Problem 17 - Calculus III

Question

Find the arc length of the curve on the given interval.Find the arc length of the vector-valued function \(\mathbf{r}(t)=-t \mathbf{i}+4 t \mathbf{j}+3 t \mathbf{k}\) over \([0,1]\).

Step-by-Step Solution

Answer

The arc length of the curve over the interval \([0, 1]\) is \(\sqrt{26}\).

Step 1: Recall the formula for arc length of a vector-valued function

The arc length of a vector-valued function \(\mathbf{r}(t)\) from \(t = a\) to \(t = b\) is given by \(L = \int_{a}^{b} \| \mathbf{r}'(t) \| \, dt\), where \(\mathbf{r}'(t)\) is the derivative of \(\mathbf{r}(t)\) and \(\| \mathbf{r}'(t) \|\) is the magnitude of the derivative.

Step 2: Differentiate the vector-valued function

Given \(\mathbf{r}(t) = -t \mathbf{i} + 4t \mathbf{j} + 3t \mathbf{k}\), find the derivative \(\mathbf{r}'(t)\). The derivative is \(\mathbf{r}'(t) = -\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}\).

Step 3: Calculate the magnitude of the derivative

The magnitude of \(\mathbf{r}'(t)\) is \(\| \mathbf{r}'(t) \| = \sqrt{(-1)^2 + 4^2 + 3^2} = \sqrt{1 + 16 + 9} = \sqrt{26}\).

Step 4: Setup the integral for the arc length

Now substitute \(\| \mathbf{r}'(t) \|\) into the integral formula. The arc length is \(L = \int_{0}^{1} \sqrt{26} \, dt\).

Step 5: Evaluate the integral

The integral \(L = \int_{0}^{1} \sqrt{26} \, dt = \sqrt{26} [t]_{0}^{1}\) evaluates to \(\sqrt{26} (1 - 0) = \sqrt{26}\).

Key Concepts

Understanding Vector-Valued Functions

A vector-valued function, such as \( \mathbf{r}(t) = -t \mathbf{i} + 4t \mathbf{j} + 3t \mathbf{k} \), represents a curve in three-dimensional space where each component (\( \mathbf{i}, \mathbf{j}, \mathbf{k} \)) is a function of the parameter \( t \). This allows us to trace a path in three dimensions by varying \( t \) over a given interval. The curve is composed of a series of vectors, each defined by the particular value of \( t \). In this case, the function describes how the vector changes direction and length as \( t \) moves from 0 to 1. This change is what lets us calculate the arc length, the total distance traveled along the path of the vector.

The Derivative of a Vector Function

To find the arc length, we first need to compute the derivative of the vector-valued function, denoted as \( \mathbf{r}'(t) \). The derivative can be thought of as the vector of rates of change for each component of the original vector function. For each component function (e.g., \( -t, 4t, 3t \)), we take the standard derivative with respect to \( t \). Thus, the derivative of our function \( \mathbf{r}(t) = -t \mathbf{i} + 4t \mathbf{j} + 3t \mathbf{k} \) is \( \mathbf{r}'(t) = -\mathbf{i} + 4\mathbf{j} + 3\mathbf{k} \).

• \( -t \) differentiates to \( -1 \)
• \( 4t \) differentiates to \( 4 \)
• \( 3t \) differentiates to \( 3 \)

This derivative tells us the instantaneous rate of change or the direction and magnitude of movement at any point \( t \) along the curve.

Calculating the Magnitude of a Vector

Once we have the derivative \( \mathbf{r}'(t) \), we need to compute its magnitude, sometimes called its length. The magnitude of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \) is given by the formula \( \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \). This represents the "length" of the vector, providing a scalar quantity. In our example for \( \mathbf{r}'(t) = -\mathbf{i} + 4\mathbf{j} + 3\mathbf{k} \):

• Calculate \( (-1)^2 + 4^2 + 3^2 \)
• Add the results: \( 1 + 16 + 9 = 26 \)
• The magnitude is \( \sqrt{26} \)

This magnitude \( \sqrt{26} \) will be used in the integral to find the arc length.

Performing Integral Evaluation

The arc length of the curve defined by the vector-valued function is computed through an integral that accounts for the length of \( \mathbf{r}'(t) \) over the interval. For our specific problem, this means finding the length from \( t = 0 \) to \( t = 1 \).The formula for the arc length is:\[ L = \int_{a}^{b} \| \mathbf{r}'(t) \| \, dt \]Using the previously calculated magnitude of \( \mathbf{r}'(t) \), the integral setup becomes:\[ L = \int_{0}^{1} \sqrt{26} \, dt \]Since \( \sqrt{26} \) is a constant, the integral simplifies to:\[ L = \sqrt{26} \times [t]_{0}^{1} = \sqrt{26} \times (1 - 0) \]The length of the arc traced by the parameter \( t \) over the interval is \( \sqrt{26} \). This is the total distance traveled along the curve from start to finish.