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 is \(-\frac{15}{32} (e^{64} - e^{16})\).

Step 1: Parameterization of the Curve

The curve is given by the vector function \( \mathbf{r}(t)=4t \mathbf{i} -3t \mathbf{j} \). We will use the parametric representation of this curve, where \( x(t) = 4t \) and \( y(t) = -3t \). Now, compute \( \frac{dx}{dt} = 4 \) and \( \frac{dy}{dt} = -3 \).

Step 2: Evaluate the Scalar Field at \\(\mathbf{r}(t)\\)

Substitute the parametric equations into the scalar field \(f(x, y) = ye^{x^{2}}\). This results in \(f(t) = (-3t)e^{(4t)^{2}} = -3t e^{16t^{2}}\).

Step 3: Set Up the Integral

The line integral of a scalar field along a curve is given by \( \int_{a}^{b} f(\mathbf{r}(t)) \|\mathbf{r}'(t)\| \, dt \), where \( \mathbf{r}'(t) = 4 \mathbf{i} - 3 \mathbf{j} \). Calculate the magnitude as \( \|\mathbf{r}'(t)\| = \sqrt{(4)^{2} + (-3)^{2}} = 5 \). The integral becomes \( \int_{-1}^{2} -3t e^{16t^{2}} \times 5 \, dt \).

Step 4: Simplify the Integral

Simplify the integral expression to \(-15 \int_{-1}^{2} t e^{16t^{2}} \, dt \). This integral requires substitution letting \(u = 16t^{2}\), then \( du = 32t \, dt \). Therefore, \( t \, dt = \frac{1}{32} du \). Rewriting our integral, we have \(-\frac{15}{32} \int e^{u} \, du \).

Step 5: Evaluate the Integral Using Limits

Change the bounds based on \(u = 16t^{2} \). For \(t = -1\), \(u = 16\), and for \(t = 2\), \(u = 64\). So the integral is \( -\frac{15}{32} \left[e^{u} \right]_{16}^{64} = -\frac{15}{32} (e^{64} - e^{16}) \).

Step 6: Calculate the Final Result

Thus, the line integral evaluates to \(-\frac{15}{32} (e^{64} - e^{16})\). The result is quite large due to the exponential function, and it's typically expressed in terms of \(e^{64} - e^{16}\).

Key Concepts

Parameterization of the Curve

When we talk about parameterization of a curve, we are referring to expressing the curve using a parameter, often represented as \( t \). This is very useful in calculus because it allows us to simplify complex equations by dealing with functions of a single variable.

In the given problem, the curve is represented by the vector function \( \mathbf{r}(t) = 4t \mathbf{i} - 3t \mathbf{j} \). Here, \( t \) is the parameter that traces the path of the curve in the plane. For this problem, \( x(t) = 4t \) and \( y(t) = -3t \). These equations link the parameter \( t \) to the position \( (x, y) \) on the curve.

The next part is to find the derivatives \( \frac{dx}{dt} = 4 \) and \( \frac{dy}{dt} = -3 \). These derivatives represent how quickly and in what direction the curve moves as \( t \) changes. They are crucial for understanding the curve's geometry and are essential in setting up the line integral.

Scalar Field

A scalar field is a function that assigns a scalar value to every point in a particular space. In simple terms, for every point \((x, y)\), a scalar field gives us a number. In the exercise, the scalar field is \( f(x, y) = y e^{x^{2}} \).

To integrate this scalar field over our curve, we evaluate it along the parameterized curve. This means substituting \( x(t) = 4t \) and \( y(t) = -3t \) into \( f(x, y) \), resulting in \( f(t) = -3t e^{(4t)^{2}} \), which simplifies to \( -3t e^{16t^{2}} \).

Understanding scalar fields is important because they allow us to measure quantities like temperature, pressure, or potential energy along a curve in a given field. In this problem, it transforms a static expression into one that changes as we move along the curve.

Integral Evaluation

Evaluating the integral involves setting up and solving the integral of a function over a curve. This is called a line integral, and it helps us measure accumulated quantities along the curve.

The line integral formula for a scalar field \( f(x, y) \) along a curve \( \mathbf{r}(t) \) is \( \int_{a}^{b} f(\mathbf{r}(t)) \|\mathbf{r}'(t)\| \, dt \). The vector \( \mathbf{r}'(t) \) is the derivative vector \( 4\mathbf{i} - 3\mathbf{j} \), and its magnitude is \( 5 \), determined using the Pythagorean theorem: \( \|\mathbf{r}'(t)\| = \sqrt{4^{2} + (-3)^{2}} = 5 \).

The integral setup becomes: \( \int_{-1}^{2} -3t e^{16t^{2}} \times 5 \, dt \) or \( -15 \int_{-1}^{2} t e^{16t^{2}} \, dt \). We use substitution \( u = 16t^{2} \) and \( du = 32t \, dt \) to rewrite this integral as: \( -\frac{15}{32} \int e^{u} \, du \).

Finally, you change the bounds according to \( u \) from \( 16 \) to \( 64 \), resulting in the expression \( -\frac{15}{32} [e^{u}]_{16}^{64} \). This yields \( -\frac{15}{32} (e^{64} - e^{16}) \). This calculation helps us see how the line integral measures the total change in the scalar field along the curve over its entire path.