Question

Arc length parameterization Determine whether the following curves use arc length as a parameter. If not, find a description that uses arc length as a parameter. $$\mathbf{r}(t)=\left\langle\frac{t}{\sqrt{3}}, \frac{t}{\sqrt{3}}, \frac{t}{\sqrt{3}}\right\rangle, \text { for } 0 \leq t \leq 10$$

Short answer

Question: Determine if the given curve uses arc length as a parameter: $$\mathbf{r}(t) = \left\langle\frac{t}{\sqrt{3}}, \frac{t}{\sqrt{3}}, \frac{t}{\sqrt{3}}\right\rangle$$ Answer: Yes, the given curve uses arc length as a parameter.

Step-by-step solution

Find the derivative of the given vector function

First, let's find the derivative of the given vector function with respect to t. $$\mathbf{r}'(t) = \left\langle\frac{\mathrm{d}(\frac{t}{\sqrt{3}})}{\mathrm{d}t}, \frac{\mathrm{d}(\frac{t}{\sqrt{3}})}{\mathrm{d}t}, \frac{\mathrm{d}(\frac{t}{\sqrt{3}})}{\mathrm{d}t}\right\rangle = \left\langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\rangle$$

Calculate the magnitude of the derivative

Now, calculate the magnitude of the derivative of the vector function. $$||\mathbf{r}'(t)|| = \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2} = \sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}} = \sqrt{1} = 1$$ Since the magnitude of the derivative is constant and equal to 1, the given curve uses arc length as a parameter, and no new description is needed.

Key concepts

Vector Functions

A vector function is a mathematical function that takes one or more variables as input and returns a vector as an output. These functions are incredibly useful in describing physical phenomena such as the trajectory of an object in space or other similar dynamic processes.

In this exercise, the vector function is represented by \( \mathbf{r}(t) = \left\langle \frac{t}{\sqrt{3}}, \frac{t}{\sqrt{3}}, \frac{t}{\sqrt{3}} \right\rangle \). This function maps a single variable \( t \) to a vector in three-dimensional space.

Key features of vector functions include:

• Components: Each component of the vector function (in our example, there are three: \( \frac{t}{\sqrt{3}} \)) corresponds to one axis in the coordinate system.
• Parameterization: Variable \( t \) acts as a parameter that changes the position of the vector throughout the curve.
• Curve Representation: Vector functions represent curves whose direction and position can be fully described by the function itself.

Understanding vector functions is crucial because they help in visualizing how a point moves in space, which can be paramount when dealing with problems involving motion, fields, or pathways.

Derivative

The derivative of a vector function provides key insights into the behavior of the curve, specifically, its rate of change. By differentiating the vector function with respect to its parameter (such as \( t \)), we can determine the rate at which the vector components are changing.

In the provided solution, the derivative is taken by differentiating each component of \( \mathbf{r}(t) \) separately:
\[ \mathbf{r}'(t) = \left\langle \frac{\mathrm{d}(\frac{t}{\sqrt{3}})}{\mathrm{d}t}, \frac{\mathrm{d}(\frac{t}{\sqrt{3}})}{\mathrm{d}t}, \frac{\mathrm{d}(\frac{t}{\sqrt{3}})}{\mathrm{d}t} \right\rangle = \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle \]
This differentiation results in a constant vector, indicating that the change in position is uniform along the curve.

The act of differentiating highlights:

• The velocity of the curve: The derivative vector effectively represents velocity, showing how fast the point is moving in each direction.
• Consistency: As shown here, vectors with constant derivatives suggest uniform motion. This is a crucial aspect, especially if we are working with arc length parameterization.
• Slope Interpretation: In simpler single-variable contexts, the derivative indicates the slope, whereas with vector functions, it gives a directional vector showing change magnitudes.

In summary, understanding derivatives of vector functions allows us to analyze speed, direction, and nature of motion along paths represented by vector functions.

Magnitude Calculation

The magnitude, or length, of a vector gives an idea of how large the vector is. For vector functions, finding the magnitude of the derivative can tell us about the 'speed' at which a point is moving along the curve.

When we calculate the magnitude of the derivative \( \mathbf{r}'(t) \) in this exercise, we follow these steps:
\[ ||\mathbf{r}'(t)|| = \sqrt{\left( \frac{1}{\sqrt{3}} \right)^2+\left( \frac{1}{\sqrt{3}} \right)^2+\left( \frac{1}{\sqrt{3}} \right)^2} = \sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}} = \sqrt{1} = 1 \]
The calculations show that the magnitude is 1, which provides an essential clue: the parameterized curve uses arc length as its parameter.

Main points in magnitude calculation include:

• Sum of Squares: For a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \), its magnitude is \( \sqrt{a_1^2 + a_2^2 + a_3^2} \).
• Consistency with Arc Length: The fact that our magnitude equals 1 shows that the curve intrinsically matches the definition of arc length, meaning it travels one unit per unit change of the parameter \( t \).
• Velocity Interpretation: Magnitude represents speed in physics, asserting how fast an object is moving irrespective of direction.

Mastering magnitude calculations equips students with an analytical tool for exploring vector lengths and behaviors on multidimensional paths.