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.