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Plane curves are fascinating because they offer a two-dimensional view of curves, typically represented in the Cartesian coordinate system. These are curves that lie flat on a plane, which means all the points that form the curve have coordinates dependent only on two variables, often denoted as x and y.
In mathematical terms, plane curves can be expressed through vectors. For example, in an exercise where the curve is given by a vector function like \( \mathbf{r}(t) = t \mathbf{i} + \ln(\cos t) \mathbf{j} \), with t as the variable of differentiation, we observe a representation of a plane curve.

• Coordinate Representation: Each component of the vector (such as \( t \mathbf{i} \) and \( \ln(\cos t) \mathbf{j} \)) defines the position of the curve at any given parameter \( t \).
• Path: The function describes a pathway within a plane, which generally changes direction and can twist based on the parameter t.

Working with plane curves involves finding various types of vectors, such as tangent and normal vectors, and also understanding curvature, which relates to how sharply a curve bends at any point.

A normal vector plays a crucial role in understanding the geometry of curves. It is perpendicular to the unit tangent vector, helping describe how the curve behaves at a point. For a given tangent vector \( \mathbf{T}(t) \), to find the normal vector \( \mathbf{N}(t) \), you differentiate \( \mathbf{T}(t) \) with respect to t and normalize the resulting vector.
Here’s the process to find \( \mathbf{N}(t) \):

• Start by differentiating the unit tangent vector: For \( \mathbf{T}(t) = \cos(t) \mathbf{i} - \sin(t) \mathbf{j} \), we have \( \mathbf{T}'(t) = -\sin(t) \mathbf{i} - \cos(t) \mathbf{j} \).
• Normalize the derivative: Find the magnitude of \( \mathbf{T}'(t) \) (in this case, it's already 1) and divide by this magnitude to get the unit normal vector.

The resulting normal vector \( \mathbf{N}(t) = -\sin(t) \mathbf{i} - \cos(t) \mathbf{j} \) describes the direction that is perpendicular to the tangent at any point on the curve. Using the normal vector, you can better understand the orientation and angle at which the curve might be bending.

Curvature is a measure of how fast a curve is changing direction at any point. For plane curves, the curvature \( \kappa \) provides insight into how "bendy" the curve is. A high curvature means that the curve turns sharply, while a low curvature indicates a gentle bend. The curvature is computed using the derivative of the unit tangent vector and the derivative of the original curve.
Specifically, for a curve \( \mathbf{r}(t) \), curvature is calculated as:

• First, determine \( \| \mathbf{T}'(t) \| \), the magnitude of the derivative of the tangent vector. In many cases, this magnitude turns out to be 1, as seen in this exercise.
• Next, find the magnitude \( \| \mathbf{r}'(t) \| \), which includes information about how the original curve evolves. Here, it turns out to be \( \sec(t) \).
• Divide to find curvature: \( \kappa = \frac{\| \mathbf{T}'(t) \|}{\| \mathbf{r}'(t) \|} \).

In this example, the curvature is \( \kappa = \cos(t) \), meaning that the way the curve bends is directly related to the cosine of the parameter t. Understanding curvature helps in applications like path planning in robotics or understanding motion dynamics.