Problem 31 - Thomas Calculus

Question

Evaluating a work integral two ways Let \(F=\nabla\left(x^{3} y^{2}\right)\) and let \(C\) be the path in the \(x y\) -plane from \((-1,1)\) to \((1,1)\) that consists of the line segment from \((-1,1)\) to \((0,0)\) followed by the line segment from \((0,0)\) to \((1,1)\) . Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) in two ways. a. Find parametrizations for the segments that make up \(C\) and evaluate the integral. b. Use \(f(x, y)=x^{3} y^{2}\) as a potential function for \(\mathbf{F} .\)

Step-by-Step Solution

Answer

The total work done is 2.

Step 1: Calculate the Gradient of the Function

First, compute the gradient of the scalar function \(f(x, y) = x^3 y^2\). The gradient, denoted as \(abla f\), is a vector containing the partial derivatives with respect to \(x\) and \(y\).\[abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) = \left(3x^2 y^2, 2x^3 y\right)\] Therefore, the vector field is \(\mathbf{F} = (3x^2 y^2, 2x^3 y)\).

Step 2: Parametrize the Path Segments

The path \(C\) consists of two segments. First, parametrize the segment from \((-1,1)\) to \((0,0)\):- Let \(\mathbf{r}_1(t) = (-1 + t, 1 - t)\), where \(t\) ranges from \(0\) to \(1\).Next, parametrize the segment from \((0,0)\) to \((1,1)\):- Let \(\mathbf{r}_2(t) = (t, t)\), where \(t\) ranges from \(0\) to \(1\).

Step 3: Evaluate the Line Integral Along Each Segment

Evaluate the line integral \(\int_{C} \mathbf{F} \cdot d\mathbf{r}\) for each segment. For \(\mathbf{r}_1(t)\):- Compute \(d\mathbf{r}_1 = (1, -1)dt\).- Substitute the parametrization into \(\mathbf{F}\): \(\mathbf{F}(-1+t,1-t) = (3(-1+t)^2 (1-t)^2, 2(-1+t)^3 (1-t))\).- Evaluate the integral:\[\int_{0}^{1} \left(3(-1+t)^2 (1-t)^2, 2(-1+t)^3 (1-t)\right) \cdot (1, -1) \, dt\]For \(\mathbf{r}_2(t)\):- Compute \(d\mathbf{r}_2 = (1, 1)dt\).- Substitute the parametrization into \(\mathbf{F}\): \(\mathbf{F}(t,t) = (3t^4, 2t^4)\).- Evaluate the integral:\[\int_{0}^{1} (3t^4, 2t^4) \cdot (1, 1) \, dt\]

Step 4: Add the Results of the Two Integrals

Add the results of the two integrals from Step 3 to get the total work done over path \(C\). The total work is the sum of the integrals of the two line segments.

Step 5: Evaluate Integral Using the Potential Function

Since \(\mathbf{F}\) is the gradient of a scalar field, the line integral can also be computed using the potential function \(f(x, y) = x^3 y^2\):\[\int_{C} \mathbf{F} \cdot d\mathbf{r} = f(1,1) - f(-1,1)\]Calculate the values of the potential function at \((1,1)\) and \((-1,1)\):\[f(1,1) = 1^3 \cdot 1^2 = 1\]\[f(-1,1) = (-1)^3 \cdot 1^2 = -1\]Thus, the potential difference is \[1 - (-1) = 2\].

Key Concepts

Line Integral

Line integrals are a fundamental concept in calculus. They allow us to find the accumulated total of a field along a specified path. Think of a line integral as walking along a path and collecting measurements at various points, such as measuring the temperature or force experienced. In the context of our exercise, we are evaluating the work done by a vector field along a particular path.

The path (C) in our exercise is composed of two straight segments in the plane. To compute the line integral, we need to parametrize each path segment and then evaluate the integral of the vector field along these segments.

• The first segment runs from ((-1,1)to ((0,0).This is parametrized as \ \mathbf{r}_1(t) = (-1 + t, 1 - t).
• The second segment continues from ((0,0) to ((1,1), with a parametrization of extrm{\textrm{\mathbf{r}_2(t) = (t, t)}}.

These parametrizations allow us to express the line integral in terms of a single parameter, \(t\). We then substitute them into the vector field \(\mathbf{F}\), calculate \(d\mathbf{r}\) for each segment, and integrate over \(t\) from 0 to 1 for each segment. The results are summed to find the total work done along path \(C\).

Potential Function

A potential function is a scalar function whose gradient is equal to the given vector field. If a vector field \(\mathbf{F}\) is conservative, meaning it can be expressed as the gradient of some scalar function \(f\), we can use the potential function to evaluate line integrals more easily.

This is because the line integral of a conservative vector field over a path \(C\) depends solely on the potential values at the endpoints of \(C\). In our exercise, we identified the potential function as \(f(x, y) = x^3 y^2\). To compute the work integral using the potential function, we simply determine \(f(1, 1)\) and \(f(-1, 1)\).

• Calculating \(f(1, 1)\) gives \(1^3 \cdot 1^2 = 1\).
• Meanwhile, \(f(-1, 1)\) results in \((-1)^3 \cdot 1^2 = -1\).

The work integral is then the difference between these two values: \(1 - (-1) = 2\). This method is efficient as it bypasses the need to parameterize and integrate along individual segments.

Path Parametrization

Path parametrization involves converting a path, typically represented as sequence of geometric points, into a mathematical form that can be manipulated using calculus. This transformation is essential for evaluating line integrals because it provides a way to express the path in terms of a single variable.

In our exercise, we perform path parametrization for both line segments composing the path \(C\). Each segment is described with a vector function, \(\mathbf{r}(t)\), where \(t\) changes continuously from 0 to 1.

• The first segment is turned into \(\mathbf{r}_1(t) = (-1+t, 1-t)\), allowing us to trace the line from \((-1, 1)\) to \((0, 0)\).
• The second segment is \(\mathbf{r}_2(t) = (t, t)\), which takes us from \((0, 0)\) to \((1, 1)\).

These parametrized equations are crucial for expressing both the vector field's influence and moving increment \(d\mathbf{r}\) on the path segments, ultimately enabling us to set up and solve the line integrals we encountered. By using \(t\) as a unified parameter, it simplifies the calculations of complicated shapes into manageable mathematical expressions.