91Ó°ÊÓ

Vector calculus is a branch of mathematics that deals with vector fields and operations on vectors. It combines calculus and linear algebra to handle multi-dimensional quantities.
In our exercise, the function \( \mathbf{r}(t) = \ln t \mathbf{i} + t \mathbf{j} \) represents a vector-valued function mapping the real number \( t \) to a vector in two-dimensional space.
Key operations in vector calculus include:

• Derivatives of vector functions: Used in problems like finding velocity or acceleration as derivatives of position vectors.
• Vector magnitudes: Calculating the size or length of vectors, important for normalization.
• Securing unit vector directions: Important in distinguishing paths, such as in the unit tangent vector.

Understanding these concepts helps to interpret and manage vectors in both abstract and real-world contexts, like in physics and engineering.

Differentiation is a fundamental concept in calculus used to find the rate at which a function is changing. In the context of vector functions, differentiation yields a new vector representing the derivative of each component—a crucial step in our exercise.
For a vector function \( \mathbf{r}(t) = \ln t \mathbf{i} + t \mathbf{j} \), we differentiate each component separately.
Here's how it looks:

• The derivative of \( \ln t \) with respect to \( t \) is \( \frac{1}{t} \).
• The derivative of \( t \) is \( 1 \).

This gives a derivative function \( \mathbf{r}'(t) = \frac{1}{t} \mathbf{i} + 1 \mathbf{j} \), representing a vector tangent to the curve described by \( \mathbf{r}(t) \) at any point \( t \). Differentiation in vector calculus helps in analyzing dynamic changes and is vital for computing tangent and normal vectors.

The unit tangent vector, \( \mathbf{T}(t) \), provides a direction at any point along a curve and is crucial for understanding the curve's behavior. It is a normalized version of the derivative of the vector function, ensuring that the resultant vector has a magnitude of one.
To get \( \mathbf{T}(t) \), we need:

• The derivative, \( \mathbf{r}'(t) \), of the original vector function to get the tangent direction.
• The magnitude of \( \mathbf{r}'(t) \), which involves calculating \( \sqrt{ \left(\frac{1}{t}\right)^2 + 1^2 } \).

By dividing \( \mathbf{r}'(t) \) by its magnitude, we obtain \( \mathbf{T}(t) \), which maintains directionality while scaling the vector to a unit size. In our problem, this ensures the tangent vector at \( t=e \) is unit-sized and tangential to the curve.

The unit normal vector, \( \mathbf{N}(t) \), is perpendicular to the unit tangent vector and provides insight into the curve's curvature. This vector indicates how the curve deviates at a given point by bending in space. It achieves this by focusing on the rate of change of the tangent vector itself.
To find \( \mathbf{N}(t) \), follow these steps:

• Differentiate the unit tangent vector \( \mathbf{T}(t) \) to find its rate of change \( \mathbf{T}'(t) \).
• Calculate the magnitude of \( \mathbf{T}'(t) \) to determine its size.
• Normalize \( \mathbf{T}'(t) \) by dividing it by its magnitude to get the unit normal vector.

This vector serves as a critical tool in analyzing the curve's path and behavior, furnishing a deeper understanding of the shape and direction of curves along a path.