91Ó°ÊÓ

To understand motion along a curve, we first need to find the velocity vector. The velocity vector is the derivative of the position vector with respect to time, representing the rate of change of position. In the case of the curve specified by \( \mathbf{r}(t) = t \mathbf{i} + \frac{1}{2} t^2 \mathbf{j} + t^2 \mathbf{k} \), the velocity vector \( \mathbf{v}(t) \) is obtained by differentiating \( \mathbf{r}(t) \) with respect to \( t \). Thus, we find that the velocity vector is:

• \( \mathbf{v}(t) = \frac{d}{dt}(t \mathbf{i} + \frac{1}{2} t^2 \mathbf{j} + t^2 \mathbf{k}) = \mathbf{i} + t \mathbf{j} + 2t \mathbf{k} \)

This vector gives us the direction and speed of a particle moving along the curve at any time \( t \). Evaluating it at a specific time, such as \( t = 1 \), gives \( \mathbf{v}(1) = \mathbf{i} + \mathbf{j} + 2 \mathbf{k} \), which describes the instantaneous velocity of the particle at that moment.

The unit tangent vector, \( \mathbf{T}(t) \), describes the direction of the curve at a specific point and is found by normalizing the velocity vector. To find the unit tangent vector, we take the velocity vector and divide it by its magnitude, which ensures the vector has a length of one. Let's find the magnitude of \( \mathbf{v}(1) = \mathbf{i} + \mathbf{j} + 2 \mathbf{k} \), which is calculated as \( \| \mathbf{v}(1) \| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6} \).

• Therefore, the unit tangent vector is \( \mathbf{T}(1) = \frac{1}{\sqrt{6}}(\mathbf{i} + \mathbf{j} + 2 \mathbf{k}) \).

This vector points in the same direction as the velocity vector, but with a standardized length. It is crucial for understanding how the curve behaves at each point along its path.

The unit normal vector, \( \mathbf{N}(t) \), provides insights into how the curve is bending at any point. It is perpendicular to the unit tangent vector, indicating the direction in which the curve is turning. To find \( \mathbf{N}(t) \), we need the derivative of the unit tangent vector, \( \mathbf{T}'(t) \). Once we have this derivative, we normalize it to get \( \mathbf{N}(t) \), ensuring it has a magnitude of one. This involves evaluating \( \mathbf{T}'(t) \) at \( t = 1 \) and then dividing by its magnitude.

• This vector is orthogonal to \( \mathbf{T}(t) \), and together they form a plane showing the instantaneous direction of bending of the curve.

Understanding \( \mathbf{N}(t) \) helps in analyzing the curvature and torsion of the curve based on its geometry.

Sketching a 3D curve involves plotting the path described by the position vector, highlighting key points and directions such as the unit tangent and normal vectors. For our curve \( \mathbf{r}(t) = t \mathbf{i} + \frac{1}{2} t^{2} \mathbf{j} + t^{2} \mathbf{k} \), we outline its path in 3D space.Begin by evaluating the position vector at the specific value of \( t \), such as \( t = 1 \). This gives us a point \( (1, \frac{1}{2}, 1) \) on the curve. Visualize how the curve travels through this space by sketching the path, emphasizing how it rises and turns based on the function components. At each point, you can add the vectors \( \mathbf{T}(1) \) and \( \mathbf{N}(1) \):

• The tangent vector, shown with an arrow, follows the path's direction.
• The normal vector, also illustrated with an arrow, shows orthogonality to \( \mathbf{T}(1) \), reflecting the curvature direction.

This approach aids in providing a visual understanding of the curve's shape and its directional properties.