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 \( \frac{\mathbf{i} - \mathbf{j} + \mathbf{k}}{\sqrt{3}} \) and the length of the curve is \( 3\sqrt{3} \).

Step 1: Differentiate the Vector Function

To find the unit tangent vector, we first need the derivative of the position vector, \( \mathbf{r}(t) = (2+t) \mathbf{i} - (t+1) \mathbf{j} + t \mathbf{k} \). The derivative is \( \mathbf{r}'(t) = \frac{d}{dt}((2+t) \mathbf{i} - (t+1) \mathbf{j} + t \mathbf{k}) = \mathbf{i} - \mathbf{j} + \mathbf{k} \).

Step 2: Compute the Magnitude of the Derivative

Next, compute the magnitude of \( \mathbf{r}'(t) \), which is \( ||\mathbf{r}'(t)|| = \sqrt{(1)^2 + (-1)^2 + (1)^2} = \sqrt{3} \).

Step 3: Find the Unit Tangent Vector

The unit tangent vector \( \mathbf{T}(t) \) is given by \( \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||} \). Thus, \( \mathbf{T}(t) = \frac{\mathbf{i} - \mathbf{j} + \mathbf{k}}{\sqrt{3}} \).

Step 4: Calculate the Length of the Curve

The length of the curve from \( t = 0 \) to \( t = 3 \) is given by the integral \( \int_{0}^{3} ||\mathbf{r}'(t)|| \, dt \), which simplifies to \( \sqrt{3} \int_{0}^{3} \, dt \). Evaluating this gives \( \sqrt{3} \cdot 3 = 3\sqrt{3} \).

Key Concepts

Derivative of Vector Function

When dealing with vector functions, such as those representing curves in 3D space, the derivative of the vector function is key. Think of the vector function as a path traced out by a point as time advances. The derivative of this function, noted as \( \mathbf{r}'(t) \), gives you the velocity vector of the point along the path at any given instant. Essentially, it tells you how fast and in which direction your point is moving.To differentiate a vector function like \( \mathbf{r}(t) = (2+t) \mathbf{i} - (t+1) \mathbf{j} + t \mathbf{k} \), you need to find the derivatives of its component functions separately:

• The derivative of \( 2+t \) is \( 1 \).
• The derivative of \( -(t+1) \) is \( -1 \).
• The derivative of \( t \) is \( 1 \).

Putting these together, we find \( \mathbf{r}'(t) = \mathbf{i} - \mathbf{j} + \mathbf{k} \). This indicates that at any moment along the curve, the point is moving consistently in the direction \( \mathbf{i} - \mathbf{j} + \mathbf{k} \), illustrating both speed and direction.

Length of Curve

The length of a curve provides a measure of the "distance traveled" by a point moving along the curve over a given parameter range. To compute the length of a portion of a curve, you will integrate the magnitude of its derivative over the desired interval.For the curve represented by \( \mathbf{r}(t) \), the length from \( t = 0 \) to \( t = 3 \) can be found by evaluating the integral:\[\int_{0}^{3} ||\mathbf{r}'(t)|| \, dt\]Based on step 2 of the solution, we know the magnitude of \( \mathbf{r}'(t) \) is \( \sqrt{3} \). Thus, the integral becomes:\[\sqrt{3} \int_{0}^{3} \, dt\]This simplifies to \( \sqrt{3} \times 3 \), giving us the value \( 3\sqrt{3} \). This result represents the total length of the curve from start to finish on the given interval. It's like you are unwinding the curve into a straight line to measure it!

Magnitude of Vector

The magnitude of a vector is a measure of its length. For a vector \( \mathbf{v} \) represented in 3D space as \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), its magnitude \( ||\mathbf{v}|| \) is calculated using the formula:\[||\mathbf{v}|| = \sqrt{a^2 + b^2 + c^2}\]This formula is a direct extension of the Pythagorean theorem to three dimensions. Each component squared and under a square root gives the "length" of the vector.In the context of finding a unit tangent vector, calculating the magnitude of the derivative of the vector function, \( \mathbf{r}'(t) \), is crucial. In our example, \( \mathbf{r}'(t) = \mathbf{i} - \mathbf{j} + \mathbf{k} \) produces:\[||\mathbf{r}'(t)|| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}\]This magnitude shows how much the vector extends in space, laying the foundation for creating a unit vector by normalizing \( \mathbf{r}'(t) \) to have a magnitude of 1.