Question

Given the force field \(\mathbf{F},\) find the work required to move an object on the given oriented curve. \(\mathbf{F}=\langle x, y\rangle\) on the line segment from (-1,0) to (0,8) followed by the line segment from (0,8) to (2,8)

Short answer

Answer: The total work required to move the object along the given oriented curve is 1.5 units.

Step-by-step solution

Parameterize the line segments

To parameterize the line segments, we need to find a vector-valued function for each segment. Let's start with the first line segment, which connects the points \((-1,0)\) and \((0,8)\). Line segment 1: Let \(C_1(t) = \langle x(t), y(t) \rangle\), where \(t\) ranges from \(0\) to \(1\). \(x(t) = -1 + t(0 - (-1)) = -1 + t\) \(y(t) = 0 + t(8 - 0) = 8t\) So the parameterization of the first line segment is \(C_1(t) = \langle -1 + t, 8t \rangle\). Now, let's parameterize the second line segment, connecting the points \((0,8)\) and \((2,8)\). Line segment 2: Let \(C_2(t) = \langle x(t), y(t) \rangle\), where \(t\) ranges from \(0\) to \(1\). \(x(t) = 0 + t(2 - 0) = 2t\) \(y(t) = 8 + t(8 - 8) = 8\) So the parameterization of the second line segment is \(C_2(t) = \langle 2t, 8 \rangle\).

Calculate the definite integrals

Now we need to calculate the definite integral \(\int_{0}^{1} \mathbf{F}(C(t)) \cdot C'(t) \, dt\) for each line segment. We will first calculate \(C'(t)\) for both line segments. Line segment 1: \(C_1'(t) = \langle \frac{d}{dt} (-1 + t), \frac{d}{dt} (8t) \rangle = \langle 1, 8 \rangle\) Now, we will find the dot product \(\mathbf{F}(C_1(t)) \cdot C_1'(t)\): \(\mathbf{F}(C_1(t)) = \langle -1 + t, 8t \rangle\) \((\mathbf{F}(C_1(t)) \cdot C_1'(t)) = \langle -1 + t, 8t \rangle \cdot \langle 1, 8 \rangle = (-1 + t) + 64t^2\) Line segment 2: \(C_2'(t) = \langle \frac{d}{dt} (2t), \frac{d}{dt} (8) \rangle = \langle 2, 0 \rangle\) Now we will find the dot product \(\mathbf{F}(C_2(t)) \cdot C_2'(t)\): \(\mathbf{F}(C_2(t)) = \langle 2t, 8 \rangle\) \((\mathbf{F}(C_{2}(t)) \cdot C_{2}'(t)) = \langle 2t, 8 \rangle \cdot \langle 2, 0 \rangle = 4t\)

Calculate the total work by adding the integrals

Now, we have to calculate the definite integrals for both line segments and then add them to get the total work. Line segment 1: \(\int_{0}^{1} (-1 + t + 64t^2) \, dt = \left[-t + \frac{1}{2}t^2 + \frac{64}{3}t^3\right]_0^1 = -1 + \frac{1}{2} + \frac{64}{3} = \frac{65}{2} - \frac{64}{3}\) Line segment 2: \(\int_{0}^{1} (4t) \, dt = \left[2t^2\right]_0^1 = 2\) Now, let's add the definite integrals to find the total work: \(W = \frac{65}{2} - \frac{64}{3} + 2 = \frac{65}{2} - \frac{58}{3} = \frac{9}{6} = 1.5\) The total work required to move the object along the given oriented curve is 1.5 units.

Key concepts

Parameterization of Line Segments

When we talk about the parameterization of line segments in vector calculus, we refer to the process of describing a path (or curve) using a parameter, typically denoted as t. It is akin to giving the GPS coordinates for every point on a path, but instead of latitude and longitude, we use mathematical functions of t to define positions on the line segment. This is crucial because we need a precise mathematical description to perform operations like line integrals.

The parameter t customarily runs from 0 to 1, representing the start and end of the path respectively. For instance, in the solution provided, the segment from (-1,0) to (0,8) was described with the vector-valued function C₁(t). By assigning values to t, we can describe every point along the segment, including the endpoints when t is 0 or 1. Such a parameterization allows us then to proceed with calculating quantities like work or circulation along the path.

Dot Product in Vector Calculus

The dot product, also known as the scalar product, in vector calculus is a fundamental operation that takes two vectors and returns a single number (a scalar). This operation is central to many applications in physics and engineering, particularly when it comes to calculating work done by a force field.

In mathematical terms, the dot product of two vectors A = \ and B = \ is calculated as A \. B = a_1b_1 + a_2b_2. What it effectively measures is the magnitude of one vector in the direction of the other. So, when we calculate the work done by a force field along a path, we're really adding up how much of that force is going in the direction of movement at each point along the path. A higher dot product means more work is done.

In the solution you're looking at, the dot product \(\mathbf{F}(C_1(t)) \. C_1'(t)\) represents the instantaneous work being done by the force field along the first line segment parameterized by C₁(t).

Definite Integrals in Vector Calculus

The concept of definite integrals in vector calculus is analogous to tallying up an infinite number of infinitesimally small quantities to determine a total amount. In the context of work and force fields, this means adding up all the little bits of work done along a curve to find the total work.

Mathematically, when we compute the definite integral of a function, we're looking at the sum of the values of that function at an infinite number of points, multiplied by an infinitesimally small width (representing a small segment of the path), over a particular interval. In practical terms, this involves finding the antiderivative of the function and evaluating it at the bounds of the interval.

In the provided solution, you calculated the definite integrals for both line segments' dot product functions to find the work done along each. By evaluating \(\int_{0}^{1} (-1 + t + 64t^2) \. dt\) and \(\int_{0}^{1} (4t) \. dt\), and summing them up, you determined the total work for moving the object along the two line segments in the given force field.