Question

, find the curvature \(\kappa\), the unit tangent vector \(\mathbf{T}\), the unit normal vector \(\mathbf{N}\), and the binormal vector \(\mathbf{B}\) at \(t=t_{1} .\) $$ x=\ln t, y=3 t, z=t^{2} ; t_{1}=2 $$

Short answer

At \( t=2 \), \( \kappa \approx 0.051 \), \( \mathbf{T} \approx \langle 0.1, 0.6, 0.8 \rangle \), \( \mathbf{N} \approx \langle -0.25, 0, 0.1 \rangle \), \( \mathbf{B} \approx \langle 0, 0, 1 \rangle \).

Step-by-step solution

Find the First Derivative

Calculate the derivative of each component with respect to time, for the given position vector \( \mathbf{r}(t) = \langle \ln t, 3t, t^2 \rangle \). - \( \frac{d}{dt}[x] = \frac{d}{dt}[\ln t] = \frac{1}{t} \) - \( \frac{d}{dt}[y] = \frac{d}{dt}[3t] = 3 \) - \( \frac{d}{dt}[z] = \frac{d}{dt}[t^2] = 2t \)Thus, \( \mathbf{r}'(t) = \left\langle \frac{1}{t}, 3, 2t \right\rangle \).

Evaluate First Derivative at t_1

Substitute \( t_1 = 2 \) into \( \mathbf{r}'(t) \) to find \( \mathbf{r}'(t_1) \):\[ \mathbf{r}'(2) = \left\langle \frac{1}{2}, 3, 4 \right\rangle \]

Find the Magnitude of the First Derivative

Compute the magnitude of \( \mathbf{r}'(2) \):\[ \| \mathbf{r}'(2) \| = \sqrt{\left(\frac{1}{2}\right)^2 + 3^2 + 4^2} = \sqrt{\frac{1}{4} + 9 + 16} = \sqrt{25.25} \approx 5.025 \]

Determine the Unit Tangent Vector \(\mathbf{T}\)

Divide each component of \( \mathbf{r}'(2) \) by its magnitude:\[ \mathbf{T}(2) = \frac{1}{5.025}\left\langle \frac{1}{2}, 3, 4 \right\rangle = \left\langle \frac{1}{10.05}, \frac{3}{5.025}, \frac{4}{5.025} \right\rangle \]

Find the Second Derivative

Calculate the second derivative \( \mathbf{r}''(t) \):- \( \frac{d}{dt}[\frac{1}{t}] = -\frac{1}{t^2} \) - \( \frac{d}{dt}[3] = 0 \) - \( \frac{d}{dt}[2t] = 2 \)Thus, \( \mathbf{r}''(t) = \left\langle -\frac{1}{t^2}, 0, 2 \right\rangle \).

Evaluate Second Derivative at t_1

Substitute \( t_1 = 2 \) into \( \mathbf{r}''(t) \) to find \( \mathbf{r}''(2) \):\[ \mathbf{r}''(2) = \left\langle -\frac{1}{4}, 0, 2 \right\rangle \]

Find Curvature \(\kappa\)

Use the formula for curvature: \( \kappa = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} \).First, find the cross product \( \mathbf{r}'(2) \times \mathbf{r}''(2) \):\[ \mathbf{r}'(2) \times \mathbf{r}''(2) = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{1}{2} & 3 & 4 \ -\frac{1}{4} & 0 & 2 \end{array} \right| = \left\langle 6, -2.5, \frac{3}{4} \right\rangle \]Next, calculate its magnitude:\[ \| \langle 6, -2.5, \frac{3}{4} \rangle \| = \sqrt{6^2 + (-2.5)^2 + \left(\frac{3}{4}\right)^2} = \sqrt{36 + 6.25 + 0.5625} = \sqrt{42.8125} \approx 6.54 \]Finally, compute the curvature:\[ \kappa = \frac{6.54}{(5.025)^3} \approx 0.051 \]

Determine the Unit Normal Vector \(\mathbf{N}\)

The unit normal vector \( \mathbf{N} \) is found by taking the derivative of \( \mathbf{T}(t) \) and then normalizing:\[ \mathbf{T}'(t) = \left\langle -\frac{1}{t^2}, 0, \frac{2}{5.025} \right\rangle \] evaluated at \( t = 2 \) gives:\[ \mathbf{T}'(2) = \left\langle -\frac{1}{4}, 0, \frac{2}{5.025} \right\rangle \] Normalize this to get \( \mathbf{N} \approx \langle -0.25, 0, 0.1 \rangle \).

Compute the Binormal Vector \(\mathbf{B}\)

The binormal vector is the cross product of the unit tangent and normal vectors:\[ \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \]. Calculating the cross product for our evaluated vectors:\[ \mathbf{B} = \left\langle \frac{3}{5.025}, \frac{4}{5.025}, -\frac{1}{10.05} \right\rangle \times \left\langle -0.25, 0, 0.1 \right\rangle \] results in:\[ \mathbf{B} \approx \langle 0, 0, 1 \rangle \]

Key concepts

Unit Tangent Vector

The unit tangent vector, often denoted as \( \mathbf{T} \), plays a crucial role in understanding curves in three-dimensional space. It provides a direction that is tangent to the curve at any given point and has a magnitude (or length) of 1. This makes it extremely useful for describing the direction of the curve without being influenced by how fast one might be moving along the curve.

To find the unit tangent vector, you start with the derivative of the position vector, \( \mathbf{r}(t) \), often notated as \( \mathbf{r}'(t) \). This derivative essentially gives us the vector pointing in the direction of the curve's movement and its rate of change at any point \( t \).

However, the derivative \( \mathbf{r}'(t) \) is not necessarily a unit vector, meaning its magnitude might not be 1. Therefore, to form \( \mathbf{T} \), you divide \( \mathbf{r}'(t) \) by its own magnitude, \( \| \mathbf{r}'(t) \| \), scaling it down to a unit length:

• \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} \)

This calculation results in a new vector, \( \mathbf{T}(t) \), which retains the direction of \( \mathbf{r}'(t) \) but is stretched or compressed to have a length of exactly 1. This forms the foundation for analyzing curves with respect to their orientation in space without concern for speed or distance traveled.

Unit Normal Vector

The unit normal vector, designated as \( \mathbf{N} \), is another fundamental component for understanding the behavior of curves. It is orthogonal (at a right angle) to both the unit tangent vector and the curve. It helps in describing how the curve is bending at a particular point.

To determine \( \mathbf{N} \), you must first find the derivative of the unit tangent vector \( \mathbf{T}(t) \); this gives a vector that shows how \( \mathbf{T} \) is changing. This derivative, \( \mathbf{T}'(t) \), points towards the direction of the curve's turn. However, similarly to \( \mathbf{r}'(t) \), \( \mathbf{T}'(t) \) is not necessarily a unit vector.

You take the following steps to calculate the unit normal vector:

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

This process adjusts \( \mathbf{T}'(t) \) such that \( \mathbf{N}(t) \) is a unit vector, ensuring that its length is exactly 1 while maintaining its direction.

The unit normal vector is vital for visualizing how a curve twists and turns in space.

Binormal Vector

The binormal vector, represented as \( \mathbf{B} \), completes the TNB (tangent-normal-binormal) frame, which is a set of three mutually orthogonal vectors at any given point on a curve in three-dimensional space. \( \mathbf{B} \) is perpendicular to both the unit tangent and unit normal vectors, thus completing the three-dimensional orthogonal coordinate system.

The binormal vector is found using a simple but effective operation known as the cross product. By taking the cross product of the unit tangent vector \( \mathbf{T} \) and the unit normal vector \( \mathbf{N} \), you get \( \mathbf{B} \):

• \( \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \)

This operation ensures \( \mathbf{B} \) is also a unit vector (having a magnitude of 1) and perpendicular to both \( \mathbf{T} \) and \( \mathbf{N} \).

Understanding \( \mathbf{B} \) is crucial as it provides insight into the "twisting" behavior of the curve — how it's spinning around the axis defined by \( \mathbf{T} \) as you move along the path. The combined TNB frame, with \( \mathbf{T} \), \( \mathbf{N} \), and \( \mathbf{B} \), thus gives a complete picture of the curve's spatial orientation, similar to a moving "roadmap" at each point along the curve. This can be particularly important in fields like physics and engineering, where understanding the dynamics of motion along a curved path is vital.