The Computer Laboratory

Practice Paper 2 Question 12

A point traces a unit circle if its coordinates satisfy \((x,y) = (\cos t, \sin t)\) as time \(t\) varies from \(0\) to \(2\pi.\) Give an equation for a point that traces a spiral centred at \((0,0)\) and that crosses the positive \(x\)-axis at \(x=1,2,3,\ldots\) at times \(t = 2\pi,4\pi,6\pi,\ldots\) and find its speed \(v(t)\) at time \(t.\)

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Warm-up Questions

  1. Give \(y=\tan x\) as a parametric equation \((x,y)\) where \(x=\arcsin t\) and \(y\) contains no trigonometric functions.
  2. Sketch the parametric curve \((x,y)=(4\cos t,2\sin t).\) Hint: Try forming \(\cos^2t+\sin^2t.\)
  3. A boat is heading North East traveling North at \(12\) kilometres per hour and East \(35\) kilometres per hour. What is the overall speed of the boat?

Hints

  • Hint 1
    A point on a spiral gets (continuously) further away from its centre with time. How would you transform the circle equation to achieve that?
  • Hint 2
    The above means that the magnitude of the position vector must increase with time for the spiral (whereas it's fixed for the circle).
  • Hint 3
    The \(x\) and \(t\) values in the question are proportional (with ratio \(2\pi\)). What does that tell you about the equation for the \(x\) component?
  • Hint 4
    The above means that time must multiply \(\cos t.\) What about the \(y\) component?
  • Hint 5
    How about resolving the speed vector into components?
  • Hint 6
    This means extracting the \(x\) and \(y\) components of the speed vector. How does one do that knowing the \(x\) and \(y\) components of the position vector?
  • Hint 7
    The question asks for the speed \(v(t),\) i.e. not a vector.

Solution

The question does not specify any conditions for \(y(t)\) while for \(x(t)\) it only specifies the crossing points with the \(x\)-axis. There are multiple solutions satisfying these conditions: any spiral is acceptable as long as the crossing points with the \(x\)-axis are satisfied. Since the given crossing points are linearly spaced, we can modify the circle equation by multiplying both \(\cos t\) and \(\sin t\) by a linear function; such a spiral (called the Archimedean spiral) is: \[ \textstyle (x,y) = \left(\frac{1}{2\pi}\,t\cos t, \frac{1}{2\pi}\,t\sin t\right). \] To find its total speed, extract the \(x\) and \(y\) components of the speed vector by differentiating the position components, and then \(v(t)\) is the vector's absolute value: \[ \begin{align} \dot x &= \frac{1}{2\pi}(\cos t - t \sin t) \\[2pt] \dot y &= \frac{1}{2\pi}(\sin t + t \cos t) \end{align} \] and \(v(t) = \sqrt{\dot x^2 + \dot y^2}\) which after some easy algebra becomes \(v(t) = \frac{1}{2\pi}\sqrt{1+t^2}.\)

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