91Ó°ÊÓ

The position vector, denoted as \( \mathbf{r}(t) \), is used to describe the location of a point \( P \) in space at any given time \( t \). This vector can be expressed in terms of the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), which correspond to the x, y, and z axes respectively. In the given exercise, the position vector is \( \mathbf{r}(t) = 4 \sin t \mathbf{i} + 2t \mathbf{j} + 9 \cos t \mathbf{k} \). Each component of the vector represents the position of the point along the respective axis:

• \( 4 \sin t \mathbf{i} \) means the x-coordinate varies as the sine function changes.
• \( 2t \mathbf{j} \) shows a linear change in the y-coordinate as time progresses.
• \( 9 \cos t \mathbf{k} \) indicates the z-coordinate follows a cosine function.

To find the position at \( t=\frac{3\pi}{4} \), simply substitute the value into the vector equation. This results in \( \mathbf{r}\left(\frac{3\pi}{4}\right) = 2\sqrt{2} \mathbf{i} + \frac{3\pi}{2} \mathbf{j} - \frac{9\sqrt{2}}{2} \mathbf{k} \). The calculation shows the precise location in three-dimensional space at this specific time.

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time. This vector represents the rate of change of position, or how fast and in what direction the point \( P \) is moving.

The procedure involves differentiating each component of the position vector:

• Differentiate \( 4 \sin t \mathbf{i} \) to get \( 4 \cos t \mathbf{i} \).
• Differentiate \( 2t \mathbf{j} \) to get \( 2 \mathbf{j} \).
• Differentiate \( 9 \cos t \mathbf{k} \) to get \( -9 \sin t \mathbf{k} \).

Therefore, the velocity vector is \( \mathbf{v}(t) = 4 \cos t \mathbf{i} + 2 \mathbf{j} - 9 \sin t \mathbf{k} \).

Evaluating this at \( t=\frac{3\pi}{4} \) gives \( \mathbf{v}\left(\frac{3\pi}{4}\right) = -2\sqrt{2} \mathbf{i} + 2 \mathbf{j} - \frac{9\sqrt{2}}{2} \mathbf{k} \). This determines the velocity, indicating both the speed and the direction at that moment.

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector with respect to time. It indicates how the velocity of the point \( P \) is changing - essentially, it describes the point's acceleration along the path.

Similar to what we did with the velocity vector, we differentiate each component of the velocity vector:

• Differentiate \( 4 \cos t \mathbf{i} \) to get \( -4 \sin t \mathbf{i} \).
• Differentiate \( 2 \mathbf{j} \) to get \( 0 \mathbf{j} \), since it's constant.
• Differentiate \( -9 \sin t \mathbf{k} \) to get \( -9 \cos t \mathbf{k} \).

Thus, the acceleration vector at time \( t \) is \( \mathbf{a}(t) = -4 \sin t \mathbf{i} + 0 \mathbf{j} - 9 \cos t \mathbf{k} \).

Evaluating it at \( t=\frac{3\pi}{4} \) provides \( \mathbf{a}\left(\frac{3\pi}{4}\right) = -2\sqrt{2} \mathbf{i} + \frac{9\sqrt{2}}{2} \mathbf{k} \). This result reveals the acceleration's direction and magnitude at that point in time.

Trigonometric functions like sine and cosine play a crucial role in describing the position, velocity, and acceleration vectors in problems involving periodic motion in space. They help define how each component of a vector changes over time.

These functions have specific values at given angles, which are essential for calculating precise vector components in exercises:

• The sine of \( \frac{3\pi}{4} \) is \( \frac{\sqrt{2}}{2} \), a value commonly encountered in trigonometry.
• The cosine of \( \frac{3\pi}{4} \) is \( -\frac{\sqrt{2}}{2} \), indicating a negative direction.

The periodic nature of sine and cosine functions is instrumental in problems like this, where repetitive, circular, or oscillatory paths are being analyzed.

Their derivatives also give insight into motion dynamics, with sine turning into cosine upon differentiation and vice versa, with a sign change for their derivative functions.