Problem 29 - 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 \(t=t_{1} .\) $$ x=7 \sin 3 t, y=7 \cos 3 t, z=14 t, t_{1}=\pi / 3 $$

Step-by-Step Solution

Answer

Find the vectors and curvature by computing derivatives and evaluating at \( t = \frac{\pi}{3} \).

Step 1: Find the position vector

Given the parametric equations, we express the position vector as \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle = \langle 7\sin(3t), 7\cos(3t), 14t \rangle \).

Step 2: Compute the velocity vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of \( \mathbf{r}(t) \) with respect to \( t \). Thus, \( \mathbf{v}(t) = \langle 21\cos(3t), -21\sin(3t), 14 \rangle \).

Step 3: Compute the unit tangent vector

The unit tangent vector \( \mathbf{T}(t) \) is \( \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} \). First, find \( \|\mathbf{v}(t)\| = \sqrt{(21\cos(3t))^2 + (-21\sin(3t))^2 + 14^2} = \sqrt{441 + 196} = \sqrt{637} \). Therefore, \( \mathbf{T}(t) = \left\langle \frac{21\cos(3t)}{\sqrt{637}}, \frac{-21\sin(3t)}{\sqrt{637}}, \frac{14}{\sqrt{637}} \right\rangle \).

Step 4: Compute the acceleration vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of \( \mathbf{v}(t) \) with respect to \( t \). So, \( \mathbf{a}(t) = \langle -63\sin(3t), -63\cos(3t), 0 \rangle \).

Step 5: Compute the normal vector

The normal vector \( \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|} \), where \( \mathbf{T}'(t) \) is the derivative of \( \mathbf{T}(t) \). Calculate \( \mathbf{T}'(t) \) and normalize by its length.

Step 6: Compute the curvature

The curvature \( \kappa \) is given by \( \kappa = \frac{\| \mathbf{v}(t) \times \mathbf{a}(t) \|}{\| \mathbf{v}(t) \|^3} \). Compute \( \mathbf{v}(t) \times \mathbf{a}(t) \) and find its magnitude, then divide by \( \|\mathbf{v}(t)\|^3 \).

Step 7: Compute the binormal vector

The binormal vector \( \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \). Use the computed vectors from previous steps to find this cross product.

Step 8: Evaluate at \( t = t_1 \)

Substitute \( t = \frac{\pi}{3} \) into all computed expressions for \( \mathbf{T}(t), \mathbf{N}(t), \mathbf{B}(t), \text{ and } \kappa \) to find their values specifically at \( t = \frac{\pi}{3} \).

Key Concepts

Unit Tangent Vector

In calculus, understanding the unit tangent vector is key to analyzing the behavior of curves. The unit tangent vector, often denoted as \( \mathbf{T}(t) \), provides the direction of the curve at a specific point. This vector is crucial for determining how a curve behaves and changes direction.

To find the unit tangent vector, start by computing the velocity vector \( \mathbf{v}(t) \), which is the derivative of the position vector \( \mathbf{r}(t) \). The velocity vector indicates the direction and speed at which a point moves along the curve. However, to obtain the unit tangent vector, you need to normalize \( \mathbf{v}(t) \) by dividing it by its magnitude \( \| \mathbf{v}(t) \| \).

The magnitude is calculated as:\[ \| \mathbf{v}(t) \| = \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} \]
Normalize the velocity vector as follows:

• Find the components of the velocity vector.
• Compute its magnitude.
• Divide each component by this magnitude.

This results in a vector tangent to the curve with a length of one, hence the name "unit" tangent vector.

Parametric Equations

Parametric equations are a way to express the coordinates of a point on a plane or in space as a function of a parameter, often denoted as \( t \). This is a powerful tool in calculus because it allows modeling and understanding complex curves and motions that aren't easily described by one equation.

Imagine drawing a path in space with each position described by a different parameter value. For example, consider these parametric equations:

• \( x = f(t) \)
• \( y = g(t) \)
• \( z = h(t) \)

These equations describe how each coordinate changes with \( t \), creating a smooth and continuous motion that might describe anything from a pendulum's swing to an electron's path.

Using parametric equations allows you to work with curves in space effectively, providing insights into their shape and orientation. They also simplify the differentiation and integration processes because you can deal directly with functions \( f(t) \), \( g(t) \), and \( h(t) \).

Differentiation in Calculus

Differentiation in calculus is the process of finding the derivative of a function. It is a fundamental tool for understanding how functions behave and change. Differentiation reveals the rate of change of quantities, which is essential not only in theoretical math but also in applications like physics and engineering.

In the context of curves described by parametric equations, differentiation allows you to find velocity and acceleration vectors, crucial for analyzing motion. The velocity vector, \( \mathbf{v}(t) \), is derived by differentiating the position vector \( \mathbf{r}(t) \) with respect to the parameter \( t \).

• Find the derivative of each component of the position vector.
• This gives you the velocity, illustrating instant rates of change.

Further, the acceleration vector is found by differentiating the velocity vector. Differentiation lets you peel back the layers of a curve, understanding not just the path itself, but the nuances of its speed and direction. This comprehension is key to solving problems involving curvature, a measure of how sharply a curve bends, and is inherently tied to the first and second derivatives of the parametric equations.