Problem 55 - Calculus 3

Question

Find the unit tangent vector for the following parameterized curves. $$\mathbf{r}(t)=6 \mathbf{i}+\cos (3 t) \mathbf{j}+3 \sin (4 t) \mathbf{k}, \quad 0 \leq t<2 \pi$$

Step-by-Step Solution

Answer

The unit tangent vector is \( \frac{-3 \sin(3t) \mathbf{j} + 12 \cos(4t) \mathbf{k}}{\sqrt{9 \sin^2(3t) + 144 \cos^2(4t)}} \).

Step 1: Find the Derivative of the Position Vector

To find the unit tangent vector, we first need the derivative of the position vector \( \mathbf{r}(t) \). Given \( \mathbf{r}(t) = 6 \mathbf{i} + \cos(3t) \mathbf{j} + 3 \sin(4t) \mathbf{k} \), we differentiate each component with respect to \( t \).\[\mathbf{r}'(t) = 0 \mathbf{i} - 3 \sin(3t) \mathbf{j} + 12 \cos(4t) \mathbf{k}\].

Step 2: Calculate the Magnitude of the Derivative

The next step is to find the magnitude of \( \mathbf{r}'(t) \). The magnitude is calculated as:\[\| \mathbf{r}'(t) \| = \sqrt{0^2 + (-3 \sin(3t))^2 + (12 \cos(4t))^2} = \sqrt{9 \sin^2(3t) + 144 \cos^2(4t)}.\]

Step 3: Simplify the Magnitude Expression

Further simplify the expression under the square root. Although the expression can generally be complicated, in most cases an exact evaluation isn’t needed unless specifically requested; we would typically be satisfied with the expression in its radical form. So, the simplified magnitude is \( \| \mathbf{r}'(t) \| = \sqrt{9 \sin^2(3t) + 144 \cos^2(4t)} \).

Step 4: Find the Unit Tangent Vector

Finally, determine the unit tangent vector by dividing the derivative \( \mathbf{r}'(t) \) by its magnitude \( \| \mathbf{r}'(t) \| \):\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} = \frac{0 \mathbf{i} - 3 \sin(3t) \mathbf{j} + 12 \cos(4t) \mathbf{k}}{\sqrt{9 \sin^2(3t) + 144 \cos^2(4t)}}.\]

Key Concepts

Parameterized Curves

Parameterized curves are fundamental in understanding vector calculus and are used to describe the motion of objects in space. These curves represent paths traced by a particle as it moves through space and depend on a parameter, often time denoted by \( t \). Each of these paths can be expressed as a vector function \( \mathbf{r}(t) \), where different components of the vector are functions of \( t \).
As \( t \) varies, the vector generates a curve in 3D space, allowing us to study the geometry and physics of the curve. This representation is critical for analyzing motion, as it allows us to see how position, velocity, and acceleration change over time. In essence, parameterized curves are a way of "drawing" a curve in space using mathematical functions.
Key points include:

• The parameter \( t \) can represent time or another variable.
• Different components of the curve change with \( t \), reflecting changes in direction.
• It provides a comprehensive method for defining curves in 3D space.

Derivative of Position Vector

The derivative of the position vector is crucial in analyzing how a curve behaves. Just like the derivative in single-variable calculus signifies the rate of change of a function, the derivative \( \mathbf{r}'(t) \) in vector calculus shows how the position vector \( \mathbf{r}(t) \) changes with respect to the parameter \( t \).
This derivative is often referred to as the velocity vector, as it indicates the speed and direction of a particle moving along the curve. To find \( \mathbf{r}'(t) \), you differentiate each component of \( \mathbf{r}(t) \) with respect to \( t \).
In this example, since \( \mathbf{r}(t) = 6 \mathbf{i} + \cos(3t) \mathbf{j} + 3 \sin(4t) \mathbf{k} \), its derivative is calculated by:

• Recognizing constants (like 6 in this example) become zero upon differentiation.
• Applying standard differentiation rules to trigonometric functions in each component.
• Resulting in \( \mathbf{r}'(t) = 0 \mathbf{i} - 3 \sin(3t) \mathbf{j} + 12 \cos(4t) \mathbf{k} \).

Magnitude of Derivative

The magnitude of the derivative vector \( \mathbf{r}'(t) \) measures the length or size of this vector. It tells us how fast the particle travels along the curve, disregarding the direction. Calculating this magnitude involves using the formula for the length of a vector\[\| \mathbf{r}'(t) \| = \sqrt{(x')^2 + (y')^2 + (z')^2}\]where \( x', y', z' \) are the components of the derivative \( \mathbf{r}'(t) \).
In this context:

• The function under the square root, \( \sqrt{9 \sin^2(3t) + 144 \cos^2(4t)} \), stems from squaring and adding the components of the derivative vector.
• Although the expression might look complex, simplifying it is later crucial for finding the unit tangent vector.

This magnitude can also give insights into how dynamic or slow a particle's movement is, depending on how the speed varies with the parameter \( t \).
In practical applications, this calculation helps in determining the rate of travel along the defined path.

Vector Calculus

Vector calculus combines mathematical operations with vectors, forming the basis for many physics and engineering applications. It studies how vectors point in different directions and how they can represent changes precisely over a space.
This discipline uses several operations:

• Derivatives to study rates of change.
• Integrals to accumulate quantities like distance or area.
• Vector fields to represent force directions and magnitudes in space.

In the context of unit tangent vectors, vector calculus provides tools like differentiation to analyze the shape and orientation of parameterized curves. Calculating unit tangent vectors involves finding the direction of motion at a point on the curve, independent of speed by normalizing the derivative of the position vector.
With these versatile tools, vector calculus enables detailed examination of physical phenomena, offering ways to quantify and simulate movements through space and time.