Velocity, Acceleration, Torsion, and Curvature

Trivium

Determine a parameterization of the line tangent to a curve at a point.

Calculate the unit tangent, unit normal, and unit binormal vectors to a parameterization of a curve at a point.

Calculate the curvature and the torsion of a parameterization of a curve at a point.

Exercises

  1. Remember to replace h4 with span class="h5".

  2. TK TK TK TK TK TK TK TK TK TK TK TK TK TK TK TK TK TK TK TK TK TK TK TK TK

    \(\displaystyle y = mx+b\)
    \(\displaystyle y = mx+b\)

Problems & Challenges

  1. James Stewart A projectile is fired from the origin with angle of elevation \(\alpha\) and initial speed \(v_0.\) Assuming that air resistance is negligible and that the only force acting on the projectile is gravity \(g,\) the position vector of the projectile is \[\mathbf{r}(t) = \bigl(v_0\cos(\alpha)t\bigr)\mathbf{i} + \bigl(v_0\sin(\alpha)t - \frac{1}{2}gt^2\bigr) \mathbf{j}.\] The maximum horizontal distance the projectile can travel is \(R = v_0^2/g\) and is achieved when \(\alpha = 45°.\)

    1. At what angle should the projectile be fired to achieve maximum height and what is the maximum height?
    2. Fix the initial speed \(v_0\) and consider the parabola \(x^2 + 2Ry - R^2 = 0\) whose graph is shown in the figure. Show that the projectile can hit any target inside or on the boundary of the region bounded by the parabola and the \(x\)-axis, and that it can’t hit any target outside this region.
    3. Suppose that the gun is elevated to an angle of inclination \(\alpha\) in order to aim at a target that is suspended at a height \(h\) directly over a point \(D\) units downrange. The target is released at the instant the gun is fired. Show that the projectile always hits the target, regardless of the value \(v_0,\) provided the projectile does not hit the ground “before” \(D.\)
    figure pending
  2. James Stewart A projectile is fired from the origin down an inclined plane that makes an angle \(\theta\) with the horizontal. The angle of elevation of the gun and the initial speed of the projectile are \(\alpha\) and \(v_0\) respectively.
    1. Find the position vector of the projectile and the parametric equations of the path of the projectile as functions of the time \(t.\) (Ignore air resistance.)
    2. Show that the angle of elevation \(\alpha\) that will maximize the downhill range is the angle halfway between the plane and the vertical.
    3. Suppose the projectile is fired up an inclined plane whose angle of inclination is \(\theta\). Show that, in order to maximize the (uphill) range, the projectile should be fired in the direction halfway between the plane and the vertical.
    4. In a paper presented in 1686, Edmond Halley summarized the laws of gravity and projectile motion and applied them to gunnery. One problem he posed involved firing a projectile to hit a target a distance \(R\) up an inclined plane. Show that the angle at which the projectile should be fired to hit the target but use the least amount of energy is the same as the angle in the previous part. (Use the fact that the energy needed to fire the projectile is proportional to the square of the initial speed, so minimizing the energy is equivalent to minimizing the initial speed.)
    Figure Pending

  3. James Stewart A projectile of mass \(m\) is fired from the origin at an angle of elevation \(\alpha\). In addition to gravity, assume that air resistance provides a force that is proportional to the velocity and that opposes the motion. Then, by Newton’s Second Law, the total force acting on the projectile satisfies the equation \[ m\frac{\mathrm{d}^2\mathbf{R}}{\mathrm{d}t^2} = -mg\mathbf{j} - k\frac{\mathrm{d}\mathbf{R}}{\mathrm{d}t} \] where \(\mathbf{R}\) is the position vector and \(k \gt 0\) is the constant of proportionality.

    1. Show that this equation can be integrated to obtain the equation \[ \frac{\mathrm{d}\mathbf{R}}{\mathrm{d}t} + \frac{k}{m}\mathbf{R} = \mathbf{v}_0-gt\mathbf{j} \] where \(\mathbf{v}_0 = \mathbf{v}(0) = \frac{\mathrm{d}\mathbf{R}}{\mathrm{d}t}(0).\)
    2. Multiply both sides of the equation in the last part by \(\mathrm{e}^{(k/m)t}\) and show that the left-hand side of the resulting equation is the derivative of the product \(\mathrm{e}^{(k/m)t}\mathbf{R}(s).\) Then integrate to find an expression for the position vector \(\mathbf{R}(r).\)
  4. James Stewart Find the curvature of the curve defined parametrically by these equations: \[ x = \int\limits_0^t \sin\biggl(\frac{\pi}{2}\theta^2\biggr)\,\mathrm{d}\theta \qquad y = \int\limits_0^t \cos\biggl(\frac{\pi}{2}\theta^2\biggr)\,\mathrm{d}\theta \]

  5. James Stewart In designing transfer curves to connect sections of straight railroad tracks, it’s important to realize that the acceleration of the train should be continuous so that the reactive force exerted by the train on the track is also continuous. This will be the case if the curvature varies continuously.

    1. A logical candidate for a transfer curve to join existing tracks given by lines \(y=1\) for \(x\leq0\) and \(y = \sqrt{2} - x\) for \(x \gt 1/\sqrt{2}\) might be the function \(f(x) = \sqrt{1-x^2}\) for \(0 \lt x \lt 1/\sqrt{2},\) whose graph is the arc of the circle shown in the figure. It looks reasonable at first glance. Show that the function \[ F(x) = \begin{cases} 1 \quad&\text{ if } x \leq 0 \\ f(x) \quad&\text{ if } 0 \lt x \lt 1/\sqrt{2} \\ \sqrt{2}-x \quad&\text{ if } x \geq 1/\sqrt{2} \end{cases} \] is continuous and has continuous slope, but does not have continuous curvature. Therefore, \(f\) is not an appropriate transfer curve.
    2. Find a fifth-degree polynomial to serve as a transfer curve between the following straight line segments: \(y=0\) for \(x\leq0\) and \(y=x\) for \(x \geq 1.\) Could this be done with a fourth-degree polynomial? Use a graphing calculator or computer to sketch the graph of the “connected” function and check to see that it looks like the one in the figure.
    Figure Pending
  6. For a given a smooth planar curve \(\bm{r}(t)\) parameterized for \(t \in \mathbf{R}\) write down a parameterization of the curve that is the locus of the centers of curvature along each point of \(\bm{r}(t)\). Call this resulting curve \(\operatorname{C}(\bm{r})\). Does there exist a curve \(\bm{r}\) such that \(\operatorname{C}\bigl(\operatorname{C}(\bm{r})\bigr) = \bm{r}?\)