(a) Sketch the plane curve with the given vector equation. r(t) = <3t-1, t^2 + 1> (b) Find **r '**(*t*).
**r '**(*t*) = <  , >

Let's go through the problem step by step.

### (a) Sketch the plane curve given by the vector equation:  
\[
\mathbf{r}(t) = \langle 3t - 1, t^2 + 1 \rangle
\]
This describes a parametric curve in the plane. 

1. **First component (x-coordinate):**  
   \[
   x(t) = 3t - 1
   \]
   This is a linear equation, meaning the x-coordinate moves linearly as \( t \) changes.

2. **Second component (y-coordinate):**  
   \[
   y(t) = t^2 + 1
   \]
   This is a quadratic equation, so the y-coordinate moves in a parabolic manner.

To sketch the curve, you can plot points for different values of \( t \), such as \( t = -2, -1, 0, 1, 2 \), and then connect the points smoothly. The x-coordinate will increase linearly, while the y-coordinate will follow the shape of a parabola opening upwards, but shifted by +1.

### (b) Find the derivative \( \mathbf{r}'(t) \):

To find \( \mathbf{r}'(t) \), differentiate each component of \( \mathbf{r}(t) \) with respect to \( t \).

1. **First component:**
   \[
   \frac{d}{dt}[3t - 1] = 3
   \]

2. **Second component:**
   \[
   \frac{d}{dt}[t^2 + 1] = 2t
   \]

Thus, the derivative \( \mathbf{r}'(t) \) is:
\[
\mathbf{r}'(t) = \langle 3, 2t \rangle
\]

### Summary:

- The sketch will show a curve with a linearly increasing x-coordinate and a parabolic y-coordinate.
- The derivative of the vector function is \( \mathbf{r}'(t) = \langle 3, 2t \rangle \).

Do you have any questions or want further details on any step?

### Related Questions:
1. What is the significance of the derivative in terms of the tangent to the curve?
2. How would the sketch change if the coefficients of \( t \) were different in the parametric equation?
3. Can you find the second derivative \( \mathbf{r}''(t) \)?
4. What is the speed of the particle at any time \( t \)?
5. What is the direction of motion at \( t = 0 \)?

**Tip:** The first derivative of a parametric curve gives the velocity vector, which points in the direction of the tangent at any given time.

Sketch the plane curve with the given vector equation r(t) = <3t-1, t^2 + 1> and find r'(t).

This page explains how to sketch the parametric curve given by r(t) = <3t-1, t^2+1> and provides a step-by-step guide to finding its derivative r'(t) = <3, 2t>. Learn key concepts of parametric equations and derivatives, suitable for students in Grades 11-12 and early college.

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