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.

  • $\begingroup$ You need an equation for $\theta$ as well (the angle between the line that connects the origin to a point on the curve and the x-axis). And that's not the same as $t$! $\endgroup$ – NickD Apr 1 '17 at 1:39

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.

Plot of Curve

  • $\begingroup$ Thank You very much for your answer. I realize what You have done. And I see where I went wrong. This was very nice to realize. Tank You again. Is it only this I have missed? $\endgroup$ – user431265 Apr 1 '17 at 1:46
  • $\begingroup$ To sketch the parametric corve, is it not necessary to know where the tangent lines are are vertical and horizontal? $\endgroup$ – user431265 Apr 1 '17 at 1:51
  • $\begingroup$ Thank You very much. This look so nice, awsome... $\endgroup$ – user431265 Apr 1 '17 at 1:56
  • $\begingroup$ My approach to sketching the curve would be similar to yours, except that I would note that $\theta(t)$ is a nice linear function, so we truly do get a spiral. I have added a graph of what the curve looks like. $\endgroup$ – erfink Apr 1 '17 at 1:56
  • $\begingroup$ Would You not bother about the tangent lines? $\endgroup$ – user431265 Apr 1 '17 at 2:00

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


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

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


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


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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.