Question

In Exercises \(13-16,\) find the work performed by the force field \(\vec{F}\) moving a particle along the path \(C\). \(\vec{F}=\left\langle y, x^{2}\right\rangle \mathrm{N} ; C\) is the segment of the line \(y=x\) from (0,0) to \((1,1),\) where distances are measured in meters.

Short answer

The work done is \(\frac{5}{6}\) Joules.

Step-by-step solution

Understand the Problem

The problem asks us to find the work done by the force field \(\vec{F} = \langle y, x^2 \rangle\) as a particle moves along the path \(C\), which is the segment from \((0, 0)\) to \((1, 1)\) along the line \(y = x\).

Define the Path C

The path \(C\) is given by the equation \(y = x\). This implies that any point on the path can be represented as \((t, t)\), where \(t\) ranges from 0 to 1. Thus, the vector parametrization of \(C\) is \(\vec{r}(t) = \langle t, t \rangle\).

Calculate the Derivative of the Path

The derivative of the path \(\vec{r}(t) = \langle t, t \rangle\) with respect to \(t\) is \(\frac{d\vec{r}}{dt} = \langle 1, 1 \rangle\). This represents the tangent vector to the path.

Evaluate the Force Field on the Path

Substitute \(y = t\) and \(x = t\) into the force field \(\vec{F}(x, y) = \langle y, x^2 \rangle\). This yields \(\vec{F}(t, t) = \langle t, t^2 \rangle\).

Calculate the Dot Product

Compute the dot product \(\vec{F}(t, t) \cdot \frac{d\vec{r}}{dt} = \langle t, t^2 \rangle \cdot \langle 1, 1 \rangle = t + t^2\).

Integrate the Dot Product to Find the Work

Integrate the dot product over the interval from 0 to 1: \[ W = \int_{0}^{1} (t + t^2) \, dt \].

Perform the Integration

Compute the integral: \[\int_{0}^{1} (t + t^2) \, dt = \left[ \frac{t^2}{2} + \frac{t^3}{3} \right]_{0}^{1} = \left( \frac{1}{2} + \frac{1}{3} \right) - \left( 0 + 0 \right) = \frac{5}{6}.\]

Conclude the Work Calculation

The work done by the force field \(\vec{F}\) in moving the particle from \((0, 0)\) to \((1, 1)\) is \(\frac{5}{6}\, \text{Joules}.\)

Key concepts

work done in a force field

Work done in a force field involves calculating the energy required to move an object along a certain path under the influence of a vector field, which represents a force. The force field given in this exercise, \(\vec{F} = \langle y, x^2 \rangle\), describes a field where the force depends on both the \(x\) and \(y\) coordinates. To find the work done by this force field as the particle moves, we need to compute a line integral.

The concept of work done can be understood by thinking of it as the integral of the force along the path of motion, quantifying how much energy is used to move through the field. In simple terms, if you're pushing an object, the work you do depends on both the force applied and the distance moved in the field direction.

This exercise demonstrates that work is not just a scalar product of force and distance but involves the direction of movement along a curve. It is crucial to evaluate the force as a particle moves along a specific path to grasp this concept fully.

parametrization of curves

Parametrization of curves is a vital skill in vector calculus, allowing us to describe a curve in terms of a single parameter, usually \(t\). This approach makes calculation simpler when working with vector fields. In our exercise, the path \(C\) is the line segment from \((0, 0)\) to \((1, 1)\) along the line \(y = x\).

The parametrization chosen here uses \(t\) to describe every point \((t, t)\) on the path as \(t\) moves from 0 to 1. This defines our curve \(\vec{r}(t) = \langle t, t \rangle\). Parametrization provides a structured way to track the position of a particle traveling along the curve, which is essential for performing operations like differentiation and integration.

For this path, the derivative of the position, \(\frac{d\vec{r}}{dt} = \langle 1, 1 \rangle\), represents the direction and speed of movement along the curve. It provides the tangent vector necessary to evaluate how the force field acts on the particle all along its path. Understanding parametrization is key to solving problems that involve curves in vector calculus.

line integrals

Line integrals extend the concept of integrals to paths in vector fields, measuring cumulative quantities like work, flux, or mass along a curve. In this exercise, the line integral calculates the total work done by the force field \(\vec{F} = \langle y, x^2 \rangle\) as the particle travels along the parametrized path \(C\).

The integral we evaluate is \(\int_{0}^{1} (t + t^2) \, dt\). This expression arises from the dot product of the force field evaluated on the curve and the derivative of the path. The calculation involves integrating the dot product across the range of \(t\), simplifying through anti-derivatives, and evaluating the result at the boundaries provided by the parameter.

Line integrals require an understanding of both the path formula and the vector field action. By evaluating \(\int_{0}^{1} (t + t^2) \, dt\), we find that the total work done is \(\frac{5}{6}\) Joules, showcasing how line integrals allow us to quantify the influence of dynamic systems across spatial paths.