Question

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. Work Consider a particle that moves through the force field \(\mathbf{F}(x, y)=(y-x) \mathbf{i}+x y \mathbf{j}\) from the point (0,0) to the point (0,1) along the curve \(x=k t(1-t), y=t .\) Find the value of \(k\) such that the work done by the force field is 1

Short answer

The value of \(k\) is obtained by solving the equation that comes from setting the result of the work integral equal to 1. You will have to compute the integral, leading to an equation with \(k\) as the unknown, and then solve this equation for \(k\).

Step-by-step solution

Calculate the line integral

We are given the path as parametric equations, \( x = kt(1-t) \) and \( y = t \). The line integral of the force field along this curve is given by \(\int_{A}^{B} \mathbf{F}\cdot d\mathbf{r} \). Here \(d\mathbf{r} = x'dt\mathbf{i} + y'dt\mathbf{j}\), where \(x'\) and \(y'\) are the derivatives of x and y with respect to t respectively. So we need to find \(x'\) and \(y'\). For the path, we have \( x' = k - 2kt \) and \(y' = 1\) (since y=t). Substitute these values and \(\mathbf{F}(x, y)\) into the line integral.

Evaluate the integral

Now, we have to compute the integral \(\int_{0}^{1} [(y-x)\cdot x' + xy\cdot y']dt = \int_{0}^{1} [(t-k t(1-t))(k - 2kt) + k t^2(1-t)\cdot 1] dt\). This integral can be solved using standard calculus techniques, resulting in a single equation with \(k\) as the unknown.

Solve for k

Set the result of the integral to be equal to 1, as the work done by the force field is given to be 1. Solve this equation to find the value of \(k\) to satisfy the work done is 1.

Key concepts

Work Done by a Force Field

In physics, the work done by a force field is a measure of energy transfer that occurs when an object moves through a force field. Think of work as the result of a force moving an object a certain distance. In calculus, specifically in vector calculus, we calculate this work by evaluating a line integral. When dealing with work, we typically consider a force field which can be represented by a vector function \( \mathbf{F}(x, y) \). The amount of work done by this force field on a particle moving along a path from point A to point B is expressed as the line integral \( \int_{A}^{B} \mathbf{F}\cdot d\mathbf{r} \), where \( \mathbf{r} \) is the position vector describing the path, and \( d\mathbf{r} \) is an infinitesimal segment of the path.

In the given problem, we investigate the work done by the force field \( \mathbf{F}(x, y) = (y-x) \mathbf{i} + xy \mathbf{j} \) as a particle moves through it. The goal is to find the specific value of k such that this integral equals 1, which corresponds to one unit of work being done. This requires us to parametrize the curve, find the differential elements, and evaluate the line integral along the given path.

Parametric Equations

When it comes to describing paths in mathematics, parametric equations are a powerful method to represent curves. Unlike the typical y=f(x) format which defines a function explicitly, parametric equations describe a curve using a third variable, often 't', to define both x and y coordinates as separate functions of t: \( x(t) \) and \( y(t) \). This is especially useful for describing complex paths that may not function well with traditional y=f(x) due to issues like having multiple y values for a single x value.

In our exercise, the parametric equations for the path are \( x = kt(1-t) \) and \( y = t \) which task us to compute the work done by the force field along this specific path. By utilizing these parametric equations, we are able to compute the derivatives \( x' \) and \( y' \), which are crucial for finding the differential element \( d\mathbf{r} \) and hence, solving the line integral.

Line Integrals in Vector Fields

Finally, the exercise deals with the concept of line integrals in vector fields. A line integral is a type of integral where a function is evaluated along a curve or path. When dealing with vector fields, this involves integrating a vector function along a curve, which, intuitively, can mean accumulating the effect of the field along the path. The result of a line integral can represent various physical concepts, such as work done by a force field.

The line integral is calculated by taking the dot product of the vector field and the derivative of the position vector, resulting in a scalar function which is then integrated with respect to the parameter that defines the path, which in our example is 't'. The formal definition for the line integral over a vector field \( \mathbf{F} \) along a curve C is given by \( \int_{C} \mathbf{F}\cdot d\mathbf{r} \).

In our provided solution, we applied this concept by setting up and calculating such a line integral, taking into account the dot product of the force field and the derivatives of the parametric equations along the curve described by those equations, to determine the value of k which results in the work done being equal to 1.