91Ó°ÊÓ

In vector calculus, a **position vector** is a fundamental concept used to describe the location of a point in space relative to an origin. For the given problem, the position vector is defined as \( \mathbf{r}(t) = t \mathbf{i} + (t^2 + 1) \mathbf{j} \). This expression tells us how the particle moves along the parabola described by \( y = x^2 + 1 \) in the xy-plane.

• The \( t \mathbf{i} \) component represents movement along the x-axis. As time, \( t \), changes, so does the x-coordinate of the particle.
• The \((t^2 + 1) \mathbf{j}\) component represents movement along the y-axis. Here, the position depends on the square of \( t \) plus one, showing how the particle's height changes over time.

Understanding the position vector helps in visualizing the particle's path as it moves over time. This vector ultimately serves as the basis for finding other related vectors, like velocity and acceleration.

A **velocity vector** gives the rate of change of the position vector with respect to time. It shows both the speed and direction of the particle's motion. To find this, we take the derivative of the position vector \( \mathbf{r}(t) = t \mathbf{i} + (t^2 + 1) \mathbf{j} \) with respect to time \( t \).

Deriving the Velocity Vector

We compute:

• \( \frac{d}{dt}(t \mathbf{i}) = \mathbf{i} \) indicates a constant unit movement along the x-axis.
• \( \frac{d}{dt}((t^2 + 1) \mathbf{j}) = 2t \mathbf{j} \) shows a change in y-component, which varies with \( t \).

Thus, the velocity vector becomes \( \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} \).
This vector provides crucial insights into how quickly and in which direction the particle is moving at any given time.

The **acceleration vector** indicates the rate of change of the velocity vector over time. It captures how the particle's speed and direction are changing. To find the acceleration vector, differentiate the velocity vector \( \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} \) with respect to \( t \).

Calculating the Acceleration Vector

Differentiating yields:

• \( \frac{d}{dt}(\mathbf{i}) = 0 \) as the x-component doesn't change with time.
• \( \frac{d}{dt}(2t \mathbf{j}) = 2 \mathbf{j} \) is the constant rate of change in the y-direction.

This results in an acceleration vector \( \mathbf{a}(t) = 2 \mathbf{j} \), which is constant. This tells us that the particle experiences a constant upward acceleration at every point along its path.

At the heart of vector calculus is the concept of the **derivative**, which is essential for understanding how quantities change. When we discuss derivatives in this context, we're usually referring to the derivative of vector functions, such as position vectors.

Importance of Derivatives

• The derivative helps us find the velocity vector by differentiating the position vector. It tells us how fast the position is changing over time.
• By further differentiating the velocity vector, we derive the acceleration vector, indicating how the velocity is changing at each moment.

The derivative is a critical tool in calculus, allowing us to analyze and predict the motion of particles across various contexts. Understanding how to take and apply derivatives is key to solving real-world problems involving motion.