Find an arc length parametrization of (cos(t) + (t)sin(t), sin(t)-(t)cos(t)) with reference point (i.e., starting point) (1,0), in the positive direction.

Solution:

We can write:

\begin{cases}x = cos(t)+tsin(t)\\y=sin(t)-tcos(t)\end{cases}

So we have:

\frac{\text{d}x}{\text{d}t} = -sin(t)+sin(t)+tcost(t) = tcos(t)

\frac{\text{d}y}{\text{d}t} = cos(t)-cos(t)+tsin(t) = tsin(t)

Let s(t) represent arclength, let t=a, and the reference point is (1,0) which means t = 0;

Use the arc length formula and substitute the values in to get:

s(a)=\int_{0}^{a} \sqrt{t^{2}cos^{2}(t)+t^{2}sin^{2}(t)} \text{d}t

=\int_{0}^{a}(t)\text{d}t

=\frac{a^{2}}{2}

In general, we have:

s(t)=\frac{t^{2}}{2}

t(s)=\sqrt{2s}

Thus, we can write:

\begin{cases}x = sin\sqrt{2s}+\sqrt{2s}cos\sqrt{2s}\\y=cos \sqrt{2s}-\sqrt{2s}sin\sqrt{2s}\end{cases}

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