Problem 9 - Essential 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 curve reparametrized in terms of arc length is \(\mathbf{r}(s) = \frac{2s}{\sqrt{29}}\mathbf{i} + (1 - \frac{3s}{\sqrt{29}})\mathbf{j} + (5 + \frac{4s}{\sqrt{29}})\mathbf{k}\).

Step 1: Find the Derivative of the Curve

To reparametrize with respect to arc length, we first need the derivative of the curve \(\mathbf{r}(t) = 2t\mathbf{i} + (1-3t)\mathbf{j} + (5+4t)\mathbf{k}\). The derivative, \(\mathbf{r}'(t)\), is calculated as follows:\[\mathbf{r}'(t) = \frac{d}{dt}[2t\mathbf{i} + (1-3t)\mathbf{j} + (5+4t)\mathbf{k}] = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}.\]

Step 2: Compute the Magnitude of the Derivative

Next, find the magnitude of \(\mathbf{r}'(t)\) to use in calculating arc length:\[||\mathbf{r}'(t)|| = \sqrt{(2)^2 + (-3)^2 + (4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29}.\]

Step 3: Express Arc Length as a Function of \(t\)

The arc length \(s\) from \(t=0\) is given by the integral of \(||\mathbf{r}'(t)||\) with respect to \(t\). Since the magnitude is constant, the arc length \(s(t)\) is directly proportional to \(t\):\[s(t) = \int_0^t ||\mathbf{r}'(u)|| \, du = \int_0^t \sqrt{29} \, du = t\sqrt{29}.\]

Step 4: Solve for \(t\) as a Function of Arc Length \(s\)

We have the equation for arc length \(s = t\sqrt{29}\). To reparametrize the curve in terms of arc length \(s\), solve for \(t\):\[t = \frac{s}{\sqrt{29}}.\]

Step 5: Substitute \(t\) in Terms of \(s\) Back into \(\mathbf{r}(t)\)

Finally, substitute \(t = \frac{s}{\sqrt{29}}\) into the original parameterization to express the curve in terms of \(s\):\[\mathbf{r}(s) = 2\left(\frac{s}{\sqrt{29}}\right)\mathbf{i} + \left(1 - 3\left(\frac{s}{\sqrt{29}}\right)\right)\mathbf{j} + \left(5 + 4\left(\frac{s}{\sqrt{29}}\right)\right)\mathbf{k}.\]

Key Concepts

Vector Calculus

Vector calculus is an important branch of mathematics focusing on differentiation and integration of vector fields. These operations are essential in describing physical phenomena, such as fluid flow and electromagnetic fields.
In this context, we work with vector functions like \( \mathbf{r}(t) = 2t\mathbf{i} + (1-3t)\mathbf{j} + (5+4t)\mathbf{k} \), which describe paths or curves in three-dimensional space. A vector function maps each parameter \( t \) to a vector, giving the location of a point in the curve at any time \( t \).

To tackle vector calculus problems, especially those involving curve reparametrization, we use techniques such as:

• Finding derivatives of vector functions to get velocity vectors at given points on the curve.
• Computing magnitudes of these vectors to help in calculating lengths between points, also known as arc length.

Understanding these basic principles helps in solving more complex problems in physics and engineering.

Derivatives of Vector Functions

Derivatives in vector calculus extend the concepts of single-variable calculus to situations involving multiple dimensions and directions. By taking the derivative of a vector function, we essentially determine the rate and direction at which a point moves along the curve.
When given a vector function like \( \mathbf{r}(t) = 2t\mathbf{i} + (1-3t)\mathbf{j} + (5+4t)\mathbf{k} \), we find its derivative \( \mathbf{r}'(t) \) by differentiating each component separately.

Steps to Differentiate:

• Differentiate the \( i \)-component, \( 2t \), which results in \( 2 \).
• Differentiate the \( j \)-component, \( 1 - 3t \), resulting in \( -3 \).
• Differentiate the \( k \)-component, \( 5 + 4t \), resulting in \( 4 \).

After combining these results, the derivative of the vector function is \( \mathbf{r}'(t) = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k} \), representing the rate of change of position along the curve. This derivative is key in computing the arc length, which is essential for reparametrizing the curve.

Integration for Arc Length

Arc length, an important measure in vector calculus, tells us how far an object has traveled along a curve. Calculating it involves integrating the magnitude of the derivative of the vector function over the given interval.
To find the magnitude of \( \mathbf{r}'(t) = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k} \), we use the formula \( ||\mathbf{r}'(t)|| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{29} \).
This magnitude is constant, which simplifies our integration process considerably.

Integral Calculation:

• The arc length \( s \) from \( t = 0 \) can be expressed as: \( s(t) = \int_0^t \sqrt{29} \, du = t\sqrt{29} \).

This formula expresses arc length as a function of \( t \). Knowing \( s \) allows us to solve for \( t \), giving us the reparametrization \( t = \frac{s}{\sqrt{29}} \).
Thus, the curve \( \mathbf{r}(t) \) becomes reparametrized in terms of arc length \( s \) rather than \( t \), yielding \( \mathbf{r}(s) = 2\left(\frac{s}{\sqrt{29}}\right) \mathbf{i} + \left(1 - 3\left(\frac{s}{\sqrt{29}}\right)\right) \mathbf{j} + \left(5 + 4\left(\frac{s}{\sqrt{29}}\right)\right) \mathbf{k} \).
This transformation enhances the understanding of the curve's geometry by making its traversal length clear and consistent, irrespective of the chosen parameterization.