t.   Show that if   F X dF/dt = 0   then the particle must always move in the same direction.

      F = < x(t), y(t), z(t) > , differentiable for all t
      F X dF/dt = 0 ==> (i) F = 0 or (ii) dF/dt = 0 or (iii) F || dF/dt
      (i)   Þ   (ii)
      (ii)   Þ   F   is a constant vector and the particle is stationary.
      But we are told that the particle is moving along a curve C.
      Therefore cases (i) and (ii) can be dismissed and
      (iii) must be true   Þ   the velocity vector is parallel to   F(t)   at all times
      Þ     F   never changes direction.
      Therefore F X dF/dt = 0   Þ   the particle stays on a straight line path for all time.

    1. Let   r = < alpha*sin(t), alpha*cos(t), beta*t >   be the position vector of a particle.   You may assume that   alpha   and   beta   are positive constants.

      1. vector v, v, vector a, a, aT, aN, kappa, T-hat, N-hat, B-hat

        Let   s = sin t ,   c = cos t.
        a, b   are positive constants.
        r = < alpha*s, alpha*c, beta*t >

        v = dr/dt = < alpha*c, -alpha*s, beta >
        |v| = sqrt{alpha^2 + beta^2}
        a = dv/dt = -alpha* < s, c, 0 >
        |a| = alpha
        aT = d|v|/dt = 0   (The speed v is constant)
        or
        aT = v.a/|v| = < alpha*c, -alpha*s, beta > dot -alpha* < s, c, 0 > / |v| = 0
        aN = sqrt{a^2 - aT^2} = |a| = alpha (or find | v cross a | / |v| )
        aN = kappa |v|^2 --> kappa = alpha / (alpha^2 + beta^2)
        That = v / |v| = < alpha*c, -alpha*s, beta > / sqrt{alpha^2 + beta^2}
        Nhat = dT/dt / |dT/dt| = - < s, c, 0 >
        or
        a = aT That + aN Nhat ; aT = 0, aN = alpha ==> Nhat = a / |a| = - < s, c, 0 >
        Bhat = That cross Nhat = ... = < beta*c, -beta*s, -alpha > / sqrt{alpha^2 + beta^2}
        and it is easy to check that   B dot T = T dot N = N dot B = 0
        and that   || B || = || N || = || T || = 1.

      2. Does the curve lie in one plane?

        B   is a non-constant function of   t (for the x and y components).   Therefore

          NO, the curve does not lie in one plane.  

      3. What type of curve is it?

          It is a helix, centred on the z-axis.  

        [The constants   a, b   in the parametric equation of this curve represent the [constant] magnitude of the acceleration and a measure of the degree to which the helix is stretched out along its axis, respectively.]

    1. There are three formulæ connecting arc length and the unit tangent, unit principal normal and unit binormal vectors, known as the Frenet-Serret formulæ.

      1. Show that the first Frenet-Serret formula is dT/ds = k N ,
        where   s   is arc length and   kappa   is the curvature.

        By the definition of N-hat, Nhat = rho dT/ds and rho = 1/kappa
        The first formula follows immediately:

        dT/ds = k N
      2. Use B = T X N to show that dB/ds is perpendicular to T-hat.

        B = T X N ==> dB/ds = dT/ds X N + T X dN/ds
        [Note that the cross product is anti-commutative, so the order in which the terms of the derivative by the product rule are written down is important.]
        But   (1)     Þ
        dT/ds = kappa N ==> dB/ds = k N X N + T X dN/ds
        But N X N = 0 ==> dB/ds = 0 + T X dN/ds,
        which is perpendicular to T-hat.

      3. Use B . B = 1 to show that dB/ds is perpendicular to B-hat.

        B-hat is a unit vector     Þ
        B.B = 1 ==> d/ds(B.B) = d.ds(1) = 0 --> 2 B . dB/ds = 0 ==> dB/ds perp. to B

      4. Hence prove the second Frenet-Serret formula   dB/ds = - tau*N, where the torsion,   tau, is some scalar function of   s.   [The torsion is a measure of how much the curve is twisting out of the plane of   T-hat and N-hat.]

        Any vector perpendicular to both B-hat and T-hat must be || N-hat,
        (because   T-hat, N-hat, B-hat   form an orthonormal basis).
        dB/ds must be some multiple of N
        Define that multiple to be   -t   (where   t = torsion).   Then

        dB/ds = -tau N

      5. Use the fact that   T-hat, N-hat, B-hat   form a mutually orthogonal right-handed triad of unit vectors, (from which it necessarily follows that   B-hat = T-hat X N-hat, T-hat = N-hat X B-hat and N-hat = B-hat X T-hat),   to establish the third Frenet-Serret formula   dN/ds = tau*B - kappa*T

        N = B X T     [Cyclic system: Cyclic: T --> N --> B --> T]

        ==> dN/ds = dB/ds X T + B X dT/ds
        Substitute (1) and (2):
        dN/ds = -tau*N X T + B X kappa*N
        = -tau(-T X N) + kappa(-N X B)     (cross products are anti-commutative)
        = +tau*B - kappa*T

        Therefore the third Frenet-Serret formula is

        d(N-hat)/ds = tau*(B-hat) - kappa*(T-hat)

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    Created 2002 01 03 and most recently modified 2008 12 27 by Dr. G.H. George