Problem 13 - Calculus Early Transcendentals

Question

Reparametrize the curve with respect to arc length measured from the point where \(t=0\) in the direction of increasing \(t.\) \(\mathbf{r}(t)=2 t \mathbf{i}+(1-3 t) \mathbf{j}+(5+4 t) \mathbf{k}\)

Step-by-Step Solution

Answer

The reparameterized curve is \(\frac{2s}{\sqrt{29}} \mathbf{i} + \left(1 - \frac{3s}{\sqrt{29}}\right) \mathbf{j} + \left(5 + \frac{4s}{\sqrt{29}}\right) \mathbf{k}.\)

Step 1: Compute the Derivative of the Curve

First, find the derivative of the vector function with respect to parameter \(t\). Given the vector function \(\mathbf{r}(t) = 2t\mathbf{i} + (1-3t)\mathbf{j} + (5+4t)\mathbf{k}\), compute \(\mathbf{r}'(t)\):\[\mathbf{r}'(t) = \frac{d}{dt}(2t)\mathbf{i} + \frac{d}{dt}(1-3t)\mathbf{j} + \frac{d}{dt}(5+4t)\mathbf{k} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}.\]

Step 2: Find the Magnitude of the Derivative

Next, calculate the magnitude (norm) of the derivative found in Step 1. This will give us the speed of the parameterized curve:\[||\mathbf{r}'(t)|| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}.\]

Step 3: Set Up the Arc Length Function

The arc length \(s(t)\) from the point where \(t=0\) is given by the integral of the speed:\[s(t) = \int_0^t ||\mathbf{r}'(u)|| \, du = \int_0^t \sqrt{29} \, du = \sqrt{29} t.\]

Step 4: Solve for Parametric Variable in Terms of Arc Length

We need to express \(t\) in terms of \(s\). From Step 3, we have \(s = \sqrt{29}t\). Solving for \(t\) gives:\[t = \frac{s}{\sqrt{29}}.\]

Step 5: Reparameterize the Curve

Substitute \(t = \frac{s}{\sqrt{29}}\) into the original vector function \(\mathbf{r}(t)\):\[\begin{align*}\mathbf{r}(s) & = 2\left(\frac{s}{\sqrt{29}}\right) \mathbf{i} + \left(1 - 3\left(\frac{s}{\sqrt{29}}\right)\right) \mathbf{j} + \left(5 + 4\left(\frac{s}{\sqrt{29}}\right)\right) \mathbf{k} \& = \frac{2s}{\sqrt{29}} \mathbf{i} + \left(1 - \frac{3s}{\sqrt{29}}\right) \mathbf{j} + \left(5 + \frac{4s}{\sqrt{29}}\right) \mathbf{k}.\end{align*}\]

Key Concepts

Arc Length

In vector calculus, arc length is a fundamental concept when dealing with curves. It is the measure of the distance traveled along a curve from one point to another.

To compute the arc length of a parametric curve defined by a vector function, we use an integral to sum up infinitesimally small line segments along the curve.

• First, we find the derivative of the vector function. This gives the velocity vector of the curve, indicating how the position changes with the parameter.
• Next, we calculate the magnitude of this velocity vector, which represents the speed of the curve at any point along its path.
• Finally, we integrate this speed over the desired interval to find the total arc length.

In our example, the arc length function is derived from the constant speed along the curve, thus simplifying the reparametrization process. Arc length helps in understanding real-world problems, such as finding the distance a particle travels over a period of time.

Vector Calculus

Vector calculus extends the concepts of ordinary calculus into multiple dimensions, allowing for the exploration of curves, surfaces, and manifolds.

In vector calculus, curves are often defined by parametric equations, where each coordinate of a point on the curve is expressed as a function of one or more variables called parameters.

• The derivative of a vector function is an essential tool. It helps in determining the behavior of curves and their respective properties, like tangent vectors and rate of change.
• The magnitude of this derivative represents the speed of a moving point along a curve, crucial for calculating arc length and for understanding the kinematics of the motion.
• The reparametrization of curves helps in transforming the parameter from time to be proportional to arc length, simplifying calculations and interpretations.

Vector calculus is a powerful tool used in physics and engineering to solve problems related to motion, heat, electromagnetism, and more.

Parametric Equations

Parametric equations describe curves by expressing each coordinate as a function of a parameter, typically denoted by variables such as \(t\) or \(s\). These equations provide a flexible way to model curves that might be cumbersome to express using standard Cartesian equations.

• A parametric representation of a curve allows for more control over the curve's construction, as each coordinate's behavior can be independently defined, reflecting changes over time or another variable.
• Parametric equations are especially useful for representing curves that double back on themselves or can't be described by a single function, like circles or ellipses.
• Reparametrization with respect to arc length, as seen in the exercise, is an important technique to achieve a uniform representation, where the parameter directly relates to the distance along the curve.

By converting a parameter, like time, into arc length, we can simplify complex integrals and foster a deeper understanding of the curve's geometry and dynamics.