Problem 57 - Calculus Volume 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 \(\mathbf{T}(t)=\left(-\frac{12}{13}\sin(4t)\right)\mathbf{i}+\left(\frac{12}{13}\cos(4t)\right)\mathbf{j}+\frac{5}{13}\mathbf{k}\).

Step 1: Differentiate the Vector Function

Begin by finding the derivative \( \mathbf{r}'(t) \) of the given vector function \( \mathbf{r}(t) = 3 \cos (4t) \mathbf{i} + 3 \sin (4t) \mathbf{j} + 5t \mathbf{k} \). Differentiate each component separately:- The derivative of \(3 \cos(4t)\) with respect to \(t\) is \(-12 \sin(4t)\).- The derivative of \(3 \sin(4t)\) with respect to \(t\) is \(12 \cos(4t)\).- The derivative of \(5t\) with respect to \(t\) is \(5\).Thus, \( \mathbf{r}'(t) = -12 \sin(4t) \mathbf{i} + 12 \cos(4t) \mathbf{j} + 5 \mathbf{k} \).

Step 2: Find the Magnitude of \(\mathbf{r}'(t)\)

The next step is to find the magnitude of the derivative \( \mathbf{r}'(t) \): \[|\mathbf{r}'(t)| = \sqrt{(-12 \sin(4t))^2 + (12 \cos(4t))^2 + 5^2}\]Simplify each term:- \((-12 \sin(4t))^2 = 144 \sin^2(4t)\)- \((12 \cos(4t))^2 = 144 \cos^2(4t)\)- \(5^2 = 25\)Combine and simplify under the square root:\[|\mathbf{r}'(t)| = \sqrt{144(\sin^2(4t) + \cos^2(4t)) + 25} = \sqrt{144 + 25} = \sqrt{169} = 13\]

Step 3: Calculate the Unit Tangent Vector

To find the unit tangent vector \( \mathbf{T}(t) \), divide \( \mathbf{r}'(t) \) 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, the unit tangent vector is:\[\mathbf{T}(t) = \left(-\frac{12}{13} \sin(4t)\right) \mathbf{i} + \left(\frac{12}{13} \cos(4t)\right) \mathbf{j} + \frac{5}{13} \mathbf{k}\]

Key Concepts

Vector Differentiation

Differentiating vectors involves taking the derivative of each component of a vector function with respect to a parameter such as \( t \).
In the context of a parameterized curve \( \mathbf{r}(t) = 3 \, \cos(4t) \, \mathbf{i} + 3 \, \sin(4t) \, \mathbf{j} + 5t \, \mathbf{k} \):

• 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 \(5\).

This is how you arrive at \( \mathbf{r}'(t) = -12 \, \sin(4t) \, \mathbf{i} + 12 \, \cos(4t) \, \mathbf{j} + 5 \, \mathbf{k} \). Differentiation helps us understand the rate of change in each direction along a curve at a given point.

Parameterized Curves

Parameterized curves are a way to represent a curve using a parameter rather than a traditional function \( y = f(x) \).
In our example, \( \,\mathbf{r}(t) = 3 \, \cos(4t) \, \mathbf{i} + 3 \, \sin(4t) \, \mathbf{j} + 5t \, \mathbf{k} \, \), each component is expressed in terms of \( t \).

• The \( x \)-component is \( 3 \, \cos(4t) \): makes it oscillate due to the cosine function.
• The \( y \)-component is \( 3 \, \sin(4t) \): adds another oscillating direction.
• The \( z \)-component is \( 5t \): results in a constant increase/decrease along the z-direction.

Parameterized curves can describe complex paths in space that are difficult to convey using simple coordinates \((x, y, z)\). They provide flexibility to create dynamic and diverse motion by adjusting the parameter.

Magnitude Calculation

Calculating the magnitude of a vector is essential when you need to understand its length or size in space.
For the vector \( \mathbf{r}'(t) = -12 \, \sin(4t) \, \mathbf{i} + 12 \, \cos(4t) \, \mathbf{j} + 5 \, \mathbf{k} \), the magnitude is calculated using:

• \([-12 \, \sin(4t)]^2 = 144 \sin^2(4t)\)
• \([12 \, \cos(4t)]^2 = 144 \cos^2(4t)\)
• \([5]^2 = 25\)

Add them together: \(144(\sin^2(4t) + \cos^2(4t)) + 25\). Since \( \sin^2(4t) + \cos^2(4t) = 1 \), the simplified magnitude is \(\sqrt{169} = 13\).
This calculation confirms the continuous length of the derivative vector, giving us a scalar value crucial in forming the unit tangent vector.

Calculus

Calculus is integral in finding elements like tangents and derivatives of curves.
Understanding vector calculus concepts, such as differentiating a parameterized curve or calculating magnitudes, is vital for analyzing motion and change.

• Differentiation: Helps find the rate of change or the slope of a function.
• Magnitude: Determines the vector's length at any point.
• Tangent Vectors: Give direction and speed of change along the curve.

These basic principles are the foundation of calculus, allowing us to explore complex dynamic systems in mathematics, physics, and engineering with greater precision and insight.