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 $t.$
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?

A point on a spiral gets (continuously) further away from its centre with time. How would you transform the circle equation to achieve that?

The above means that the magnitude of the position vector must increase with time for the spiral (whereas it’s fixed for the circle).

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?

The above means that time must multiply $\cos t.$ What about the $y$ component?

How about resolving the speed vector into components?

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?

The question asks for the speed $v(t),$ i.e. not a vector.

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{aligned} \dot x &= \frac{1}{2\pi}(\cos t - t \sin t) \\[2pt] \dot y &= \frac{1}{2\pi}(\sin t + t \cos t) \end{aligned} and $v(t) = \sqrt{\dot x^2 + \dot y^2}$ which after some easy algebra becomes $v(t) = \frac{1}{2\pi}\sqrt{1+t^2}.$