Question

A particle moves along the path \(y=x^{2}\) from the point (0,0) to the point (1,1) . The force field \(\mathbf{F}\) is measured at five points along the path and the results are shown in the table. Use Simpson's Rule or a graphing utility to approximate the work done by the force field. $$ \begin{array}{|l|c|c|c|c|c|} \hline(x, y) & (0,0) & \left(\frac{1}{4}, \frac{1}{16}\right) & \left(\frac{1}{2}, \frac{1}{4}\right) & \left(\frac{3}{4}, \frac{9}{16}\right) & (1,1) \\ \hline \mathbf{F}(x, y) & \langle 5,0\rangle & \langle 3.5,1\rangle & \langle 2,2\rangle & \langle 1.5,3\rangle & \langle 1,5\rangle \\ \hline \end{array} $$

Short answer

The approximate work done by the force field along the given path, calculated using Simpson's Rule, is approximately 2.2083 units.

Step-by-step solution

Understanding the Problem

The particle is moving along the path defined by the function \(y=x^{2}\). The force at various points along this path is given in the table. The force \(\mathbf{F}(x, y)\) is a vector field, and is given by \(\langle F_x, F_y \rangle\). Calculating the work done by the force field can be achieved through integrating the force along the path.

Applying Simpson's Rule

Simpson's Rule is a numerical method for approximating the definite integral of a function. It is applied here to approximate the integral representing the work done by the force field. We have five points of data, therefore we can apply Simpson's rule with \(n=4\) subintervals. Simpson's Rule for this case is given by \( S = \frac{1}{6n} [y_0 + 4y_1 + 2y_2 + 4y_3 + y_4] \). The 'y' values represent the dot product between force and the derivative of path at the corresponding points.

Calculating the Dot Product

The dot product of the force and the derivative of the path at a certain point is simply the magnitude of the force in the direction of the path motion (x-axis). \(y_i = \mathbf{F}(x, y) \cdot \frac{d}{dx}(x,y) = F_x \cdot dx = F_x\). Now, we substitute the values from the table into Simpson's Rule formula.

Substituting Values and Simplifying

Substitute \(n=4\), \(y_0 = 5\), \(y_1 = 3.5\), \(y_2 = 2\), \(y_3 = 1.5\), and \(y_4 = 1\) into the Simpson's Rule formula, then simplify to get the approximate work done. After substitution, we get \( S = \frac{1}{6 * 4} [5 + 4 * 3.5 + 2 * 2 + 4 * 1.5 + 1] = 2.2083 \) units of work.

Key concepts

Simpson's Rule

When it comes to solving real-life problems like calculating the work done by a force field along a path, we often have to deal with integrating over complex functions or data points. This is where Simpson's Rule comes into play – it's a numerical method that helps us approximate the value of a definite integral, especially when exact integration is impossible or impractical. The elegance of Simpson's Rule lies in its use of parabolic arcs to approximate the curve of a function.

Simply put, imagine you’re drawing smooth, parabolic segments between a series of points along a curve to estimate the area underneath it. That’s essentially what Simpson’s Rule does. It is particularly accurate when the function to be integrated is smooth and can be well approximated by quadratic functions across each segment. No wonder it is a favorite tool among engineers and scientists for tasks like finding the approximate work done by a variable force along a given path.

Vector Fields

The concept of a vector field is central to the calculation of work done by a force along a path. A vector field is like a map that assigns a vector to every point in space. Imagine standing in a field with wind blowing around you. At every point in that field, the wind has a direction and a strength, or speed. This is akin to a vector field, where instead of wind, we discuss force vectors. These vectors can represent gravity, electric or magnetic fields, fluid velocity, or any other quantity that has both magnitude and direction.

Understanding vector fields is crucial when calculating work, as it represents the spatial variation of the forces that do work on objects moving through the field. When a particle travels from one point to another, it essentially collects the influence of the field along its path, leading to either an increase or decrease in energy – that's the work being done.

Integrating Force

Integrating force over a path gives us the work done by that force, and it ties together our understanding of vector fields with the actual mechanics involved. To compute work, we assume a particle moves along a path through a vector field and we integrate the force exerted along this trajectory. This integration is not just about summation— it's about summing up the effects of force, considering both magnitude and direction, at every infinitesimal step along the path.

In the case of a straight path, this would simply be a force multiplied by distance. However, for paths that curve and twist through space, as defined by a vector function, the integration needs to account for the changing direction and magnitude of both the path and the force field, hence we resort to dot products to extract the work component along the path's tangent.

Dot Product

The dot product, also known as scalar product, is a way to multiply two vectors that gives us a scalar – a single number. This operation is fundamental in understanding work done by a force field because it measures how much one vector acts along the direction of another.

Imagine you’re pushing a box along a floor. The force of your push has both magnitude (how hard you're pushing) and direction. The dot product tells you how much of your force is going in the direction of the movement, which ultimately is the part doing work. In mathematical terms, the dot product of two vectors is calculated by multiplying their corresponding components and then adding those products together. A key aspect of the dot product is that it includes the cosine of the angle between vectors, which encapsulatsedirectionality – exactly what's needed when calculating work involving a force exerted at an angle to the direction of movement.