Thank you for the image. Let's analyze the problem and solve it step-by-step.

Problem Description:

The pair of parametric equations given is: $x = \sqrt{t}, \quad y = 1 - t.$

Tasks:

1. (a) Sketch the curve and indicate the direction of the curve as $t$ increases.
2. (b) Eliminate the parameter $t$ to find the rectangular-coordinate equation of the curve, where $x \geq 0$.

Solution:

Part (a): Sketch the Curve and Indicate Direction

1. Observations:

   • $x = \sqrt{t}$: Since the square root is defined only for $t \geq 0$, this sets the domain of $t$ as $t \geq 0$.
   • $y = 1 - t$: This linearly decreases as $t$ increases.

2. Plot Points: Evaluate $x$ and $y$ for different $t$ values:

   • When $t = 0$: $x = \sqrt{0} = 0$, $y = 1 - 0 = 1$. So, the point is $(0, 1)$.
   • When $t = 1$: $x = \sqrt{1} = 1$, $y = 1 - 1 = 0$. So, the point is $(1, 0)$.
   • When $t = 4$: $x = \sqrt{4} = 2$, $y = 1 - 4 = -3$. So, the point is $(2, -3)$.

3. Curve Direction:

   • As $t$ increases, $x = \sqrt{t}$ increases, and $y = 1 - t$ decreases.
   • Arrows should indicate motion in the positive $x$-direction (rightward), while $y$ decreases (downward).

The graph in the first box (bottom-left graph) matches the correct direction.

Part (b): Eliminate the Parameter $t$

1. From $x = \sqrt{t}$, solve for $t$: t = x^2, \quad \text{(since x \geq 0)}.

2. Substitute $t = x^2$ into $y = 1 - t$: $y = 1 - x^2.$

Rectangular Equation: y = 1 - x^2, \quad \text{where x \geq 0}.

Final Answer:

• (a) The correct graph is the bottom-left graph.
• (b) The rectangular-coordinate equation is: $y = 1 - x^2, \quad x \geq 0.$

Would you like me to elaborate on any of the steps? Let me know!

Related Questions:

1. What is the domain and range of the curve $y = 1 - x^2$ under the restriction $x \geq 0$?
2. How do you find the vertex of the parabola $y = 1 - x^2$?
3. Can this parametric equation represent a physical trajectory? If so, what does it represent?
4. How would the parametric graph change if $x = -\sqrt{t}$ instead of $x = \sqrt{t}$?
5. How do parametric equations differ from standard Cartesian equations?

Tip:

When eliminating the parameter in parametric equations, always ensure to preserve any constraints (like $x \geq 0$) given in the problem.