91Ó°ÊÓ

The cross product, a fundamental operation in vector algebra, is a way to multiply two vectors in three-dimensional space. Imagine it as a tool that helps you find a new vector that is perpendicular to both of the original vectors. This is incredibly useful in many fields, including physics and engineering.

To perform a cross product between two vectors, you can use the determinant of a 3x3 matrix. The top row of this matrix consists of unit vectors (i, j, k), while the second and third rows are the components of the vectors being multiplied. For example, the cross product \(\mathbf{v} \times \mathbf{a}\) in the given exercise involves:

• First row: \(\mathbf{i}, \mathbf{j}, \mathbf{k} \)
• Second row: Components of the velocity vector \(\mathbf{v}(t) \)
• Third row: Components of the acceleration vector \(\mathbf{a}(t) \)

By solving this matrix, you find the new vector resulting from their cross product. The direction of this vector is determined by the right-hand rule, which helps in defining the orientation of the perpendicular vector.

The velocity vector represents the rate of change of position of an object along a parameterized curve. Imagine tracing a path in space; the velocity vector points in the direction you are moving at any given point on this path and indicates how fast you are moving.

To find the velocity vector from a parameterized curve like \(\mathbf{r}(t) = \langle e^t \cos t, e^t \sin t, e^t \rangle\), we differentiate each component of the position vector with respect to time \(t\). This operation produces the velocity vector \(\mathbf{v}(t)\).

• The x-component: Derivative of \(e^t \cos t\)
• The y-component: Derivative of \(e^t \sin t\)
• The z-component: Derivative of \(e^t\)

By calculating these derivatives, you obtain \( \mathbf{v}(t) = \langle e^t(\cos t - \sin t), e^t(\sin t + \cos t), e^t \rangle \). This vector gives you both the speed and direction of movement along the curve at any point \(t\).

The acceleration vector describes how the velocity vector changes over time. It gives us information about how the speed and direction of a moving object are altering.

If you consider the velocity vector \(\mathbf{v}(t)\) as our starting point, obtaining the acceleration vector requires taking its derivative with respect to time \(t\). This differentiation reveals how the velocity vector's components change over time.

• For the x-component, differentiate \(e^t(\cos t - \sin t)\)
• For the y-component, differentiate \(e^t(\sin t + \cos t)\)
• For the z-component, differentiate \(e^t\)

The outcome is the acceleration vector \(\mathbf{a}(t) = \langle e^t ( -2 \sin t - \cos t ), e^t ( 2\cos t - \sin t ), e^t \rangle \). This vector indicates how rapidly the velocity vector is changing, providing insights into any forces that might be acting on the object to change its motion.

Parameterized curves are a powerful way to describe paths or trajectories in space. They use a parameter, often \(t\), to define the position of a point along the curve. By employing parameterized functions, one can capture the essence of complex curves and movements.

Each component function in a parameterized curve represents a coordinate in 2D or 3D space. For the given curve \(\mathbf{r}(t) = \langle e^t \cos t, e^t \sin t, e^t \rangle\), each part \(e^t \cos t\), \(e^t \sin t\), and \(e^t\) defines a distinct trajectory in space.

• The x-component: \(e^t \cos t\)
• The y-component: \(e^t \sin t\)
• The z-component: \(e^t\)

The parameter \(t\) allows the entire curve to be expressed through these interconnected functions. As \(t\) varies, it traces out the path or shape of the curve in space. By understanding parameterized curves, you can analyze complex paths and determine properties like velocity, acceleration, and curvature, as in this exercise.