Parametric Equation Problem 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:

Why are the arrows pointing in both directions? Is there no orientation? Also, they describe the curve as not being smooth. Why is this?
EDIT:
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 $0<y\le2$.
