I have: $\mathbf{r}(t) = (t^2 - 1)\mathbf{j} + 2t\mathbf{k}$, $-1\leq t \leq 1$ and $\mathbf{r}(t) = (2\cos t)\mathbf{i} + 2(\sin t)\mathbf{k}$, $0 \leq t \leq \pi$.
I've already graphed them but I don't understand why they are semi circles on the graph. Why aren't they just straight line vectors connecting the two points?
Thank you in advance!
The second one has the property that its $x$ component squared plus its $y$ component squared is constant, $$x^2+y^2=4\cos^2t+4\sin^2t=2^2$$ and thus is a circle (of radius $r=2$). What makes it a semicircle is the fact that $t$ ranges only over $0\le t\le \pi$ rather than, say, $0\le t\le 2\pi$.
The first one is not a circle/semicircle since (as you can check), $$x^2+y^2=t^4+2t^2+1\not=\text{ constant}.$$
Edit: Some pictures to help see things... Here's the sequence of vectors as well as the paramteric curve they trace out for the first function
and here's the same type of picture for the second function