91Ó°ÊÓ

In calculus, understanding the concept of a velocity vector is crucial, particularly when dealing with motion in a plane. The velocity vector \(\mathbf{v}\) represents the rate of change of the position vector \(\mathbf{r}\) with respect to time. To find it, one must differentiate each component of the position vector separately with respect to time \(t\). For example, if a position vector is given by \(\mathbf{r} = 12 \sqrt{t} \mathbf{i} + t^{3/2} \mathbf{j}\), the corresponding velocity vector is calculated as \(\mathbf{v} = \frac{d}{dt}(12 \sqrt{t}) \mathbf{i} + \frac{d}{dt}(t^{3/2}) \mathbf{j}\).
To break it down further:

• The derivative of \(12 \sqrt{t}\) is \(\frac{6}{\sqrt{t}}\), calculated using the power rule by recognizing \(\sqrt{t} = t^{1/2}\).
• The derivative of \(t^{3/2}\) is \(\frac{3}{2}t^{1/2}\) derived similarly.

Thus, the velocity vector becomes \(\mathbf{v} = \frac{6}{\sqrt{t}} \mathbf{i} + \frac{3}{2}t^{1/2}\mathbf{j}\). This vector describes how quickly and in what direction the position of the particle is changing at any given time \(t\).

The minimum speed of a particle is of great interest, especially in optimization problems or when analyzing the particle's movement. To find this, one must look at the magnitude of the velocity vector. The speed is the magnitude, expressed as \|\mathbf{v}| = \sqrt{\left(\frac{6}{\sqrt{t}}\right)^2 + \left(\frac{3}{2}t^{1/2}\right)^2}\.
The goal is to find when this magnitude is at its minimum. To simplify calculations, the derivative of the squared magnitude \(\|\mathbf{v}|^2\) is used instead. For our example, this would be:

• Squared magnitude \(= \frac{36}{t} + \frac{9}{4}t\).
• Take the derivative with respect to \(t\), resulting in \(-\frac{36}{t^2} + \frac{9}{4}\).

Set this derivative to zero to find the critical points: solving \(-\frac{36}{t^2} + \frac{9}{4} = 0\) yields \(t = 4\).
Substitute \(t = 4\) into the original magnitude function to find the minimum speed: \|\mathbf{v}(4)| = 3\sqrt{2}\. This method effectively identifies when and how the particle reaches its slowest speed.

A position vector describes a particle's location in space at any given time. It is given in terms of its components along \(\mathbf{i}\) and \(\mathbf{j}\) axes, which represent the horizontal and vertical directions, respectively. In our scenario, the position vector is \(\mathbf{r} = 12 \sqrt{t} \mathbf{i} + t^{3/2} \mathbf{j}\).
At the minimum speed, which we calculated occurs at \(t = 4\), we substitute back into the position vector to get:

• \(\mathbf{r}(4) = 12 \sqrt{4} \mathbf{i} + 4^{3/2} \mathbf{j}\).
• This simplifies to \(24 \mathbf{i} + 8 \mathbf{j}\).

Therefore, when the particle is moving slowest, it is located at the point \((24, 8)\) in the plane. This means the particle is 24 units along the x-axis and 8 units along the y-axis from the origin. Understanding how these vectors describe the particle's motion helps visualize the path taken and the position maintained over time.