The parametric curve *C* is given by; $x(t)={t}^{2}\cos \left( 4\,t \right)$ , $y(t)={t}^{2}\sin \left( 4\,t \right)$

Question

The parametric curve C is given by $$x(t)={t}^{2}\cos \left( 4\,t \right)$$ $$y(t)={t}^{2}\sin \left( 4\,t \right)$$ $$\,0≤t≤π\,.$$

1. To convert the parametric curve C into a polar representation, what would be the correct polar representation and what would be the correct way to reach that conclusion?
2. Find the points where the curve C intersects the x axis and sketch the curve.

My attempt Part 1. $$({r(t)})^2={(x(t))}^2+{(y(t))}^2={({t}^{2}\cos \left( 4\,t \right))}^2+{({t}^{2}\sin \left( 4\,t \right))}^2={t}^4$$ so that ${r(t)}={t}^2$ for $\,0≤t≤π\,.$ In a polar diagram this does not look like a spiral for $\,0≤t≤π\,$, but it does for $\,0≤t≤10π\,$ and the radius seem to increase when going from $t=0\,\,$ to $\,t=10{π}.$ I find the length of the polar curve to be $$L=\int_0^π \sqrt {{t}^2 (4+{t}^2)}\,dt≈14,5.$$

My attempt Part 2.

I find the points where C intersects the $x$ and $y$ axis when $t$ goes from $0$ to $π\,.$

$y(t)={t}^{2}\sin \left( 4\,t \right)=0\,; t=0\,, \, t={\frac {π}{4}}n\,\,$for$\,\,n=0, \,1,\,2,\,3,4.$

$x(t)={t}^{2}\cos \left( 4\,t \right)=0\,; t=0\,, \, t={\frac {π}{8}} +{\frac {π}{4}}n\,\,$for$\,\,n=0, \,1,\,2,\,3.$

Without knowing where the tangent lines are vertical or horizontal I plot the points where the parametric curve C intersects the axis. With a little imagination this surely looks like a spiral. I continue to find the length of the parametruc curve C $$L=\int_0^π 2\sqrt {{t}^2+4 {t}^4}\,dt≈42,8.$$

These are the points where the parametric curve C intersects the x axis $$t_{{0}}=0 →\,\,x(0)=0$$ $$t_{{1}}={\frac {π}{4}}\,\, →\,\,x({\frac {π}{4}})=-{\frac {{π}^{2}}{16}}$$ $$t_{{2}}={\frac {π}{2}}\,\, →\,\,x({\frac {π}{2}})={\frac {{π}^{2}}{4}}$$ $$t_{{3}}={\frac {3π}{4}}\,\, →\,\,x({\frac {3π}{4}})=-{\frac {{9π}^{2}}{16}}$$ $$t_{{4}}={π}\,\, →\,\,x({π})={{π}^{2}}$$ From this information only, is it possible to draw enough conclusions about the parametric curve to be able to sketch it? I find it difficult. What would be the correct way to solve this problem? All help appreciated.

Answer 1

As you have noted, $r(t)=t^2$. However, the part you're missing is that there is a $\cos (4t)$ and a $\sin (4t)$. As such, consider the pair of parametric polar equations: \begin{align*} r(t) = t^2 \\ \theta(t) = 4t \end{align*} which generate the same curve as $(x(t), y(t))$ on $0 \leq t \leq \pi$. Eliminating the parameter gives $$r(\theta) = (\theta/4)^2, \quad 0 \leq \theta \leq 4 \pi.$$

The locations of points intersecting the $x$-axis should be easy to find in this representation ($r=0$ or $\theta =n \pi$ for some $n \in \mathbb{Z}$).

Arc Length

This also gives the arc length calculation:

\begin{align*} L &= \int_0^{4 \pi} \sqrt{[r(\theta)]^2 + [r'(\theta)]^2} ~\mathrm{d} \theta \\ &= \int_0^{4\pi} \sqrt{ \frac{\theta^4}{256} + \frac{\theta^2}{64}} ~\mathrm{d} \theta \\ &= \frac{1}{6} \left(\left(1+4 \pi ^2\right)^{3/2}-1\right) \end{align*} Note that it is possible to evaluate this integral in closed form using a trig sub.

Answer 2

I present an alternative solution in the complex plane. Clearly,

$$z=t^2e^{i4t}$$

Let $\theta=4t$ so that

$$z=\frac{1}{16}\theta^2e^{i\theta}, \ \ \ 0\le\theta\le 4\pi$$

This is in the general family of Archimedean spirals, although it has no particular name. Since the exponent on $\theta$ is greater than 1, the rings increase in spacing as the spiral evolves.

Now, the arc length in the complex plane is given by

$$s=\int |\dot z(u)| du$$

Thus

$$ \dot z=\frac{1}{16}[i\theta^2+2\theta]e^{i\theta}\\ |\dot z|=\frac{1}{16}\sqrt{\theta^4+4\theta^2} $$

Then,

$$\begin{align} s & =\frac{1}{16}\int_0^{4\pi}\sqrt{\theta^4+4\theta^2}\ d\theta\\ & =\frac{1}{16}\int_0^{4\pi}\theta\sqrt{\theta^2+4}\ d\theta\\ & =\frac{1}{32}\int_0^{16\pi^2}\sqrt{x+4}\ dx,\ \ \ x=\theta^2\\ & =\frac{1}{32}\frac{(x+4)^{3/2}}{3/2}\large|_0^{16\pi^2}\\ & = \frac{1}{6} \left[\left(4 \pi ^2+1\right)^{3/2}-1\right] \end{align} $$

This is in agreement with the previous result. It just seems to me that there's a lot less fuss working in the complex plane.