Question

Determine whether the following curves use arc length as a parameter. If not, find a description that uses arc length as a parameter. $$\mathbf{r}(t)=\left\langle\frac{\cos t}{\sqrt{2}}, \frac{\cos t}{\sqrt{2}}, \sin t\right\rangle, \text { for } 0 \leq t \leq 10$$

Short answer

Answer: Yes, the given parametric curve uses arc length as a parameter.

Step-by-step solution

Compute the derivative of the given curve

To find if the given curve uses arc length as a parameter, first compute the derivative of \(\mathbf{r}(t)\) with respect to \(t\). $$\frac{d\mathbf{r}}{dt}=\left\langle-\frac{\sin t}{\sqrt{2}},-\frac{\sin t}{\sqrt{2}},\cos t\right\rangle$$

Calculate the magnitude of the derivative

Now, find the magnitude of the derivative \(\frac{d\mathbf{r}}{dt}\): $$\left\lVert\frac{d\mathbf{r}}{dt}\right\rVert=\sqrt{\left(-\frac{\sin t}{\sqrt{2}}\right)^2+\left(-\frac{\sin t}{\sqrt{2}}\right)^2+\left(\cos t\right)^2}$$ After simplifying, we get: $$\left\lVert\frac{d\mathbf{r}}{dt}\right\rVert=\sqrt{\frac{\sin^2t}{2}+\frac{\sin^2t}{2}+\cos^2t}$$ Further simplification: $$\left\lVert\frac{d\mathbf{r}}{dt}\right\rVert=\sqrt{\sin^2t+\cos^2t}$$ Since \(\sin^2t+\cos^2t=1\), we have: $$\left\lVert\frac{d\mathbf{r}}{dt}\right\rVert=\sqrt{1}=1$$

Conclusion

Since the derivative of the given curve has a magnitude of 1, the curve \(\mathbf{r}(t)=\left\langle\frac{\cos t}{\sqrt{2}}, \frac{\cos t}{\sqrt{2}}, \sin t\right\rangle\) for \(0 \leq t \leq 10\) already uses arc length as a parameter. There is no need to reparameterize the curve using arc length.

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and operations on these fields. Vectors are quantities that have both magnitude and direction, making them extremely useful in representing physical quantities like force and velocity. In this context, vector calculus focuses on functions that output vectors rather than scalars.
Understanding the derivative of a vector function can be quite insightful. It allows us to explore changes over time, like how an object's trajectory is modified by its speed and direction. Computing the derivative of a vector function, such as the curve given by \( \mathbf{r}(t) \), is a foundational step in determining properties like velocity and acceleration in physical spaces. For the curve \( \mathbf{r}(t)=\left\langle\frac{\cos t}{\sqrt{2}}, \frac{\cos t}{\sqrt{2}}, \sin t\right\rangle \), this concept becomes pivotal when checking if it uses arc length parameterization.

Key points to remember in vector calculus include:

• The derivative provides the rate of change and is critical for studying dynamic systems.
• Magnitude and direction are focal points that determine the behavior of vector functions.

Parameterization

Parameterization in mathematics is a strategy used to express a set of equations or a curve. This involves defining one or more variables (parameters) that describe the system or curve. In the case of the curve \( \mathbf{r}(t) \), the parameter is \( t \), which implies that each component of the curve depends on \( t \).

The choice of parameterization can significantly affect calculations and results. For example, many practical applications utilize arc length as a parameter. This is advantageous due to its straightforward association with physical distance. A curve is said to be parameterized by arc length if the derivative of the curve with respect to the parameter has a constant magnitude of 1.
Here are key aspects of parameterization:

• Choosing an appropriate parameter can simplify complex problems.
• Arc length parameterization offers a direct correlation with physical distances, enhancing intuitive understanding.
• It is important to compute the derivative's magnitude to verify arc length parameterization.

Derivative Magnitude

The concept of derivative magnitude in vector calculus involves calculating the "size" of the rate of change of a vector function. For a vector \( \mathbf{v}(t) = \left\langle v_1(t), v_2(t), v_3(t) \right\rangle \), its magnitude is given by the formula \( \sqrt{v_1(t)^2 + v_2(t)^2 + v_3(t)^2} \).
In this given problem, calculating the magnitude of the derivative \( \frac{d\mathbf{r}}{dt} \) is crucial. The goal is to verify whether \( \mathbf{r}(t) \) is already parameterized by arc length, implying that \( \left\lVert \frac{d\mathbf{r}}{dt} \right\rVert \) equals 1. Since the component calculations lead to the identity \( \sin^2 t + \cos^2 t = 1 \), this confirms that the derivative's magnitude is 1 across the specified range.
Importance of derivative magnitude includes:

• It reflects the rate at which the curve is traversed.
• Consistency in the magnitude often indicates parameterization by arc length.
• It provides insights into the geometric properties of the curve.

Curve Analysis

Curve analysis utilizes mathematical tools to explore the properties and features of a curve in space. By analyzing curves, we obtain insights about their behavior, shape, and nature.

For a curve like \( \mathbf{r}(t) \), we examine whether it uses arc length as its parameter. The analysis involves computing derivatives and their magnitudes to ensure uniform speed—that is, a constant rate of change concerning \( t \). It's an essential step in understanding motion around a curved path because it helps in optimization and accurate modeling of real-world phenomena.
Benefits of curve analysis include:

• Providing detailed information about the curvature and geometry.
• Allowing for determining the best parameterization method.
• Helping predict future behavior based on current path analysis.