91Ó°ÊÓ

Parametric equations are a powerful tool in mathematics that allow us to describe curves in a plane or in space. Unlike standard forms of equations, which define curves by fixed relationships between x and y, parametric equations introduce a third variable, often denoted as "t", which acts as a parameter. This allows for a more flexible representation of curves. In this exercise, the parametric equation is given by the expression

• \( \mathbf{r}(t) = \sin t \, \mathbf{i} + \cos t \, \mathbf{j} + \tan t \, \mathbf{k} \).

The variable \( t \) is the parameter, and as \( t \) varies, the vector

• \( \mathbf{r}(t) \) traces out the curve in 3D space.

This form of representation is crucial in applications such as computer graphics and physics, where the dynamics of a system over time need to be analyzed in a spatial context. Here, by knowing \(t = \pi/4\), we can locate the point \((\sqrt{2}/2, \sqrt{2}/2, 1)\) on the curve, connecting the parameter value directly to a specific point in space.

The tangent vector plays a key role in understanding the path of a curve at any point. It essentially provides a direction of the curve at a particular instant defined by the parameter \( t \). For a parametric curve smoothly defined as \( \mathbf{r}(t) \), the tangent vector \( \mathbf{T}(t) \) is derived by taking the derivative of \( \mathbf{r} \) with respect to \( t \). This derivative tells us how the curve is changing at any given point:

• \( \mathbf{T}(t) = \frac{d}{dt} \left[ \sin t \, \mathbf{i} + \cos t \, \mathbf{j} + \tan t \, \mathbf{k} \right] \).

This results in

• \( \mathbf{T}(t) = \cos t \, \mathbf{i} - \sin t \, \mathbf{j} + (1/\cos^2 t) \, \mathbf{k} \).

At \( t = \pi/4 \), this vector becomes

• \( \mathbf{T}(\pi/4) = \sqrt{2}/2 \, \mathbf{i} - \sqrt{2}/2 \, \mathbf{j} + 2 \, \mathbf{k} \).

This specific vector points in the direction of the curve at that parameter, indicating the curve's path through space and emphasizing the curve's steepness with respect to time.

The binormal vector \( \mathbf{B} \) is another crucial element when analyzing curves in space. It completes the triad of vectors — along with the tangent and normal vectors — that describe the geometry of the curve. The binormal vector is obtained by taking the cross product of the tangent vector \( \mathbf{T} \) and the normal vector \( \mathbf{N} \). Although in this exercise the normal vector is not explicitly used, the binormal vector is derived as follows:

• \( \mathbf{B} = \frac{\mathbf{T} \times \mathbf{N}}{||\mathbf{T} \times \mathbf{N}||} \).

For practical cases like this,

• \( \mathbf{B}(\pi/4) = -2 \sqrt{2} \, \mathbf{i} + 2 \sqrt{2} \, \mathbf{j} \).

This vector is perpendicular to both the tangent and normal vectors and provides a third dimension to the understanding of the curve's geometry by defining the periphery of the curve's surface or direction.

The cross product is an operation in vector algebra that plays a critical role in determining the normal direction to a surface. When you apply the cross product to two vectors \( \mathbf{A} \) and \( \mathbf{B} \), you obtain a third vector \( \mathbf{C} \) which is orthogonal (perpendicular) to both \( \mathbf{A} \) and \( \mathbf{B} \). For 3D vectors, this is calculated by:

• \( \mathbf{A} \times \mathbf{B} = (a_2b_3 - a_3b_2)\mathbf{i} + (a_3b_1 - a_1b_3)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k} \).

In the context of our exercise, the cross product is used to find vectors like the binormal vector and to further calculate the normal vector to the rectifying plane. It is versatile in finding the orientation and is particularly useful in physics and engineering to find torques and rotational effects. The result of these operations significantly impacts how we establish plans and visualize the spatial orientation of forces and movements.