Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \int_{C} y d x+x d y ; C \text { is the curve } y=x^{2}, 0 \leq x \leq 1 $$

Step-by-Step Solution

Answer

The line integral evaluates to 1.

Step 1: Identify the vector field and parameterize the curve

The given line integral \( \int_{C} y \, dx + x \, dy \) corresponds to a vector field \( \mathbf{F}(x, y) = (y, x) \). The curve \( C \) is described by \( y = x^2 \) for \( 0 \leq x \leq 1 \). We can parameterize \( C \) by \( \mathbf{r}(t) = (t, t^2) \), where \( t \) ranges from 0 to 1.

Step 2: Compute the derivatives of parameterization

From the parameterization, \( \mathbf{r}(t) = (t, t^2) \), calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). We find that \( \frac{dx}{dt} = 1 \) and \( \frac{dy}{dt} = 2t \).

Step 3: Substitute parameterization in the integral

Substitute \( x = t \) and \( y = t^2 \) into the line integral \( \int (y \, dx + x \, dy) \). This yields \( \int (t^2 \, 1 \, dt + t \, 2t \, dt) = \int (t^2 + 2t^2) \, dt = \int 3t^2 \, dt \).

Step 4: Integrate with respect to parameter t

Evaluate the integral \( \int_0^1 3t^2 \, dt \). This simplifies to \( 3 \int_0^1 t^2 \, dt = 3 \left[ \frac{t^3}{3} \right]_0^1 \).

Step 5: Calculate the definite integral

The expression \( \left[ \frac{t^3}{3} \right]_0^1 \) evaluates to \( \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \). Multiply this result by 3 to obtain the final result: 1.

Key Concepts

Understanding Vector Fields

A vector field is a way to assign a vector to each point in space. Imagine the wind blowing through a field where at each point, you can represent the wind direction and strength as a vector. This vector might point north, east, or any direction depending on the wind’s behavior.
In mathematics, vector fields are often represented as functions. For example, the vector field \( \mathbf{F}(x, y) = (y, x) \) given in our exercise represents a field where the vector at each point \((x, y)\) is \((y, x)\).
Understanding these fields helps us solve problems where quantities have both direction and magnitude, and makes it easier to interpret how these interact over curves like those found in line integrals.

• Vector fields are represented by functions assigning vectors (directions and magnitudes) to each point.
• They can be visualized as arrows over a plane showing these direction and strength.
• In this context, vector fields help in visualizing and computing line integrals efficiently.

Curve Parameterization Explained

Curve parameterization is a powerful tool to describe curves algebraically. It involves representing a curve by a set of equations that identify all points along the curve using a parameter, usually denoted as \( t \). In our example, the parameterization \( \mathbf{r}(t) = (t, t^2) \) describes the curve \( y = x^2 \) as \( t \) ranges from 0 to 1.
This process simplifies line integrals and other operations on curves, translating them from their standard form to a more calculable framework. By parameterizing, we can define how each point on the curve relates to the parameter \( t \), allowing us to work in terms of \( t \) rather than \( x \) or \( y \).
This conversion is highly essential in advanced calculus for setting up integrals over a curve and understanding the nature of these paths.

• Parameterization represents curves as functions of a parameter \( t \).
• It provides a simple framework for expressing complex paths mathematically.
• Parameterization enables easier calculation of line integrals and other operations.

The Role of Definite Integrals

The definite integral is a fundamental concept in calculus used to calculate quantities that accumulate over an interval. It represents the "net area" under a curve within a specific range. In line integrals, definite integrals help determine the accumulated effect of a vector field along a path.
In our scenario, we compute the definite integral \( \int_0^1 3t^2 \, dt \), which represents the cumulative value of \( 3t^2 \) as \( t \) moves from 0 to 1. This offers a practical way of summing up the contributions at each point along the curve.
Definite integrals are crucial not just in mathematics, but in fields like physics and engineering to solve real-world problems involving continuous data.

• Definite integrals calculate the accumulated value over a specific range.
• They are crucial for finding net quantities, like areas under curves or work done by a field.
• They connect deeply with line integrals, summarizing vector field effects along a path.