Problem 26 - Calculus: Early Transcendental Functions

Question

Compute the work done by the force field \(\mathbf{F}\) along the curve \(C.\) \(\mathbf{F}(x, y)=\\{2 y,-2 x\rangle, C\) is the line segment from (4,2) to (0,4)

Step-by-Step Solution

Answer

The work done by the force field \(F\) along the curve \(C\) is 0

Step 1: Parameterize the Curve

The line \(C\) connects the points (4,2) to (0,4). Any point (x,y) on this line can be expressed in parametric form using a parameter \(t\): \[x=4-4t, y=2+2t.\] where \(t\) ranges between 0 and 1.

Step 2: Substitute Parameterized Expression

Substitute the parameterized form of the function \(F(x, y) = 2y, -2x\). This becomes \(F(t) = 2(2+2t), -2(4-4t)\).

Step 3: Compute the line integral

Work done by the force is computed by taking the line integral over \(C\) of \(F\cdot dr\), where \(dr = dx, dy\). This becomes \[W= \int_{0}^{1} F(t) \cdot dr = \int_{0}^{1} (2(2+2t), -2(4-4t)) \cdot (4-4t',2+2t') dt \] Computing the derivatives with respect to \(t\) and substituting them in, we get \[W = \int_{0}^{1} (8+8t, 8-8t) \cdot (-4,-2) dt \] Finally, we evaluate this integral to get \(W =0\)

Key Concepts

Parameterization of a Curve

In order to compute the work done along a curve by a force field, understanding parameterization is essential. Parameterization involves expressing coordinates, such as \(x\) and \(y\), in terms of another variable, usually denoted by \(t\). This parameter, \(t\), typically takes on values over a defined interval to describe points along the curve.
For example, in the presented problem, the curve \(C\) is a line segment between the points (4,2) and (0,4). We express both \(x\) and \(y\) as functions of \(t\) ranging from 0 to 1.

• For \(x\): \(x = 4 - 4t\)
• For \(y\): \(y = 2 + 2t\)

This expression allows us to systematically explore every point on the curve by changing \(t\) from 0 to 1. Thus, parameterization translates a geometric path into an algebraic form that is necessary for further calculus operations, such as calculating line integrals.

Force Field in Vector Calculus

A force field in mathematics, particularly in vector calculus, can be seen as a function that assigns a vector to each point in a region of space. The vector indicates the direction and strength of the force at that specific point.
In the given exercise, the force field \(\mathbf{F}(x, y)=\{2y, -2x\}\\) is defined. This force field assigns vector directions based on the current position \(x, y\). Understanding the direction helps in visualizing how the force would act on an object at any given point along the line segment that forms the curve \(C\).

• The component \(2y\) indicates that the force's y-component increases as y increases.
• Likewise, \(-2x\) shows that the x-component, in contrast, works against an increase in x.

This understanding of force fields is crucial when we refer to work done by the field over a specific path, as the effects are contingent on how these vectors interact with the movement along the curve.

Work Done by the Force Field

Calculating work done by a force field along a curve combines the principles of physics and calculus. The work done is essentially the integral of the force field dot product with the differential element of path along the curve.
Given the parameterized line and the force field components, work done is calculated as a line integral. This involves integrating the dot product of the force field \(\mathbf{F}(t)\) and the differential path \(dr\) for each point along \(t\).
In the given problem, after substituting parameters we arrive at the expression \[W = \int_{0}^{1} (8+8t, 8-8t) \cdot (-4, -2) \, dt\]Evaluating this integral determines the total work done, equating to zero in this case. This result implies that the net work done by the force along this specified path is zero. Such results can occur in situations like this one, where the effects of the force field components precisely cancel out over the course from the start to the end of the path, highlighting the importance of direction and path in these calculations.