Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\int_{C} y d x+x^{2} d y\); \(C\) is the curve \(x=2 t, \quad y=t^{2}-1\), \(0 \leq t \leq 2\).

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{100}{3} \).

Step 1: Understand the Problem

The problem requires evaluating a line integral of the form \( \int_C y \, dx + x^2 \, dy \). The curve \( C \) is parametrized by \( x = 2t \) and \( y = t^2 - 1 \) for \( 0 \leq t \leq 2 \). We need to compute this integral over the given parametrized curve.

Step 2: Parameterize the Line Integral

Replace \( x ext{ and } y \) in the line integral with their expressions in terms of \( t \). This gives:\[ dx = \frac{d}{dt}(2t) \, dt = 2 \, dt \]\[ dy = \frac{d}{dt}(t^2 - 1) \, dt = 2t \, dt \]Substitute these into the integral:\[ \int_{C} y \, dx + x^2 \, dy = \int_{0}^{2} ((t^2 - 1) \, 2 \, dt + (2t)^2 \, 2t \, dt) \].

Step 3: Compute the Parametrized Integral

Simplify and compute the integral:\[ \int_{0}^{2} (2(t^2 - 1) \, dt + 8t^3 \, dt) = \int_{0}^{2} (2t^2 - 2 + 8t^3) \, dt \]Combine like terms:\[ = \int_{0}^{2} (8t^3 + 2t^2 - 2) \, dt \].

Step 4: Evaluate the Integral

Integrate each term separately:\[ \int 8t^3 \, dt = 2t^4 \]\[ \int 2t^2 \, dt = \frac{2}{3}t^3 \]\[ \int -2 \, dt = -2t \]Evaluating from \( t=0 \) to \( t=2 \):\[ \left[ 2t^4 + \frac{2}{3}t^3 - 2t \right]_{0}^{2} \].

Step 5: Calculate the Definite Integral

Substitute \( t = 2 \) and \( t = 0 \) into the antiderivatives:\[ = \left(2(2)^4 + \frac{2}{3}(2)^3 - 2(2)\right) - \left(2(0)^4 + \frac{2}{3}(0)^3 - 2(0)\right) \]\[ = (32 + \frac{16}{3} - 4) - 0 \]\[ = 32 - 4 + \frac{16}{3} = 28 + \frac{16}{3} \]Convert \( 28 \) to have the same denominator:\[ = \frac{84}{3} + \frac{16}{3} = \frac{100}{3} \].

Key Concepts

Parametrization

Parametrization is a technique used to describe a curve by expressing the coordinates of its points as functions of a single variable, often referred to as the parameter. In our exercise, the curve is parametrized by functions of the parameter \(t\), where \(x = 2t\) and \(y = t^2 - 1\). This means that as \(t\) varies from 0 to 2, it traces out the path of the curve \(C\) in the coordinate plane. When we perform parametrization, we're essentially transforming a complex integral along a curve into an easier, more manageable form. By substituting the expressions of \(x\) and \(y\) in terms of \(t\) into our line integral, we convert it into a definite integral with respect to \(t\). This simplification is crucial for solving line integrals, as it enables us to handle more complex paths and integrands with ease.

Definite Integral

A definite integral is a fundamental concept in calculus that represents the signed area under a curve within a certain interval. In the context of line integrals, a definite integral allows us to compute the total effect of a vector field along a specified path. In our exercise, we evaluate the line integral \( \int_C y \, dx + x^2 \, dy \), which becomes a definite integral after parametrization: \[ \int_{0}^{2} (8t^3 + 2t^2 - 2) \, dt \]Evaluating this definite integral involves finding the antiderivatives of each term, then applying the limits of integration. This results in determining the net effect or flux along the entire curve of the vector field described by \(y\) and \(x^2\). Definite integrals are powerful tools that, once evaluated correctly, provide insights into quantities like work, potential energy, or mass along curves.

Calculus Problem Solving

Calculus problem solving involves strategic application of mathematical concepts to find solutions to complex problems. In our given problem, several calculus techniques come into play:

• Parametrization: Converting the curve into a parameterized form to simplify the integration process.
• Derivatives: Computing derivatives \( dx \) and \( dy \) with respect to the parameter \(t\) to transform the integral's components.
• Integration: Evaluating the transformed definite integral, which gives the desired line integral value.

Each of these steps requires a clear understanding of the underlying principles of calculus, such as the fundamental theorem of calculus, as well as practice in manipulating algebraic expressions and applying limits. By mastering these skills, students can effectively approach and solve various calculus problems, enhancing their analytical thinking and problem-solving abilities.