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 portion of \(y=\sqrt{x}\) from (0,0) to \((1,1),\) where distances are measured in meters.

Short answer

The work done is \( \frac{13}{15} \) Joules.

Step-by-step solution

Understand the Problem

We need to find the work done by the force field \( \vec{F} = \langle y, x^2 \rangle \) on a particle moving along the path \( C \) defined by \( y = \sqrt{x} \) from \((0,0)\) to \((1,1)\). Work done by a force field along a path is determined by evaluating the line integral of the force field along the path.

Parameterize the Path

To evaluate the line integral, we need a parameterization of the path. Since \( y = \sqrt{x} \), we can let \( x = t \) and \( y = \sqrt{t} \) for \( t \) ranging from 0 to 1. Therefore, the parameterization of the path \( C \) is \( \vec{r}(t) = \langle t, \sqrt{t} \rangle \).

Calculate the Derivative of the Path

Differentiate the parameterized path \( \vec{r}(t) \) with respect to \( t \) to get \( \vec{r}'(t) = \langle 1, \frac{1}{2\sqrt{t}} \rangle \). This derivative represents the tangent vector to the path at each point \( t \).

Prepare for the Line Integral

The work done is given by the line integral \( \int_C \vec{F} \cdot d\vec{r} \). Substituting the parameterization, the line integral becomes \( \int_0^1 \vec{F}(t) \cdot \vec{r}'(t) \ dt \). Here, \( \vec{F}(t) = \langle \sqrt{t}, t^2 \rangle \).

Substitute and Simplify

Substitute \( \vec{F}(t) = \langle \sqrt{t}, t^2 \rangle \) and \( \vec{r}'(t) = \langle 1, \frac{1}{2\sqrt{t}} \rangle \) into the dot product: \( \vec{F}(t) \cdot \vec{r}'(t) = \sqrt{t} \cdot 1 + t^2 \cdot \frac{1}{2\sqrt{t}} = \sqrt{t} + \frac{t^{2}}{2\sqrt{t}} = \sqrt{t} + \frac{t^{3/2}}{2} \).

Evaluate the Integral

Evaluate the integral \( \int_0^1 \left( \sqrt{t} + \frac{t^{3/2}}{2} \right) \ dt \). This separates into two integrals: \( \int_0^1 \sqrt{t} \ dt \) and \( \frac{1}{2}\int_0^1 t^{3/2} \ dt \).

Solve the Integrals

Calculate \( \int_0^1 \sqrt{t} \ dt = \int_0^1 t^{1/2} \ dt = \left[ \frac{2}{3}t^{3/2} \right]_0^1 = \frac{2}{3} \). Now, calculate \( \frac{1}{2}\int_0^1 t^{3/2} \ dt = \frac{1}{2}\left[ \frac{2}{5}t^{5/2} \right]_0^1 = \frac{1}{5} \).

Add the Results

Add the results from the two integrals: \( \frac{2}{3} + \frac{1}{5} = \frac{10}{15} + \frac{3}{15} = \frac{13}{15} \). Hence, the work done is \( \frac{13}{15} \) Joules.

Key concepts

Line Integral

A line integral is a fundamental concept in vector calculus previously utilized to evaluate the work done by a force field along a path. Here's how it works: if you have a force field expressed as a vector function, the line integral helps us compute how this force interacts along a specific curve or path. This procedure is crucial in understanding how energy travels through varied paths within a field.

In essence, to evaluate the line integral of a force field along a path, you need:

• A force vector field \( \vec{F} \), which influences our particle.
• A parametric representation of the path \( C \), which the particle traces.

The work done by the force as the particle moves along the path is then calculated by integrating the dot product of the force field and the path's tangent vector over the entire distance traveled.

Mathematically, this integral is expressed as \( \int_C \vec{F} \cdot d\vec{r} \). Understanding line integrals not only helps with physics problems but opens doors to solving complex integrals in various fields of science.

Parameterization of Paths

Parameterizing a path is an essential step in evaluating line integrals. It simplifies the description of a curve or path by the use of a parameter, usually denoted as \( t \). By choosing a sensible parameter, the problem becomes much more manageable.

For the given path where \( y = \sqrt{x} \), you can parameterize it by setting \( x = t \) and \( y = \sqrt{t} \), with \( t \) ranging from 0 to 1. This provides a clean and organized way to represent the movement from the starting point \((0,0)\) to the endpoint \((1,1)\).

The parameterization allows us to transform the original curve into a vector function \( \vec{r}(t) = \langle t, \sqrt{t} \rangle \). Analyzing this vector function permits us to explore the path's properties more deeply by unlocking possibilities for differentiation and integration over a simplified domain.

Tangent Vectors

The concept of tangent vectors is pivotal when calculating line integrals, as these vectors provide the direction of the path at any given point. In practical terms, the derivative of the parameterized path produces the tangent vector at each point along the curve.

For the parameterized path \( \vec{r}(t) = \langle t, \sqrt{t} \rangle \), the derivative \( \frac{d\vec{r}}{dt} = \langle 1, \frac{1}{2\sqrt{t}} \rangle \) forms the tangent vector. This computation essentially breaks down the path into its intrinsic directional components.

Tangent vectors are crucial because, combined with the force field vector, they allow the calculation of the work done along the path through the dot product operation. This operation simplifies the integration process by isolating only the contribution along the path’s direction.

Force Field in Calculus

Understanding force fields in calculus is crucial for solving various physics and engineering problems. A force field exerts a force at every point in a given space, represented generally by a vector function \( \vec{F}(x, y) \). The task often involves determining how this force influences motion along a particular path.

The force field given in the exercise, \( \vec{F} = \langle y, x^2 \rangle \), implies that the force at any point depends on the coordinates \( y \) and \( x^2 \). Here, the force field becomes a dynamic entity, changing its strength and direction throughout the space.

Force fields in calculus enable us to not only calculate work done but also predict the behavior of particles and objects as they move under the influence of these compelling and variable forces. This understanding forms the basis of modeling real-world phenomena, ranging from electromagnetism to gravitational fields and beyond.