Problem 21 - Thomas Calculus

Question

Find the line integral of \(f ( x , y ) = y e ^ { x ^ { 2 } }\) along the curve \(\mathbf { r } ( t ) = 4 t \mathbf { i } - 3 t \mathbf { j } , - 1 \leq t \leq 2\)

Step-by-Step Solution

Answer

The line integral evaluates to \( -\frac{15}{32}(e^{64} - e^{16}) \).

Step 1: Determine the Integral Representation

The line integral of a vector field \( \mathbf{F}(x, y) \) along a parametrized curve \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} \) is given by \[ \int_{C} f(x, y) \, ds = \int_{a}^{b} f(x(t), y(t)) \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt. \]

Step 2: Identify the Parametric Components

From \( \mathbf{r}(t) = 4t \mathbf{i} - 3t \mathbf{j} \), we have \( x(t) = 4t \) and \( y(t) = -3t \).

Step 3: Determine the Differential Components

Find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):- \( \frac{dx}{dt} = 4 \)- \( \frac{dy}{dt} = -3 \)

Step 4: Substitute into the Line Integral Formula

Substitute \( x(t) \), \( y(t) \), and the derivatives into the integral. We need to compute \( f(x(t), y(t)) \):- \( f(x(t), y(t)) = y(t) e^{(x(t))^2} = (-3t)e^{(4t)^2} = -3t e^{16t^2}. \)

Step 5: Compute the Magnitude of Velocity

Calculate the magnitude of the derivative:\[ \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5. \]

Step 6: Evaluate the Integral

Insert all the components into the integral from \(-1\) to \(2\):\[ \int_{-1}^{2} (-3t e^{16t^2}) \cdot 5 \, dt = -15 \int_{-1}^{2} t e^{16t^2} \, dt. \]

Step 7: Solve the Integral

Use integration by substitution for \( u = 16t^2 \), \( du = 32t \, dt \); thus, \( t \, dt = \frac{1}{32} \, du \). Transform limits of integration (\( t = -1 \), \( u = 16 \); \( t = 2 \), \( u = 64\)):\[ -\frac{15}{32} \int_{16}^{64} e^{u} \, du. \]

Step 8: Calculate the Result

This integral evaluates to:\[ -\frac{15}{32} \left( e^{64} - e^{16} \right). \]

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and operations applied to them. It extends the concepts of calculus for scalar functions to vector fields, which have both magnitude and direction. A key component of vector calculus is the line integral, which computes the nature of a vector field along a path.

• Line Integral: In the context of a vector field, a line integral allows us to integrate a scalar or vector field along a curve. It's useful in calculating work done by a force field along a path. The integrals we deal with here relate to scalar fields, integrating over a curve or path in the plane.
• Applications: Line integrals are widely applied in physics, engineering, and computer graphics, among other fields. They help model phenomena like electromagnetic fields, fluid flow, and object paths through fields.

Understanding the foundation of vector calculus allows us to evaluate how vectors behave along defined paths, which is essential for the analysis of vector fields in the query.

Parametric Equations

Parametric equations offer a powerful way to describe curves in the plane by expressing the coordinates as functions of a single parameter, usually denoted as \( t \). In many cases, parametric equations provide a clearer picture of how curves are formed and manipulated, especially in vector calculus.

• Parametric Representation: In our exercise, the curve is represented parametrically by \( \mathbf{r}(t) = 4t \mathbf{i} - 3t \mathbf{j} \). This means as \( t \) varies, \( x(t) = 4t \) and \( y(t) = -3t \) describe the path.
• Simplicity and Flexibility: Parametric equations simplify handling complex curves and are essential in calculating derivatives, allowing us to analyze motion along the curve, as seen in finding \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

By understanding the use of parametric equations, students can break down complex curves into simpler parts, making it easier to perform calculations like line integrals along those paths.

Integration Techniques

Integration techniques are critical when solving line integrals as they involve integrating over a parameterized region or domain. The process often requires substitution, transformation of variables, and careful handling of boundaries.

• Integration by Substitution: In the problem, we use the substitution method, where \( u = 16t^2 \) leads to \( du = 32t \, dt \). This technique helps simplify the integral by transforming it into a more manageable form.
• Transforming Limits: After substitution, the limits of integration need conversion in terms of the new variable, \( u \). This involves recalculating whole boundaries which turned from \( t = -1 \) to \( t = 2 \) into \( u = 16 \) to \( u = 64 \).
• Evaluation and Final Result: Completing integrals involves finding antiderivatives and computing the definite integral over the new limits. Integral transforms make seemingly complex functions easier to evaluate, yielding compact results like \( -\frac{15}{32} \left( e^{64} - e^{16} \right) \).

Mastery of integration techniques is vital for tackling complex integrals found in line integral problems, allowing for a deeper understanding and ability to solve intricate calculus inquiries.