91Ó°ÊÓ

One of the first steps in finding both the unit tangent vector \( \mathbf{T}(t) \) and the normal vector \( \mathbf{N}(t) \) involves differentiating the given vector function. In this exercise, our vector function is \( \mathbf{r}(t) = \ln t \mathbf{i} + t \mathbf{j} \). When we differentiate it, we use the power of calculus to find the rate of change of each component with respect to \( t \).

For \( \ln t \) (the i-component), the derivative is \( \frac{1}{t} \), since the derivative of a natural logarithm \( \ln x \) is \( \frac{1}{x} \). For the t-component of \( \mathbf{j} \), since it's a simple linear term, its derivative is \( 1 \). By combining these, we get the derivative of the full vector:

• \( \mathbf{r}'(t) = \frac{1}{t} \mathbf{i} + 1 \mathbf{j} \)

This derivative \( \mathbf{r}'(t) \) is crucial because it represents the direction of change of the vector function at any point \( t \). When evaluating at \( t = e \), it provides the instantaneous rate of change which forms the basis of the tangent vector at that point.

Finding the normal vector, denoted as \( \mathbf{N}(t) \), is a bit more involved than finding the derivative. The normal vector is perpendicular to the unit tangent vector, \( \mathbf{T}(t) \), at the point of interest. To compute \( \mathbf{N}(t) \), we typically find the derivative of the unit tangent vector \( \mathbf{T}'(t) \).

This part of the process often requires re-normalizing the derivative after it's been calculated. The principal normal vector will point towards the center of curvature of the path at the given point. Here's a simple breakdown of steps:

• Calculate the derivative of the unit tangent vector: \( \mathbf{T}'(t) \).
• Normalize this derivative to ensure unit length: \( \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{||\mathbf{T}'(t)||} \).

It's important to remember this step can introduce additional complexity, often tailored by specific calculus techniques, depending on the vector function. Thus, \( \mathbf{N}(e) \) essentially aligns to the nature of how the curve is bending at \( t = e \).

Normalization of a vector is about adjusting its magnitude to be exactly one, while maintaining its direction. When dealing with vectors like \( \mathbf{r}'(t) \), once the derivative is computed, its normalization results in the unit tangent vector \( \mathbf{T}(t) \). Normalizing ensures we focus on direction alone, without concern for how 'strong' or long the vector originally was.

The steps involved in normalization are:

• First, calculate the length or magnitude of the vector.
• Then, divide each component of the vector by this magnitude.

For example, with the derivative \( \mathbf{r}'(e) = \frac{1}{e} \mathbf{i} + 1 \mathbf{j} \), we determine its magnitude using the formula:

\[||\mathbf{r}'(e)|| = \sqrt{\left(\frac{1}{e}\right)^2 + 1^2} = \sqrt{\frac{1}{e^2} + 1}\]

Then, divide each component of \( \mathbf{r}'(e) \) by this magnitude to obtain \( \mathbf{T}(e) \). This careful process ensures we describe the path's orientation without scaling or distortion.

Understanding these steps makes the process of finding directional vectors, such as tangent and normal vectors, much clearer and straightforward for any given vector function.