The problem is, "to determine any differences between the curves of the parametric equations. Are the graphs the same? Are the orientations the same? Are the curves smooth? Explain."

(a) $x=t;\quad y=2t+1$

(b) $x=\cos\theta;\quad y=2\cos\theta +1$

(c) $x=e^{-t};\quad y=2e^{-t}+1$

(d) $x=e^t; \quad y=2e^t+1$

The only one I am having difficulty with is (b). The graph provided in the answer key is:

enter image description here

Why are the arrows pointing in both directions? Is there no orientation? Also, they describe the curve as not being smooth. Why is this?


I have another question; the criteria given in the problem above is the same for this problem.

The parametric equations are, (a) $x=\cos\theta$, $y=2\sin^2\theta$

(b) $x=\cos(-\theta)$, $y=2\sin^2(-\theta)$

where $0<\theta<\pi$.

With the information, $-1<\cos\theta<1 \implies -1<x<1$ and $0<\sin\theta<0\implies0<2\sin^2\theta<0\implies0<y<0$

In the answer key, however, they have $-1\le x\le1$ How did they get the "equal to" component in there?


Hint for 1st part: $\cos\theta$ is a periodic function that cycles between $\pm 1$. A curve may fail to be smooth if its velocity vector is ever $0$--that is, if $x'$ and $y'$ are both $0$ at some point--as happens at the endpoints, here.

2nd part answer: As a hint for comparing the two parts, use the fact that sine is an odd function and cosine is an even function.

Assuming that the $0<\theta<\pi$ is correct, then you're correct that $-1<x<1$.

You are incorrect, though, in your claim that $0<\sin\theta<0$--for that would imply that $0$ is less than itself, which makes no sense. If you happen to know that a function is monotone (increasing or decreasing) on an entire interval--as is the case with $\cos\theta$ on the interval $0<\theta<\pi$--then you can simply check the values (or limiting behavior) at the endpoints of the interval to find the bounds of the function on that interval--as you did with $-1<\cos\theta<1$. Unfortunately, this doesn't work for all functions (as it can lead to problems like the aforementioned nonsense). On the interval $0<\theta<\pi$, we can say that $0<\sin\theta$, however, $\sin\theta$ achieves a maximum value of $1$ at $\theta=\frac\pi2$. Hence, $0<\sin^2\theta\le1$, and so $0<y\le2$.

  • $\begingroup$ I tried searching for cyclic functions in google, but I couldn't find anything that you seem to be hinting at? Also, what do you mean by velocity vectors? $\endgroup$ – Mack Nov 25 '12 at 12:43
  • $\begingroup$ My apologies for misspeaking. I've corrected the first, and clarified the second. $\endgroup$ – Cameron Buie Nov 25 '12 at 13:00
  • $\begingroup$ No, it's quite all right. I do appreciate your help. One more thing, do you know of any online article(s) I could on this topic of smoothness of a curve? I was never taught that a curve wasn't smooth when it had a zero first derivative at a particular point. $\endgroup$ – Mack Nov 25 '12 at 13:10
  • $\begingroup$ You can check out this link. It's entirely possible that you're using a different definition of "smooth" in your class, though. If so, it's probably a good idea to specify the definition in your post, so we can answer appropriately. $\endgroup$ – Cameron Buie Nov 25 '12 at 13:38
  • $\begingroup$ My post has been edited, containing a similar problem I am working on, and I was wondering if you could take a look. $\endgroup$ – Mack Nov 25 '12 at 14:48

Though,each of the parametric equation reduce to $y=2x+1,$ the ranges of the values of $x$ for the real values of $t,\theta $ will differ.

In $(a),x$ cam assume any real values.

But in $(b),-1\le x\le 1$

In $(c),(d),x>0$ for the real values of $t$.

Using this, each curve has orientation zero.


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