Problem 4 - Thomas Calculus

Question

In Exercises \(1-8,\) find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. $$ \mathbf{r}(t)=(2+t) \mathbf{i}-(t+1) \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 3 $$

Step-by-Step Solution

Answer

The unit tangent vector is \( \frac{1}{\sqrt{3}}(\mathbf{i} - \mathbf{j} + \mathbf{k}) \); length is \( 3\sqrt{3} \).

Step 1: Find the Derivative of the Vector

The first step is to find the derivative \( \mathbf{r}'(t) \) of the given vector \( \mathbf{r}(t) = (2 + t) \mathbf{i} - (t + 1) \mathbf{j} + t \mathbf{k} \). Compute the derivative component-wise:\[\mathbf{r}'(t) = \frac{d}{dt}((2+t) \mathbf{i} - (t+1) \mathbf{j} + t \mathbf{k}) = \mathbf{i} - \mathbf{j} + \mathbf{k}.\]

Step 2: Find the Magnitude of the Derivative

Next, calculate the magnitude of \( \mathbf{r}'(t) = \mathbf{i} - \mathbf{j} + \mathbf{k} \). This will help us find the unit tangent vector.\[\|\mathbf{r}'(t)\| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}.\]

Step 3: Calculate the Unit Tangent Vector

Now, use the magnitude to find the unit tangent vector \( \mathbf{T}(t) \) by dividing \( \mathbf{r}'(t) \) by its magnitude:\[\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} = \frac{1}{\sqrt{3}} ( \mathbf{i} - \mathbf{j} + \mathbf{k}).\]

Step 4: Find Length of the Curve

To find the length of the curve from \( t = 0 \) to \( t = 3 \), integrate the magnitude of the derivative over the given interval.\[\text{Length} = \int_{0}^{3} \|\mathbf{r}'(t)\| \, dt = \int_{0}^{3} \sqrt{3} \, dt = \sqrt{3} \times (3-0) = 3\sqrt{3}.\]

Step 5: Final Answer

The unit tangent vector is \( \mathbf{T}(t) = \frac{1}{\sqrt{3}} (\mathbf{i} - \mathbf{j} + \mathbf{k}) \) and the length of the curve from \( t = 0 \) to \( t = 3 \) is \( 3\sqrt{3} \).

Key Concepts

Vector Derivative

When we talk about the vector derivative, we're discussing the process of finding how a vector function changes with respect to a variable, usually time. This is similar to taking the derivative in normal calculus, but applied to functions that have components in multiple directions (like \( \mathbf{i} \), \( \mathbf{j} \), \( \mathbf{k} \)).
Calculating the vector derivative involves finding the derivative of each component separately.
For example, for the vector \( \mathbf{r}(t) = (2+t) \mathbf{i} - (t+1) \mathbf{j} + t \mathbf{k} \),

• The derivative of \( (2+t) \mathbf{i} \) with respect to \( t \) is \( \mathbf{i} \).
• The derivative of \( -(t+1) \mathbf{j} \) is \( - \mathbf{j} \).
• The derivative of \( t \mathbf{k} \) is \( \mathbf{k} \).

These result in the derivative \( \mathbf{r}'(t) = \mathbf{i} - \mathbf{j} + \mathbf{k} \).This fundamental step of calculating a vector's derivative forms the backbone of finding both the unit tangent vector and understanding motion along a path.

Curve Length

Understanding curve length is crucial in fields like physics and engineering, where it helps in determining the distance along a particular path.
The length of a curve can be found by integrating the magnitude of its derivative across the interval we're interested in.
The magnitude tells us the rate at which the function's position changes as \( t \) changes.
For instance, in this exercise, we're evaluating the path from \( t = 0 \) to \( t = 3 \).Here's how it's done:

• First, calculate \( \|\mathbf{r}'(t)\| \), which we previously found to be\( \sqrt{3} \).
• Integrate this magnitude value over the interval:\[ \int_{0}^{3} \sqrt{3} \, dt = \sqrt{3} \times (3-0) = 3\sqrt{3}. \]

This result, \( 3\sqrt{3} \), represents the total length of the specified curve segment, allowing us to see how far we've traveled along the path from starting point to end.

Magnitude Calculation

The calculation of magnitude is an essential part of understanding vector quantities. The magnitude essentially tells you how long a vector is and, in the context of motion, can indicate speed.
To find the magnitude of a vector like \( \mathbf{r}'(t) = \mathbf{i} - \mathbf{j} + \mathbf{k} \), we use the formula for the magnitude of a vector: \[ \|\mathbf{v}\| = \sqrt{v_x^2 + v_y^2 + v_z^2} \].

• Substitute each component: \( 1 \) for \( v_x \), \( -1 \) for \( v_y \), and \( 1 \) for \( v_z \).
• Calculate: \( \|\mathbf{r}'(t)\| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \).

This computation gives you a clearer picture of the velocity's size as opposed to its direction. It's a basic method, similar to the Pythagorean theorem, but applied in three dimensions.This simple but powerful calculation helps unlock understanding of more complex vector mechanics.