To identify which diagram correctly represents the parametric curve given by:

$x = -\sin t, \quad y = \cos t, \quad 0 \leq t \leq \pi,$

we can analyze the behavior of $x$ and $y$ over the interval $0 \leq t \leq \pi$ to understand the path of the curve.

1. When $t = 0$:

   • $x = -\sin(0) = 0$
   • $y = \cos(0) = 1$
   • The curve starts at the point $(0, 1)$.

2. When $t = \pi$:

   • $x = -\sin(\pi) = 0$
   • $y = \cos(\pi) = -1$
   • The curve ends at the point $(0, -1)$.

3. Intermediate values of $t$:

   • For $t$ between $0$ and $\pi$, $\sin(t)$ varies from $0$ to $1$ (positive values) and then back to $0$. Therefore, $-\sin(t)$ will range from $0$ to $-1$ and back to $0$.
   • $\cos(t)$ varies from $1$ to $-1$.

From this information, we can deduce that the curve traces a semicircle, moving from $(0, 1)$ to $(0, -1)$, with $x$ values remaining negative or zero. This matches a left-facing semicircle centered at the origin.

Diagram Choice:

• The correct option is (B), which shows a left-facing semicircle from $(0, 1)$ to $(0, -1)$.

Let me know if you'd like more details on any step.

Additional Questions:

1. How can you determine the direction of a parametric curve?
2. Why does the value of $t$ only go up to $\pi$ here?
3. What would happen if the range of $t$ extended beyond $\pi$?
4. Can parametric curves represent other shapes besides circles?
5. How would the graph change if $x = \sin t$ instead of $x = -\sin t$?

Tip: When analyzing parametric curves, checking endpoint values and the general shape by looking at intermediate values can help quickly identify the curve’s form.