Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the curvature \(\kappa\), the unit tangent vector \(\mathbf{T}\), the unit normal vector \(\mathbf{N}\), and the binormal vector \(\mathbf{B}\) at \(t=t_{1}\). \(\mathbf{r}(t)=\frac{1}{2} t^{2} \mathbf{i}+t \mathbf{j}+\frac{1}{3} t^{3} \mathbf{k} ; t_{1}=2\)

Step-by-Step Solution

Answer

At \(t=2\), \(\mathbf{T}(2) = \frac{2}{\sqrt{21}} \mathbf{i} + \frac{1}{\sqrt{21}} \mathbf{j} + \frac{4}{\sqrt{21}} \mathbf{k}, \mathbf{N}(2)= \ldots , \mathbf{B}(2)= \ldots , \kappa = \frac{\sqrt{597}}{21}.\)

Step 1: Find the First Derivative

The given vector function is \( \mathbf{r}(t) = \frac{1}{2} t^{2} \mathbf{i} + t \mathbf{j} + \frac{1}{3} t^{3} \mathbf{k} \). To find the first derivative, differentiate each component with respect to \(t\):\[\mathbf{r}'(t) = \frac{d}{dt}\left( \frac{1}{2}t^2 \right) \mathbf{i} + \frac{d}{dt}(t) \mathbf{j} + \frac{d}{dt}\left( \frac{1}{3}t^3 \right) \mathbf{k} = t \mathbf{i} + \mathbf{j} + t^2 \mathbf{k}.\]

Step 2: Compute the Unit Tangent Vector \(\mathbf{T}\)

The unit tangent vector \(\mathbf{T}\) is defined as the derivative of \(\mathbf{r}(t)\) divided by its magnitude:\[\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||}.\] For \(t_1 = 2\), \(\mathbf{r}'(2) = 2 \mathbf{i} + \mathbf{j} + 4 \mathbf{k}\). The magnitude is \[||\mathbf{r}'(2)|| = \sqrt{2^2 + 1^2 + 4^2} = \sqrt{21}.\] Thus,\[\mathbf{T}(2) = \frac{2 \mathbf{i} + \mathbf{j} + 4 \mathbf{k}}{\sqrt{21}}.\]

Step 3: Find the Second Derivative

Differentiate \(\mathbf{r}'(t)\) to find \(\mathbf{r}''(t)\):\[\mathbf{r}''(t) = \frac{d}{dt}(t) \mathbf{i} + \frac{d}{dt}(1) \mathbf{j} + \frac{d}{dt}(t^2) \mathbf{k} = \mathbf{i} + 2t \mathbf{k}.\] Substituting \(t_1 = 2\), we have \(\mathbf{r}''(2) = \mathbf{i} + 4 \mathbf{k}\).

Step 4: Calculate the Unit Normal Vector \(\mathbf{N}\)

The unit normal vector \(\mathbf{N}\) is given by\[\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{||\mathbf{T}'(t)||}.\] Calculate \(\mathbf{T}'(t)\) first: The derivative of \(\mathbf{r}'(t)\) can be found as \(\mathbf{r}''(t)\). Thus,\[\mathbf{T}'(t) = \mathbf{r}''(t) - \frac{(\mathbf{r}''(t) \cdot \mathbf{T}(t)) \mathbf{T}(t)}{||\mathbf{T}(t)||^{2}}.\] Substituting, we find \(\mathbf{T}'(2)=\mathbf{i} + \frac{10}{21}\mathbf{j} - \frac{8}{21}\mathbf{k}\). The magnitude \(||\mathbf{T}'(2)|| = \sqrt{1+\frac{100}{441} + \frac{64}{441}} = \sqrt{\frac{597}{441}}\), so \[\mathbf{N}(2) = \frac{21}{\sqrt{597}} (\mathbf{i} + \frac{10}{21}\mathbf{j} - \frac{8}{21}\mathbf{k}).\]

Step 5: Calculate the Binormal Vector \(\mathbf{B}\)

The binormal vector \(\mathbf{B}\) is the cross product of \(\mathbf{T}\) and \(\mathbf{N}\):\[\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t).\] Use the unit vectors at \(t_1 = 2\):\[\mathbf{B}(2) = \left(\frac{2}{\sqrt{21}} \mathbf{i} + \frac{1}{\sqrt{21}} \mathbf{j} + \frac{4}{\sqrt{21}} \mathbf{k}\right) \times \left(\frac{21}{\sqrt{597}} \mathbf{i} + \frac{210}{21\sqrt{597}} \mathbf{j} - \frac{168}{21\sqrt{597}} \mathbf{k}\right).\] Calculate this cross product to find \(\mathbf{B}(2)\).

Step 6: Determine the Curvature \(\kappa\)

The curvature \(\kappa\) is given by \(\kappa = \frac{||\mathbf{T}'(t)||}{||\mathbf{r}'(t)||}\). Using the magnitudes found earlier for \(t_1=2\):\[\kappa = \frac{\sqrt{\frac{597}{441}}}{\sqrt{21}}.\] Simplify this expression to find the curvature. After simplification, \[\kappa = \frac{\sqrt{597}}{21}.\]

Key Concepts

Vector Calculus

Vector calculus is a mathematical field that deals with vector fields in multi-dimensional spaces. It helps in understanding how vector fields change and interact. This branch of mathematics combines calculus and linear algebra to analyze vector functions, which are crucial in physics and engineering. In the context of this exercise, vector calculus enables us to find the curvature, tangent, normal, and binormal vectors of a given vector function. By examining these aspects, we understand the behavior of curves in three-dimensional space.

In vector calculus:

• A vector function describes a curve in space.
• The derivatives of vector functions provide information about the velocity and acceleration of particles moving along these curves.
• The dot product and cross product operations are tools for finding angles between and areas spanned by vectors, respectively.

The task of finding the curvature and various vectors in this exercise leverages these concepts to analyze the path described by the vector function.

Tangent Vector

The tangent vector is an essential concept in vector calculus, providing insight into the direction of a curve at a specific point. In simple terms, it is a vector that "touches" the curve at one point without crossing it, indicating the curve's direction at that location.

To calculate the tangent vector \(\mathbf{T}\), follow these steps:

• Differentiate the given vector function \(\mathbf{r}(t)\) with respect to \(t\) to find \(\mathbf{r}'(t)\).
• Determine the magnitude of \(\mathbf{r}'(t)\), \(||\mathbf{r}'(t)||\).
• Divide \(\mathbf{r}'(t)\) by its magnitude to obtain the unit tangent vector \(\mathbf{T}(t)\).

For instance, in our exercise:- The first derivative was calculated as \(\mathbf{r}'(t) = t \mathbf{i} + \mathbf{j} + t^2 \mathbf{k}\).- At \(t_1 = 2\), we calculated the magnitude \(||\mathbf{r}'(2)||\), resulting in \(\sqrt{21}\).- Therefore, the unit tangent vector at that point is \(\frac{2 \mathbf{i} + \mathbf{j} + 4 \mathbf{k}}{\sqrt{21}}\).

Understanding the tangent vector gives us a snapshot of a curve's trajectory at a specific instant.

Normal Vector

The normal vector represents the direction in which the curve is changing most rapidly at any given point. Unlike the tangent vector, which touches the curve, the normal vector is perpendicular to it. This perpendicularity makes it crucial in understanding the curve's curvature and how it bends.

To derive the unit normal vector \(\mathbf{N}\):

• First, compute the derivative of the unit tangent vector \(\mathbf{T}(t)\), which gives \(\mathbf{T}'(t)\).
• Calculate the magnitude of \(\mathbf{T}'(t)\).
• Divide \(\mathbf{T}'(t)\) by its magnitude to get the unit normal vector \(\mathbf{N}(t)\).

In our particular problem, the calculations led to:
- The derivative of the tangent vector \(\mathbf{T}'(2)\) turned out to be \(\mathbf{i} + \frac{10}{21}\mathbf{j} - \frac{8}{21}\mathbf{k}\).
- The magnitude \(||\mathbf{T}'(2)||\) was found as \(\sqrt{\frac{597}{441}}\).
- Therefore, the unit normal vector at \(t_1 = 2\) was calculated to be \(\frac{21}{\sqrt{597}} (\mathbf{i} + \frac{10}{21}\mathbf{j} - \frac{8}{21}\mathbf{k})\).

The normal vector is crucial for understanding the shape of a curve at a point, particularly how it curves in space.

Binormal Vector

The binormal vector is the third unit vector that, together with the tangent and normal vectors, forms an orthonormal basis for the Frenet-Serret frame of a space curve. This vector is orthogonal (at a right angle) to both the tangent and normal vectors, giving a complete three-dimensional analysis of the curve's behavior at a specific point.

The steps to find the binormal vector \(\mathbf{B}\) are:

• Compute the cross product of the unit tangent vector \(\mathbf{T}(t)\) and the unit normal vector \(\mathbf{N}(t)\).

Cross products give a vector perpendicular to two given vectors, ensuring \(\mathbf{B}\) is orthogonal to both \(\mathbf{T}\) and \(\mathbf{N}\).

For the given exercise, the calculation resulted in:

• The binormal vector was found by crossing \(\mathbf{T}(2)\) and \(\mathbf{N}(2)\).
• This provided a direction that completes the triple crucial for exploring spatial relationships along the curve.

Understanding the binormal vector helps in visualizing the orientation of the curve, allowing us to see how it is embedded in the surrounding space.