# 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 \(t.\)

## Related topics

## Warm-up Questions

- Give \(y=\tan x\) as a parametric equation \((x,y)\) where \(x=\arcsin t\) and \(y\) contains no trigonometric functions.
- Sketch the parametric curve \((x,y)=(4\cos t,2\sin t).\)
*Hint:*Try forming \(\cos^2t+\sin^2t.\) - 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 1A point on a spiral gets (continuously) further away from its centre with time. How would you transform the circle equation to achieve that?
- Hint 2The above means that the magnitude of the position vector must increase with time for the spiral (whereas it's fixed for the circle).
- Hint 3The \(x\) 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 4The above means that time must multiply \(\cos t.\) What about the \(y\) component?
- Hint 5How about resolving the speed vector into components?
- Hint 6This 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 7The 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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