Problem 57 - Calculus 3

Question

Find the unit tangent vector for the following parameterized curves. $$\mathbf{r}(t)=3 \cos (4 t) \mathbf{i}+3 \sin (4 t) \mathbf{j}+5 t \mathbf{k}, 1 \leq t \leq 2$$

Step-by-Step Solution

Answer

The unit tangent vector is \( \mathbf{T}(t) = -\frac{12}{13}\sin(4t)\mathbf{i} + \frac{12}{13}\cos(4t)\mathbf{j} + \frac{5}{13}\mathbf{k} \).

Step 1: Understand the problem

We need to find the unit tangent vector for the vector-valued function \( \mathbf{r}(t)=3 \cos (4t) \mathbf{i}+3 \sin (4t) \mathbf{j}+5t \mathbf{k} \). A unit tangent vector is obtained by first finding the derivative of the curve with respect to its parameter \( t \), then normalizing this derivative.

Step 2: Compute the derivative

Compute \( \mathbf{r}'(t) \), the derivative of \( \mathbf{r}(t) \). The derivative is obtained by taking the derivative of each component:- Derivative of \( 3\cos(4t) \) is \( -12\sin(4t) \)- Derivative of \( 3\sin(4t) \) is \( 12\cos(4t) \)- Derivative of \( 5t \) is \( 5 \)Thus, \( \mathbf{r}'(t) = -12\sin(4t) \mathbf{i} + 12\cos(4t) \mathbf{j} + 5 \mathbf{k} \).

Step 3: Find the magnitude of the derivative

The magnitude of \( \mathbf{r}'(t) \) is found using the formula:\[ \| \mathbf{r}'(t) \| = \sqrt{(-12\sin(4t))^2 + (12\cos(4t))^2 + 5^2} \].Simplify the expression:\[ \| \mathbf{r}'(t) \| = \sqrt{144\sin^2(4t) + 144\cos^2(4t) + 25} \]Using the identity \( \sin^2(x) + \cos^2(x) = 1 \), this becomes:\[ \| \mathbf{r}'(t) \| = \sqrt{144 + 25} = \sqrt{169} = 13 \].

Step 4: Normalize the derivative to find the unit tangent vector

The unit tangent vector \( \mathbf{T}(t) \) is found by dividing \( \mathbf{r}'(t) \) by its magnitude:\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} = \frac{-12\sin(4t) \mathbf{i} + 12\cos(4t) \mathbf{j} + 5 \mathbf{k}}{13} \].Thus:\[ \mathbf{T}(t) = \left( -\frac{12}{13}\sin(4t) \right) \mathbf{i} + \left( \frac{12}{13}\cos(4t) \right) \mathbf{j} + \frac{5}{13} \mathbf{k} \].

Key Concepts

Derivative of Parametric Curves

In the world of calculus, understanding parametric curves is vital. These curves represent a set of equations where one or more variables are expressed as functions of a parameter. Often, this parameter is denoted by \( t \). For example, a curve can be defined by \( x(t), y(t) \), or in our case, \( \mathbf{r}(t) = 3 \cos(4t) \mathbf{i} + 3 \sin(4t) \mathbf{j} + 5t \mathbf{k} \). The goal is to find its behavior as we vary \( t \).

The derivative of this parametric curve, denoted as \( \mathbf{r}'(t) \), provides vital information about the curve's direction at any point. To find this, differentiate each component of the curve with respect to \( t \).

• The derivative of \( 3\cos(4t) \) is \( -12\sin(4t) \).
• The derivative of \( 3\sin(4t) \) is \( 12\cos(4t) \).
• The derivative of \( 5t \) is simply \( 5 \).

Bringing these together, we get the derivative as a vector: \( \mathbf{r}'(t) = -12\sin(4t) \mathbf{i} + 12\cos(4t) \mathbf{j} + 5 \mathbf{k} \).

This vector essentially tells us the velocity of a point moving along the curve, indicating both speed and direction.

Magnitude of a Vector

Once we have the derivative vector from the parametric curve, we must understand its magnitude. The magnitude of a vector is a measure of its length in space. It is particularly important when we aim to find unit vectors. To compute the magnitude of a vector \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), we use: \ \[ \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2 + v_3^2} \] \ For our derivative vector \( \mathbf{r}'(t) = -12\sin(4t) \mathbf{i} + 12\cos(4t) \mathbf{j} + 5 \mathbf{k} \), the components are:

• \( -12\sin(4t) \)
• \( 12\cos(4t) \)
• \( 5 \)

Calculating the magnitude involves: 1. Square each component: - \( (-12\sin(4t))^2 = 144\sin^2(4t) \) - \( (12\cos(4t))^2 = 144\cos^2(4t) \) - \( 5^2 = 25 \)2. Add these squared values: \[ \| \mathbf{r}'(t) \| = \sqrt{144\sin^2(4t) + 144\cos^2(4t) + 25} \]3. Use the Pythagorean trigonometric identity \( \sin^2(x) + \cos^2(x) = 1 \) to simplify: \[ \| \mathbf{r}'(t) \| = \sqrt{144 + 25} = \sqrt{169} = 13 \] This value, 13, is the magnitude of our derivative vector.

Normalization of Vectors

Normalization of vectors is an essential process when working with parametric curves, especially when identifying unit tangent vectors. To normalize a vector means to scale it to a unit length of 1. This creates a unit vector that points in the same direction as the original vector but with a magnitude of one.To normalize, you divide each component of the vector by its magnitude. Let's apply this to our derivative vector \( \mathbf{r}'(t) = -12\sin(4t) \mathbf{i} + 12\cos(4t) \mathbf{j} + 5 \mathbf{k} \), which has a magnitude of 13.

• Divide the \( \mathbf{i} \)-component: \( -12\sin(4t) \div 13 \Rightarrow \left( -\frac{12}{13} \sin(4t) \right) \)
• Divide the \( \mathbf{j} \)-component: \( 12\cos(4t) \div 13 \Rightarrow \left( \frac{12}{13} \cos(4t) \right) \)
• Divide the \( \mathbf{k} \)-component: \( 5 \div 13 \Rightarrow \frac{5}{13} \)

By performing these divisions, we obtain the unit tangent vector:\[ \mathbf{T}(t) = \left( -\frac{12}{13} \sin(4t) \right) \mathbf{i} + \left( \frac{12}{13} \cos(4t) \right) \mathbf{j} + \frac{5}{13} \mathbf{k} \]This unit tangent vector \( \mathbf{T}(t) \) maintains directionality with a magnitude of 1, which is perfect for defining lines or directions.