Problem 14 - 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. $$ \mathbf{r}(t)=(1+2 t) \mathbf{i}+(1+3 t) \mathbf{j}+(6-6 t) \mathbf{k}, \quad-1 \leq t \leq 0 $$

Step-by-Step Solution

Answer

The arc length of the curve from \( t = -1 \) to \( t = 0 \) is 7.

Step 1: Find the velocity vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to \( t \). The position vector is given by \( \mathbf{r}(t) = (1+2t)\mathbf{i} + (1+3t)\mathbf{j} + (6-6t)\mathbf{k} \). Differentiate each component:\[ \mathbf{v}(t) = \frac{d}{dt}((1+2t)\mathbf{i} + (1+3t)\mathbf{j} + (6-6t)\mathbf{k}) = 2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}. \]

Step 2: Compute the magnitude of the velocity

To find \( |\mathbf{v}(t)| \), compute the magnitude of the velocity vector:\[ |\mathbf{v}(t)| = \sqrt{(2)^2 + (3)^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7. \]

Step 3: Set up the arc length integral

The arc length \( s \) from the point where \( t=0 \) is given by \[ s = \int_{0}^{t} |\mathbf{v}(\tau)| \, d\tau. \]Substitute the magnitude \( |\mathbf{v}(t)| = 7 \) into the integral:\[ s = \int_{0}^{t} 7 \, d\tau. \]

Step 4: Evaluate the integral

Evaluate the integral:\[ s = \int_{0}^{t} 7 \, d\tau = [7\tau]_{0}^{t} = 7t. \]

Step 5: Calculate the total arc length over the interval

Now find the total arc length from \( t = -1 \) to \( t = 0 \). Substitute \( t = -1 \) and \( t = 0 \) into \( s = 7t \), then compute the difference:\[ s(0) - s(-1) = 7(0) - 7(-1) = 0 + 7 = 7. \]

Key Concepts

Velocity Vector

When we talk about a velocity vector in the context of calculus and motion, we're referring to the rate at which the position changes with time along a given path in space. In practical terms, the velocity vector, denoted as \( \mathbf{v}(t) \), provides us with both speed and direction of an object at any instance \( t \).
For a curve defined by a position vector \( \mathbf{r}(t) \), the velocity vector is essentially the derivative \( \mathbf{v}(t) = \frac{d \mathbf{r}(t)}{dt} \).
- This derivative indicates how position changes as time progresses, giving insight into the movement dynamics of the point on the curve.
Let's break this down with an example: Given a position vector \( \mathbf{r}(t) = (1+2t)\mathbf{i} + (1+3t)\mathbf{j} + (6-6t)\mathbf{k} \), we find the velocity by differentiating each component:
- \( \mathbf{v}(t) = 2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k} \).
- This tells us that for every unit of time, the point's horizontal movement (rate of change in \( \mathbf{i} \) direction) increases by 2, vertically by 3, but it moves downward in the \( \mathbf{k} \) direction at a rate of 6.

Position Vector

The position vector is a fundamental concept in vector calculus that describes the location of a point in space as a function of time \( t \). It is essentially a vector that extends from the origin of the coordinate system to the point \( (x(t), y(t), z(t)) \).
For instance, in our example, the position vector is denoted by \( \mathbf{r}(t) = (1+2t)\mathbf{i} + (1+3t)\mathbf{j} + (6-6t)\mathbf{k} \). This tells us exactly where a point is located at any given value of \( t \):
- The \( 1+2t \) in \( \mathbf{i} \) component represents the x-coordinate as a function of time.
- The \( 1+3t \) in \( \mathbf{j} \) component represents the y-coordinate as a function of time.
- The \( 6-6t \) in \( \mathbf{k} \) component represents the z-coordinate, which decreases as \( t \) increases.
Understanding the position vector allows us to visualize and analyze the path that a point traces in 3-dimensional space over time.
- It is foundational for deriving other vectors such as the velocity and acceleration vectors, providing a base from which these are calculated.

Arc Length Integral

The arc length integral is an important tool for determining the distance along a curve. To compute the arc length, we integrate the magnitude of the velocity vector over the interval of interest.
The formula for arc length \( s \) from an initial point where \( t = 0 \) is given by the integral:
- \( s = \int_{0}^{t} |\mathbf{v}(\tau)| \, d\tau \).
- This expression computes the total length of the path traveled by a point moving along the curve, from the starting point up to time \( t \).
In our example, we first found the magnitude of the velocity vector:
- \( |\mathbf{v}(t)| = 7 \).
- The arc length integral becomes \( s = \int_{0}^{t} 7 \, d\tau \).
After evaluating this integral, we find that the arc length is \( s = 7t \).
- Specifically, for the interval \( -1 \leq t \leq 0 \), the total arc length is calculated as a straightforward difference, \( 7(0) - 7(-1) = 7 \).
- Thus, the total arc length of this portion of the curve is 7 units.
Understanding the arc length integral helps in measuring how far an object has traveled over a certain time period, providing a practical handle on the physical journey of moving along curved paths.