I am trying to parametrize the part of the curve $$ y = 2 -\sin \frac{\pi x}{2} $$ from (0, 2) to (1, 1).
I tried the difficult paramaterization $x=t$ and obtained $$ y=2-\sin \frac{\pi t}{2} $$ but am unable to see how to use $(0,2$ to $(1,1)$ and what to integrate form here.
Thanks
Setting $x(t)=t$ can often work, which would set $y(t)=2-\sin\frac{\pi t}{2}$. The only downside is that perhaps there might be a different parametrization that makes some of the algebra and arithmetic involved a little easier (or even the calculus).
As for going from the points, $(0,2)$ to $(1,1)$., the question would be at what time, $t$ is it at each point. Since $t=x$, you have that $0\leq t \leq 1$.
A followup question would be if the question asks what is the arc-length of the curve, or the area under the curve.
In the case that it asks the area under the curve, you can set it up as $\int\limits_0^1 2-\sin\frac{\pi t}{2} dt$.
If it is asking for arclength then you have $s(t) = \int\limits_a^t \sqrt{x'(\tau)^2 + y'(\tau)^2} d\tau = \int\limits_0^1 \sqrt{1 + \frac{\pi^2}{4}cos^2 \frac{\pi\tau}{2}}d\tau$ and continuing from there.