Problem 108 - Calculus 3

Question

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 is \(\sqrt{26}\).

Step 1: Calculate the Derivative of the Vector Function

To find the arc length, we first need to find the derivative of the given vector function. The vector function is \(\mathbf{r}(t) = -t \mathbf{i} + 4t \mathbf{j} + 3t \mathbf{k}\). The derivative \(\mathbf{r}'(t)\) is found by differentiating each component with respect to \(t\). This gives us \(\mathbf{r}'(t) = -\mathbf{i} + 4 \mathbf{j} + 3 \mathbf{k}\).

Step 2: Determine the Magnitude of the Derivative

Next, compute the magnitude of the derivative \(\mathbf{r}'(t)\). We calculate this as the square root of the sum of the squares of its components: \[|\mathbf{r}'(t)| = \sqrt{(-1)^2 + (4)^2 + (3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26}.\]

Step 3: Set Up the Arc Length Integral

The arc length \(L\) of the vector-valued function from 0 to 1 is given by the integral of the magnitude of the derivative: \[ L = \int_0^1 |\mathbf{r}'(t)| \, dt = \int_0^1 \sqrt{26} \, dt. \]

Step 4: Evaluate the Integral

Since \(\sqrt{26}\) is a constant, the integral becomes \(\sqrt{26} \cdot \int_0^1 dt\). Evaluating, we get \[ L = \sqrt{26} \cdot [t]_0^1 = \sqrt{26} \cdot (1 - 0) = \sqrt{26}. \]

Step 5: Conclude the Solution

The arc length of the vector-valued function \(\mathbf{r}(t)\) from \(t = 0\) to \(t = 1\) is \(\sqrt{26}\).

Key Concepts

Vector-Valued Function

A vector-valued function is essentially a function that takes a number as its input and assigns a vector as its output. It's a great way to represent movement or paths in space using mathematical expressions. In our original exercise, the vector-valued function is given as \[ \mathbf{r}(t) = -t \mathbf{i} + 4t \mathbf{j} + 3t \mathbf{k} \] This function has three components, each corresponding to a different spatial dimension:

• The x-component is \(-t\),
• The y-component is \(4t\),
• The z-component is \(3t\).

Such functions are very useful in physics and engineering because they can describe the trajectory of a moving object in 3D space. Each component of the vector gives you information about the object's position along a particular axis, based on the variable \(t\), often representing time.

Derivative

The derivative of a vector-valued function measures how the function's output, that is, a vector, changes with respect to changes in its input. Differentiation of a vector function involves differentiating each component separately with respect to the input variable. For the given function \[ \mathbf{r}(t) = -t \mathbf{i} + 4t \mathbf{j} + 3t \mathbf{k}, \] its derivative is determined by differentiating each component:

• The derivative of \(-t \mathbf{i}\) is \(-\mathbf{i}\),
• The derivative of \(4t \mathbf{j}\) is \(4 \mathbf{j}\),
• The derivative of \(3t \mathbf{k}\) is \(3 \mathbf{k}\).

Thus, the derivative vector is:\[ \mathbf{r}'(t) = -\mathbf{i} + 4 \mathbf{j} + 3 \mathbf{k}. \] This derivative provides another vector that represents the velocity of the object whose movement is described by \(\mathbf{r}(t)\).

Magnitude of a Vector

The magnitude of a vector is akin to finding the "length" or "size" of the vector. For a 3-dimensional vector, it is the square root of the sum of the squares of its components. The derivative vector \[ \mathbf{r}'(t) = -\mathbf{i} + 4 \mathbf{j} + 3 \mathbf{k} \] has components

• -1 along \(\mathbf{i}\),
• 4 along \(\mathbf{j}\),
• 3 along \(\mathbf{k}\).

To find its magnitude, compute:\[ |\mathbf{r}'(t)| = \sqrt{(-1)^2 + (4)^2 + (3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26}. \] This magnitude represents the rate at which the object is moving along its path in 3D space.

Arc Length Integral

The arc length integral is a mathematical way of calculating the length of a curve represented by a vector-valued function. Once we have the magnitude of the derivative, we set up an integral over the given interval. In our case, the interval is \[ [0, 1]. \] The arc length \(L\) is determined as follows: \[ L = \int_0^1 |\mathbf{r}'(t)| \, dt. \] For our function, substitution gives: \[ L = \int_0^1 \sqrt{26} \, dt. \] Here, \(\sqrt{26}\) is a constant because the vector's speed doesn't change, allowing us to express and solve the integral simply as: \[ L = \sqrt{26} \cdot [t]_0^1 = \sqrt{26} \cdot (1-0) = \sqrt{26}. \] The arc length integral effectively sums up the tiny bits of distances over each tiny interval throughout the path, giving the total length of the curve from \(t = 0\) to \(t = 1\).