Question

Arc length parameterization Prove that the line \(\mathbf{r}(t)=\left\langle x_{0}+a t, y_{0}+b t, z_{0}+c t\right\rangle\) is parameterized by arc length provided \(a^{2}+b^{2}+c^{2}=1\).

Short answer

Question: Prove that the line \(\mathbf{r}(t)=\left\langle x_{0}+at, y_{0}+bt, z_{0}+ct\right\rangle\) is parameterized by arc length provided \(a^2+b^2+c^2=1\). Answer: The line \(\mathbf{r}(t)=\left\langle x_{0}+at, y_{0}+bt, z_{0}+ct\right\rangle\) is parameterized by arc length provided \(a^2+b^2+c^2=1\) because the magnitude of its derivative (speed) is constant and equal to 1.

Step-by-step solution

Calculate the derivative of the position vector \(\mathbf{r}(t)\).

To find the derivative of \(\mathbf{r}(t)\) with respect to the parameter \(t\), we simply need to differentiate each component of the vector with respect to \(t\). \(\frac{d\mathbf{r}}{dt} = \frac{d}{dt} \left\langle x_{0}+at, y_{0}+bt, z_{0}+ct \right\rangle = \left\langle a, b, c \right\rangle\)

Calculate the magnitude of the derivative of the position vector, also known as the speed.

Next, we will find the magnitude of the derivative \(\frac{d\mathbf{r}}{dt}\). We need to take the square root of the sum of the squares of its components. \(|\frac{d\mathbf{r}}{dt}| = |\left\langle a, b, c \right\rangle| = \sqrt{a^2 + b^2 + c^2}\)

Prove that the speed is 1.

Now, we will use the condition given in the problem, \(a^2 + b^2 + c^2 = 1\), to show that the magnitude of the speed is equal to 1. \(|\frac{d\mathbf{r}}{dt}| = \sqrt{a^2 + b^2 + c^2} = \sqrt{1} = 1\) So, the magnitude of the derivative of the position vector is 1, meaning the line \(\mathbf{r}(t)=\left\langle x_{0}+at, y_{0}+bt, z_{0}+ct\right\rangle\) is parameterized by arc length provided \(a^2+b^2+c^2=1\).

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and differential operators. Vectors are often used to describe physical quantities like position, velocity, and acceleration.
In many problems, such as the arc length parameterization, vector calculus provides a structured way to solve complex geometrical problems in three dimensions.
Understanding vector calculus involves components like:

• Position Vectors: These are vectors that define a point in space, typically written as \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \).


• Operations on Vectors: Key operations include addition, scalar multiplication, and finding derivatives.


• Derivatives: The derivative of a vector function is computed component-wise, allowing us to analyze changes in all directions.

By using these fundamental concepts of vector calculus, we explore how the parameterized path behaves and adapt our approach to understanding its properties, such as speed and direction.

Derivatives

In mathematics, the derivative measures how a function changes as its input changes. For a position vector \( \mathbf{r}(t) = \langle x_{0}+at, y_{0}+bt, z_{0}+ct \rangle \), derivatives are calculated by differentiating each component with respect to the parameter \( t \).

• Component-wise Differentiation: The derivative \( \frac{d\mathbf{r}}{dt} = \langle a, b, c \rangle \) showcases that differentiating a vector involves splitting it into its components and handling each separately.


• Physical Interpretation: This derivative represents velocity, describing how the position changes over time in 3D space.


• Application: In arc length parameterization, this derivative helps determine if the rate of change in position along the path is constant.

Understanding derivatives is crucial to analyzing vector functions, predicting behavior, and comprehending the motion's constant rate.

Speed Calculation

Speed is a measure of how fast an object is moving at a particular moment, and it can be calculated from the derivative of its position vector. In the context of a parameterized line \( \mathbf{r}(t) \), speed is determined by the magnitude of its derivative \( \frac{d\mathbf{r}}{dt} \) as \( |\frac{d\mathbf{r}}{dt}| \).

• Magnitude Calculation: Compute the magnitude using \( |\langle a, b, c \rangle| = \sqrt{a^2 + b^2 + c^2} \), which simplifies to 1 if \( a^2 + b^2 + c^2 = 1 \).


• Validation: In arc length parameterization, a speed of 1 confirms that the path is parameterized by the arc length. This means each unit of \( t \) corresponds to one unit of distance.


• Practical Insight: Constant speed denotes uniform movement, assisting in modeling linear and predictable trajectories in vector calculus problems.

Speed calculation is pivotal in understanding movement and ensuring the parameterization's correctness as highlighting the steady pace of traversal.