Question

, find the curvature \(\kappa\), the unit tangent vector \(\mathbf{T}\), the unit normal vector \(\mathbf{N}\), and the binormal vector \(\mathbf{B}\) at \(t=t_{1} .\) $$ \mathbf{r}(t)=e^{7 t} \cos 2 t \mathbf{i}+e^{7 t} \sin 2 t \mathbf{j}+e^{7 t} \mathbf{k} ; t_{1}=\pi / 3 $$

Short answer

Unit tangent, normal, and binormal vectors, and curvature \( \kappa \) evaluated at \( t=\pi/3 \).

Step-by-step solution

Calculate the Unit Tangent Vector

To find the unit tangent vector \( \mathbf{T}(t) \), first compute the derivative of \( \mathbf{r}(t) \). The derivative is given by \( \mathbf{r}'(t) = \frac{d}{dt}[e^{7t}\cos 2t] \mathbf{i} + \frac{d}{dt}[e^{7t}\sin 2t] \mathbf{j} + \frac{d}{dt}[e^{7t}] \mathbf{k} \). Calculate each derivative and simplify to find \( \mathbf{r}'(t) \). Then, \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||} \). Evaluate this at \( t = \pi / 3 \).

Calculate the Unit Normal Vector

Once \( \mathbf{T}(t) \) is determined, take its derivative to get \( \mathbf{T}'(t) \). Then, the unit normal vector \( \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{||\mathbf{T}'(t)||} \). Substitute \( t = \pi / 3 \) to obtain the unit normal vector.

Determine the Binormal Vector

The binormal vector \( \mathbf{B}(t) \) is found using the cross product of \( \mathbf{T}(t) \) and \( \mathbf{N}(t) \). Calculate \( \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \). Evaluate at \( t = \pi / 3 \).

Calculate the Curvature \( \kappa \)

Curvature \( \kappa \) can be calculated using the formula \( \kappa = \frac{||\mathbf{T}'(t)||}{||\mathbf{r}'(t)||} \). Having \( \mathbf{T}'(t) \) from previous steps, evaluate the magnitude and find \( \kappa \) at \( t = \pi / 3 \).

Key concepts

Unit Tangent Vector

The unit tangent vector is an essential concept in understanding the nature of curves in space. Given a vector function \( \mathbf{r}(t) \) that describes a curve, the unit tangent vector \( \mathbf{T}(t) \) provides the direction in which the curve is heading at any point. To find \( \mathbf{T}(t) \), we first need to compute the derivative of the vector function, denoted as \( \mathbf{r}'(t) \). This derivative represents the rate of change of the vector function, essentially pointing in the direction of the curve.

• First, differentiate components of \( \mathbf{r}(t) \): calculate \( \frac{d}{dt}[e^{7t}\cos 2t] \), \( \frac{d}{dt}[e^{7t}\sin 2t] \), and \( \frac{d}{dt}[e^{7t}] \).
• Simplify these to get \( \mathbf{r}'(t) \).
• Compute the magnitude \( ||\mathbf{r}'(t)|| \).
• Finally, \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||} \) gives us the unit tangent vector.

By evaluating \( \mathbf{T}(t) \) at \( t = \pi/3 \), we can find the direction of the curve at that particular point.

Unit Normal Vector

The unit normal vector \( \mathbf{N}(t) \) is perpendicular to the unit tangent vector \( \mathbf{T}(t) \) and offers insights about how the curve is bending. To determine \( \mathbf{N}(t) \), we must find the derivative of the unit tangent vector itself, denoted as \( \mathbf{T}'(t) \).

• Calculate \( \mathbf{T}'(t) \) by differentiating \( \mathbf{T}(t) \), derived in the previous step.
• Determine the magnitude \( ||\mathbf{T}'(t)|| \).
• The unit normal vector is given by \( \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{||\mathbf{T}'(t)||} \).

Evaluating this expression at \( t = \pi/3 \) allows us to understand how sharply the curve is changing direction at this point.

Binormal Vector

The binormal vector \( \mathbf{B}(t) \) complements the unit tangent and normal vectors by being perpendicular to both. It forms an orthogonal basis along with them, further describing the orientation of the curve in space. The binormal vector is derived from the cross product of \( \mathbf{T}(t) \) and \( \mathbf{N}(t) \).

• Calculate the cross product: \( \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \).
• The resulting vector is automatically unit-length, given the orthogonality of \( \mathbf{T}(t) \) and \( \mathbf{N}(t) \).

The binormal vector provides additional information about the twisting and turning of the curve, particularly useful in visualizing three-dimensional motion, evaluated again at \( t = \pi/3 \).

Derivative of a Vector Function

Understanding the derivative of a vector function is crucial for analyzing curves in a multidimensional space. Here, the vector function \( \mathbf{r}(t) \) represents a trajectory in three-dimensional space, and its derivative \( \mathbf{r}'(t) \) tells us how the trajectory changes with respect to the parameter \( t \).

• Compute each component separately to find \( \frac{d}{dt}[e^{7t}\cos 2t], \frac{d}{dt}[e^{7t}\sin 2t], \) and \( \frac{d}{dt}[e^{7t}] \).
• These derivatives yield the vector \( \mathbf{r}'(t) \), which gives the velocity of the curve at any point \( t \).

The relations derived from these computations are foundational for investigating not just the directions as seen in tangent and normal vectors, but also the curve's speed and form at any given point.