$(x,y) = (\cos t, \sin t)$ 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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$y=\tan x$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?
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A point on a spiral gets (continuously) further away from its centre with time. How would you transform the circle equation to achieve that?
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The above means that the magnitude of the position vector must increase with time for the spiral (whereas it’s fixed for the circle).
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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?
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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.
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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$v(t) = \frac{1}{2\pi}\sqrt{1+t^2}.$