Problem 13 - Thomas Calculus

Question

In Exercises \(11-14,\) find the arc length parameter along the curve from the point where \(t=0\) by evaluating the integral $$ s=\int_{0}^{t}|\mathbf{v}(\tau)| d \tau $$ from Equation \((3) .\) Then find the length of the indicated portion of the curve. $$ \mathbf{r}(t)=\left(e^{t} \cos t\right) \mathbf{i}+\left(e^{t} \sin t\right) \mathbf{j}+e^{t} \mathbf{k}, \quad-\ln 4 \leq t \leq 0 $$

Step-by-Step Solution

Answer

The length of the curve is \( \frac{3\sqrt{3}}{4} \).

Step 1: Find Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of \( \mathbf{r}(t) \). Given \( \mathbf{r}(t) = \left(e^{t} \cos t\right) \mathbf{i} + \left(e^{t} \sin t\right) \mathbf{j} + e^{t} \mathbf{k} \), we find \( \mathbf{v}(t) = \frac{d}{dt}\mathbf{r}(t) = \left(e^{t}(\cos t - \sin t)\right) \mathbf{i} + \left(e^{t}(\sin t + \cos t)\right) \mathbf{j} + e^{t} \mathbf{k} \).

Step 2: Calculate Magnitude of Velocity

The magnitude \( |\mathbf{v}(t)| \) is calculated as: \(|\mathbf{v}(t)| = \sqrt{\left(e^t(\cos t - \sin t)\right)^2 + \left(e^t (\sin t + \cos t)\right)^2 + \left(e^t\right)^2}\). Simplifying gives \( \sqrt{3}e^t \).

Step 3: Set Up the Integral for Arc Length

The arc length \( s \) is given by \( s = \int_{0}^{t} |\mathbf{v}(\tau)| \, d\tau \). Substitute \(|\mathbf{v}(\tau)| = \sqrt{3}e^\tau \) to get \( s = \int_{0}^{t} \sqrt{3}e^\tau \, d\tau \).

Step 4: Evaluate the Integral

Evaluate \( s = \int_{0}^{t} \sqrt{3}e^\tau \, d\tau \):\(s = \sqrt{3} \left[ e^\tau \right]_{0}^{t} = \sqrt{3}(e^t - e^0) = \sqrt{3}(e^t - 1)\).

Step 5: Calculate the Length of the Curve

To find the length of the curve from \( t = -\ln 4 \) to \( t = 0 \), substitute these limits into the arc length expression:\( L = \sqrt{3}(e^{0} - e^{-\ln 4}) = \sqrt{3}\left(1 - \frac{1}{4}\right) = \frac{3\sqrt{3}}{4}.\)

Key Concepts

Parametric Curves

A parametric curve is a type of curve expressed by multiple functions. Each function describes a coordinate (e.g., x, y, z) in terms of a parameter, usually denoted as \( t \). This is different from regular curves where the relationship is usually between just \( x \) and \( y \).
With parametric curves, you have:

• \( x(t) \): describes the curve's x-coordinate
• \( y(t) \): describes the y-coordinate
• \( z(t) \): describes the z-coordinate when in 3D

Each of these acts together to plot points along the curve as \( t \) changes. This allows for representing more complex paths, like spirals, loops, and more, which aren't readily visible in standard Cartesian coordinates. In our example, the curve \( \mathbf{r}(t) = (e^t \cos t) \mathbf{i} + (e^t \sin t) \mathbf{j} + e^t \mathbf{k} \) is a path in three-dimensional space, showing how each unit vector \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) changes with time.

Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is a key concept in understanding the motion along a parametric curve. It is derived by taking the derivative of the position vector \( \mathbf{r}(t) \). The derivative provides us with a new vector that indicates not just the direction of movement, but how fast an object is moving along a curve at any point in time.
In simple terms:

• If the curve is \( \mathbf{r}(t) \), then \( \mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) \).
• This means computing the derivative separately for each coordinate.

In our specific exercise, the velocity vector \( \mathbf{v}(t) = \left(e^{t} ( \cos t - \sin t)\right) \mathbf{i} + \left(e^{t} (\sin t + \cos t)\right) \mathbf{j} + e^{t} \mathbf{k} \), gives insight into the path and speed as influenced by exponential and trigonometric functions. Understanding the velocity vector helps in finding the arc length parameter and assessing the curve's dynamics.

Integral Calculus

Integral calculus is crucial for calculating the arc length of parametric curves. The integral helps sum up infinitely small lengths along the curve to find the total distance traveled from one point to another.
In this context:

• Arc length \( s \) is found using the integral \( s = \int_0^t | \mathbf{v}(\tau)| \, d\tau \).
• This means you integrate the magnitude of the velocity vector over your desired interval.

Why is this important? It allows us to account for every small segment along the curve, giving an accurate measurement of distance. In exercises like this, evaluating \( s = \int_0^t \sqrt{3} e^\tau \, d\tau \) requires understanding integral properties and recognizing the antiderivative of the function within the integrand. Once integrated, the result \( s = \sqrt{3}(e^t - 1) \) gives the arc length from the starting point up to any value of \( t \).

Magnitude of Velocity

The magnitude of a velocity vector \( | \mathbf{v}(t) | \) adds a scalar dimension to understanding how fast and in what manner an object is traveling along a path. Essentially, it strips away direction to reveal sheer speed at any given moment.
The magnitude is computed using the formula:

• \( | \mathbf{v}(t) | = \sqrt{(v_x(t))^2 + (v_y(t))^2 + (v_z(t))^2} \)

For our velocity vector \( \mathbf{v}(t) \), calculating this leads to:

• \( | \mathbf{v}(t) | = \sqrt{3} e^{t} \).

This expression demonstrates how everything, even kinetic intensity, is intertwined with the parameter \( t \). The exponential terms reveal that speed changes exponentially with respect to \( t \), which is critical for determining overall path length.
This is used not only for ark length but also for comprehending dynamics such as acceleration or deceleration as one moves along a trajectory. The magnitude provides a full grasp of the energy and momentum involved as defined by the velocity function in parametric space.