Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the length of the curve with the given vector equation. $$ \mathbf{r}(t)=t^{2} \mathbf{i}-2 t^{3} \mathbf{j}+6 t^{3} \mathbf{k} ; 0 \leq t \leq 1 $$

Step-by-Step Solution

Answer

The curve length is \( \frac{2}{270} \times 272 \).

Step 1: Understand the Problem

We need to find the length of the curve given by the vector function \( \mathbf{r}(t) = t^2 \mathbf{i} - 2t^3 \mathbf{j} + 6t^3 \mathbf{k} \) over the interval \( 0 \leq t \leq 1 \). To do this, we will use the formula for the arc length of a vector-valued function, which is given by \( L = \int_{a}^{b} \left\| \mathbf{r}'(t) \right\| \, dt \).

Step 2: Find the Derivative

Calculate the derivative \( \mathbf{r}'(t) \) of the vector function. The derivative is \( \mathbf{r}'(t) = \frac{d}{dt}(t^2 \mathbf{i} - 2t^3 \mathbf{j} + 6t^3 \mathbf{k}) = 2t \mathbf{i} - 6t^2 \mathbf{j} + 18t^2 \mathbf{k} \).

Step 3: Compute the Magnitude of the Derivative

Find the magnitude of the derivative: \[\left\| \mathbf{r}'(t) \right\| = \sqrt{(2t)^2 + (-6t^2)^2 + (18t^2)^2} = \sqrt{4t^2 + 36t^4 + 324t^4}.\]Simplify this expression to obtain:\[\left\| \mathbf{r}'(t) \right\| = \sqrt{4t^2 + 360t^4}.\]

Step 4: Simplify the Magnitude Expression

Further simplify \( \left\| \mathbf{r}'(t) \right\| \):\[\left\| \mathbf{r}'(t) \right\| = \sqrt{4t^2 + 360t^4} = 2t \sqrt{1 + 90t^2}.\]

Step 5: Set Up the Integral for Arc Length

Use the expression for the magnitude to set up the integral for the arc length: \[L = \int_{0}^{1} 2t \sqrt{1 + 90t^2} \, dt.\]

Step 6: Solve the Integral

To solve the integral \( \int_{0}^{1} 2t \sqrt{1 + 90t^2} \, dt \), use a substitution such as \( u = 1 + 90t^2 \), with \( du = 180t \, dt \), hence \( t \, dt = \frac{du}{180} \). Transform the limits of integration, and calculate the integral. The substitution leads to new limits \( t = 0 \to u = 1 \) and \( t = 1 \to u = 91 \):\[L = \int_{1}^{91} \frac{2}{180} \sqrt{u} \, du = \frac{1}{90} \int_{1}^{91} u^{1/2} \, du.\]Evaluate this integral.

Step 7: Evaluate the Integral

Evaluate \( \frac{1}{90} \int_{1}^{91} u^{1/2} \, du \):\[\frac{1}{90} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{91} = \frac{2}{270} \left( 91^{3/2} - 1^{3/2} \right).\] Calculate this value to get the arc length.

Step 8: Final Calculation

Compute the final numerical value:\[L = \frac{2}{270} (273 - 1) = \frac{2}{270} \times 272.\] Simplify to find the final result.

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with differentiation and integration of vector fields, especially in three-dimensional Euclidean space. It extends the concepts of calculus, such as derivatives and integrals, to vector fields, which are functions that have vectors as their output. In this exercise, we explored the concept of arc length for a curve that is defined using a vector function. The arc length provides the actual distance traveled by a point moving along a path between two points on the curve.

• The vector function in this problem, \( \mathbf{r}(t) = t^2 \mathbf{i} - 2t^3 \mathbf{j} + 6t^3 \mathbf{k} \), describes a curve in space.
• Our goal is to determine the length of this curve over a specified interval.
• This involves calculating derivatives and integrating, as stipulated by vector calculus.

Understanding the vector function and its derivatives plays a significant role in how we calculate arc length in a multi-dimensional space.

Derivative Calculation

Derivatives are fundamental in calculus as they provide rates of change or slopes of tangent lines. For vector functions, derivatives help determine how the function changes in its vector components, or dimensions. In our exercise, we begin by calculating the derivative of the vector function \( \mathbf{r}(t) \).

• Given: \( \mathbf{r}(t) = t^2 \mathbf{i} - 2t^3 \mathbf{j} + 6t^3 \mathbf{k} \)
• Find \( \mathbf{r}'(t) \) by differentiating each component separately:
• Result: \( \mathbf{r}'(t) = 2t \mathbf{i} - 6t^2 \mathbf{j} + 18t^2 \mathbf{k} \)

Taking the derivative of vector functions involves applying standard derivative rules to each component, allowing us to analyze the instantaneous rate of change at any point along the curve.

Integration Techniques

Integration is the mathematical process of finding the integral, representing the area under a curve or the accumulation of quantities. In vector calculus, integrating the magnitude of a derivative over a specified interval gives us the arc length of a curve. We applied integration techniques to evaluate the arc length using the formula: \( L = \int_{0}^{1} 2t \sqrt{1 + 90t^2} \, dt \).

• This exercise used substitution as the preferred technique, setting \( u = 1 + 90t^2 \).
• This changed the differential, requiring adjustment to \( du = 180t \, dt \).
• Transform the original integral from \( t \) terms to \( u \) terms & evaluate between new limits.
• Integrate \( \int_{1}^{91} u^{1/2} \, du \) which yields an antiderivative, simplifying the problem.

Mastering various integration techniques like substitution is key to solving integrals, especially when they involve more complex functions or transformations.

Vector Functions

Vector functions map real numbers to vectors, often used to represent curves or paths in space. A vector function \( \mathbf{r}(t) \) assigns a vector to each point in its domain, making it a central tool in describing motion and change in physics and engineering. In this exercise, our vector function describes a curve with components in three-dimensional space.

• \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) are standard unit vectors in the Cartesian coordinate system.
• The vector equation \( \mathbf{r}(t) \) signifies position as a function of time \( t \).
• Understanding \( \mathbf{r}(t) \) helps us evaluate both the curve's shape and its physical length.

Using vector functions, we can beautifully encapsulate complex curves and trajectories in space, and derive meaningful measurements from them, like arc length.