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# 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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