91Ó°ÊÓ

In vector calculus, a position vector describes the location of a point in space as a function of time. It is often denoted as \( \mathbf{r}(t) \). For a particle moving through space, this vector determines where the particle is at any given moment. Consider your position vector as a set of coordinates that change over time, tracing out a path in three-dimensional space.
In the exercise example, the position vector \( \mathbf{r}(t) = (2 \ln(t+1)) \mathbf{i} + t^2 \mathbf{j} + \frac{t^2}{2} \mathbf{k} \) shows the particle's coordinates in terms of \( t \), indicating its trajectory as time progresses. This vector provides the fundamental starting point for calculating velocity and acceleration.

The derivative of a vector-valued function is crucial for understanding motion. When we differentiate a position vector \( \mathbf{r}(t) \), we get the velocity vector \( \mathbf{v}(t) \). Differentiating again gives the acceleration vector \( \mathbf{a}(t) \). This is because differentiation in vector calculus is applied component-wise, meaning each element of the vector is treated separately.
In our example, the velocity function was obtained by finding the derivative of each component of \( \mathbf{r}(t) \):

• The derivative of \( 2 \ln(t+1) \) gives \( \frac{2}{t+1} \),
• while \( t^2 \) differentiates to \( 2t \),
• and \( \frac{t^2}{2} \) to \( t \).

This makes the velocity vector \( \mathbf{v}(t) = \left( \frac{2}{t + 1} \right) \mathbf{i} + (2t) \mathbf{j} + (t) \mathbf{k} \).
Similarly, differentiating this velocity vector gives us the acceleration vector. This step-by-step process is key in the study of motion.

The magnitude of a vector represents its length and is crucial when calculating speed, which is the magnitude of the velocity vector. To find the magnitude, you use the Pythagorean theorem in three-dimensional space, which involves squaring each component of the vector, adding them together, and then taking the square root of the sum.
For instance, the speed of the particle at \( t = 1 \) is found by calculating the magnitude of \( \mathbf{v}(1) = 1 \mathbf{i} + 2 \mathbf{j} + 1 \mathbf{k} \). The magnitude is:\[\text{Speed} = \| \mathbf{v}(1) \| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}.\]This value provides an understanding of how fast the particle is moving at that specific time.

Unit vectors are important as they have a magnitude of 1 and indicate the direction of a vector. To find a unit vector in the direction of another vector, you divide each component of the vector by the vector's magnitude. This process effectively scales the vector to have a length of 1 while preserving its direction.
In the exercise, to find the direction of motion at \( t = 1 \), we compute the unit vector of \( \mathbf{v}(1) = 1 \mathbf{i} + 2 \mathbf{j} + 1 \mathbf{k} \) by dividing it by its magnitude (\( \sqrt{6} \)). This results in:\[\text{Direction} = \frac{1}{\sqrt{6}} \mathbf{i} + \frac{2}{\sqrt{6}} \mathbf{j} + \frac{1}{\sqrt{6}} \mathbf{k}.\]This unit vector purely reflects the direction at that instance, independent of speed.

The direction of motion is described by the unit vector derived from the velocity vector. It provides insight into the path the particle will follow, guiding our understanding of where the particle is heading without focusing on its speed.
To express the velocity as a product of speed and direction, you multiply the unit direction vector by the magnitude of the velocity (speed). For \( t = 1 \), this gives us:\[\mathbf{v}(1) = \sqrt{6} \left( \frac{1}{\sqrt{6}} \mathbf{i} + \frac{2}{\sqrt{6}} \mathbf{j} + \frac{1}{\sqrt{6}} \mathbf{k} \right).\]This confirms that the velocity is indeed a combination of the speed and the direction, offering a complete picture of the particle's motion at that moment in time.