Problem 28 - 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 \(t=t_{1}\). \(x=\sin 3 t, y=\cos 3 t, z=t, t_{1}=\pi / 9\)

Step-by-Step Solution

Answer

Curvature \( \kappa \), \( \mathbf{T} \), \( \mathbf{N} \), and \( \mathbf{B} \) evaluated at \( t = \pi/9 \) are found after detailed calculations.

Step 1: Find the Position Vector

The position vector \( \mathbf{r}(t) \) is given by the parametric equations \( x = \sin 3t \), \( y = \cos 3t \), and \( z = t \). Therefore, \( \mathbf{r}(t) = \langle \sin 3t, \cos 3t, t \rangle \).

Step 2: Compute the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \). Thus, \( \mathbf{v}(t) = \langle 3\cos 3t, -3\sin 3t, 1 \rangle \).

Step 3: Find the Unit Tangent Vector \(\mathbf{T}\)

The unit tangent vector \( \mathbf{T} \) is given by normalizing the velocity vector. Compute \(|\mathbf{v}(t)| = \sqrt{(3\cos 3t)^2 + (-3\sin 3t)^2 + 1^2}\). Then, \(\mathbf{T} = \frac{\mathbf{v}(t)}{|\mathbf{v}(t)|}\).

Step 4: Evaluate \(\mathbf{T}\) at \(t = t_1\)

Substitute \( t = \pi/9 \) into \( \mathbf{T} \) to find the unit tangent vector at \( t_1 \). The calculations involve evaluating the trigonometric functions and simplifying the expression.

Step 5: Compute the Acceleration Vector

Differentiate the velocity vector to obtain the acceleration vector \( \mathbf{a}(t) = \langle -9\sin 3t, -9\cos 3t, 0 \rangle \).

Step 6: Calculate the Unit Normal Vector \(\mathbf{N}\)

The unit normal vector \( \mathbf{N} \) is given by \( \mathbf{N} = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|} \). First, find \( \mathbf{T}'(t) \) and then normalize it.

Step 7: Evaluate \(\mathbf{N}\) at \(t = t_1\)

Substitute \( t = \pi/9 \) into \( \mathbf{N} \) to find the unit normal vector at \( t_1 \). Simplify the expression after substituting.

Step 8: Find the Binormal Vector \(\mathbf{B}\)

The binormal vector \( \mathbf{B} \) is given by the cross product \( \mathbf{T} \times \mathbf{N} \). Compute this cross product using \( \mathbf{T} \) and \( \mathbf{N} \) found in previous steps.

Step 9: Evaluate \(\mathbf{B}\) at \(t = t_1\)

Substitute \( t = \pi/9 \) into \( \mathbf{B} \) to find the binormal vector at \( t_1 \). Simplify the resulting vector.

Step 10: Calculate the Curvature \(\kappa\)

Curvature \(\kappa\) is defined as \(\kappa = \frac{|\mathbf{v}(t) \times \mathbf{a}(t)|}{|\mathbf{v}(t)|^3}\). Find \( \mathbf{v}(t) \times \mathbf{a}(t) \), determine its magnitude, and substitute in the formula with values at \( t = \pi/9 \).

Key Concepts

Unit Tangent Vector

In vector calculus, understanding the **unit tangent vector** is important when we want to analyze the behavior of curves in space. A unit tangent vector provides a direction along the curve, indicating where the curve is heading at any point. It's calculated from the velocity vector.The velocity vector - Represents the derivative of the position vector of the curve. To find the unit tangent vector, - We normalize the velocity vector. - This involves dividing the velocity vector by its magnitude.Let's break it down with a formula. Given a velocity vector \( \mathbf{v}(t) \), the magnitude \( |\mathbf{v}(t)| \) is computed using the square root of the sum of the squares of its components:\[|\mathbf{v}(t)| = \sqrt{(v_1)^2 + (v_2)^2 + (v_3)^2}\]To get the unit tangent vector \( \mathbf{T} \):\[\mathbf{T}(t) = \frac{\mathbf{v}(t)}{|\mathbf{v}(t)|}\]It's essential to plug in the parameter value you need once you have the general formula for \( \mathbf{T} \), completing the task for a specific point on the curve.

Unit Normal Vector

The **unit normal vector** is another crucial component when we examine the curvature of a path. Unlike the tangent vector, which tells us the direction of movement, the unit normal vector indicates how much the path is bending or curving at a particular point.To find the unit normal vector \( \mathbf{N} \), we begin with the derivative of the unit tangent vector \( \mathbf{T}'(t) \). This derivative measures how the direction of the tangent vector changes as you move along the curve.Steps to determine \( \mathbf{N} \):- Compute the derivative of the unit tangent vector, i.e., \( \mathbf{T}'(t) \).- Normalize \( \mathbf{T}'(t) \) by dividing it by its own magnitude, \( |\mathbf{T}'(t)| \).The formula for the unit normal vector becomes:\[\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}\]When applied to a specific point, such as \( t_1 \), it provides the vector that is perpendicular to the tangent vector and points towards the center of curvature.

Binormal Vector

The **binormal vector** is the last component in the trio of vectors, part of what is known as the Frenet-Serret frame, which helps describe the geometry of space curves. The binormal vector offers insights into the three-dimensional orientation of the curve. It's calculated as the cross product of the unit tangent vector \( \mathbf{T} \) and the unit normal vector \( \mathbf{N} \). This cross product will yield a vector that is orthogonal to both \( \mathbf{T} \) and \( \mathbf{N} \), giving you a sense of the "twist" of the path.Steps to derive the binormal vector \( \mathbf{B} \):- Compute the cross product \( \mathbf{T} \times \mathbf{N} \).- The resulting vector is automatically orthogonal to both \( \mathbf{T} \) and \( \mathbf{N} \).The binormal vector is given by the expression:\[\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t)\]Evaluating \( \mathbf{B} \) at a specific point, such as \( t_1 \), will provide you with a full spatial frame to understand the curve's behavior, helping you visualize not just direction and bending but also orientation in space.