Complete the following for the parametrization xequals2 t and yequals3 t minus 5​,  0less than or equalstless than or equals1.
​(a)
Graph the curve. What are the initial and terminal​ points, if​ any? Indicate the direction in which the curve is traced.
​(b)
Find a Cartesian equation for a curve that contains the parametrized curve. What portion of the graph of the Cartesian equation is traced by the parametrized​ curve?


Let's address each part of the problem one by one.

### Given Parametrization:
- \( x = 2t \)
- \( y = 3t - 5 \)
- \( 0 \leq t \leq 1 \)

### Part (a): Graph the curve

#### **Initial and Terminal Points**:
1. **Initial Point**: This occurs at \( t = 0 \).
   - When \( t = 0 \), we get:
     \[
     x = 2(0) = 0, \quad y = 3(0) - 5 = -5
     \]
     So, the initial point is \( (0, -5) \).

2. **Terminal Point**: This occurs at \( t = 1 \).
   - When \( t = 1 \), we get:
     \[
     x = 2(1) = 2, \quad y = 3(1) - 5 = -2
     \]
     So, the terminal point is \( (2, -2) \).

#### **Direction of the Curve**:
The direction of the curve is traced from the initial point \( (0, -5) \) to the terminal point \( (2, -2) \) as \( t \) increases from 0 to 1.

The curve is a straight line because both \( x \) and \( y \) are linear in \( t \).

### Part (b): Find a Cartesian Equation

To eliminate \( t \) and find the Cartesian equation, solve for \( t \) in terms of \( x \) from the equation \( x = 2t \):
\[
t = \frac{x}{2}
\]
Now substitute this expression for \( t \) into the equation for \( y \):
\[
y = 3t - 5 = 3\left(\frac{x}{2}\right) - 5
\]
Simplifying:
\[
y = \frac{3x}{2} - 5
\]
Thus, the Cartesian equation is:
\[
y = \frac{3x}{2} - 5
\]

#### **Portion of the Graph Traced**:
The parameter \( t \) is restricted to \( 0 \leq t \leq 1 \), which corresponds to the \( x \)-values between 0 and 2 (since \( x = 2t \)).

- For \( t = 0 \), we get \( x = 0 \).
- For \( t = 1 \), we get \( x = 2 \).

Therefore, the portion of the graph of the line \( y = \frac{3x}{2} - 5 \) that is traced by the parametrized curve is from \( (0, -5) \) to \( (2, -2) \).

### Conclusion:
- The initial point is \( (0, -5) \).
- The terminal point is \( (2, -2) \).
- The direction of the curve is from the initial to the terminal point.
- The Cartesian equation of the curve is \( y = \frac{3x}{2} - 5 \).
- The portion of the graph traced is the segment between \( x = 0 \) and \( x = 2 \).

Would you like a graph of this curve or further explanation on any part of the solution?

### 5 Related Questions:
1. How do you parametrize curves in 3D space?
2. What are the differences between parametric and Cartesian equations?
3. How can you determine the length of a parametric curve?
4. How do you compute the slope of a curve from its parametric equations?
5. How can you reparametrize a curve to trace it in reverse?

### Tip:
When converting from parametric to Cartesian form, always check if the relationship is linear or non-linear to predict the type of curve (line, parabola, etc.).

Complete the following for the parametrization x = 2t and y = 3t - 5, 0 ≤ t ≤ 1.
(a) Graph the curve. What are the initial and terminal points, if any? Indicate the direction in which the curve is traced.
(b) Find a Cartesian equation for a curve that contains the parametrized curve. What portion of the graph of the Cartesian equation is traced by the parametrized curve?

Explore how to graph the parametric equations x = 2t and y = 3t - 5 within the bounds 0 ≤ t ≤ 1. Learn to identify the initial and terminal points, the direction of the curve, and convert the parametric equations into a Cartesian form. Suitable for high school students in Grades 10-12.

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