Problem 65 - Calculus Volume 3

Question

In the following exercises, find the work done by force field \(\mathbf{F}\) on an object moving along the indicated path. \(\mathbf{F}(x, y)=-x \mathbf{i}-2 y \mathbf{j}\) \(C: y=x^{3}\) from (0,0) to (2,8)

Step-by-Step Solution

Answer

The work done is \(-34\).

Step 1: Set Up the Line Integral

To calculate the work done by the force field \( \mathbf{F}(x, y)=-x \mathbf{i}-2y \mathbf{j} \) along the path \( C \) described by \( y = x^3 \) from \((0,0)\) to \((2,8)\), we need to evaluate the line integral \[ \int_C \mathbf{F} \cdot d\mathbf{r}, \]where \( d\mathbf{r} = dx \mathbf{i} + dy \mathbf{j}. \) We'll express this in terms of the parameter \( x. \)

Step 2: Express the Path in Parametric Form

Given \( y = x^3 \), \( dy = 3x^2 \ dx. \) The vector expression is \( d\mathbf{r} = (dx)\mathbf{i} + (3x^2 \, dx)\mathbf{j}. \) The force field is \( \mathbf{F} = -x \mathbf{i} - 2y \mathbf{j} = -x \mathbf{i} - 2(x^3) \mathbf{j}. \)

Step 3: Substitute Into the Line Integral

Substitute the expressions for \( \mathbf{F} \) and \( d\mathbf{r} \) into the line integral:\[ \int_0^2 \left( (-x) \mathbf{i} - 2x^3 \mathbf{j} \right) \cdot (dx \, \mathbf{i} + 3x^2 \, dx \, \mathbf{j}). \]

Step 4: Compute the Dot Product

Compute the dot product:\[ (-x)(dx) + (-2x^3)(3x^2 dx) = -x \, dx - 6x^5 \, dx. \]

Step 5: Integrate Over the Range

Evaluate the integral over \( x \) from 0 to 2:\[ \int_0^2 (-x \, dx - 6x^5 \, dx) = \int_0^2 (-x - 6x^5) \, dx. \] Breaking the integral:\[ \int_0^2 -x \, dx - \int_0^2 6x^5 \, dx. \]

Step 6: Solve Each Integral

Calculate each integral separately:\[ \int_0^2 -x \, dx = \left[ -\frac{x^2}{2} \right]_0^2 = -2, \]\[ \int_0^2 6x^5 \, dx = \left[ \frac{6x^6}{6} \right]_0^2 = 32. \]

Step 7: Find the Total Work Done

Sum up the results of the integrals to find the total work done:\[-2 - 32 = -34. \]Hence, the work done by the force field \( \mathbf{F} \) on the object moving along path \( C \) is \(-34\).

Key Concepts

Vector Calculus

Vector Calculus is a branch of mathematics that allows us to explore vectors and perform calculations involving multiple variables. It is particularly useful when dealing with physical forces and fields. In our current problem, vector calculus helps us to analyze how the force field \(\mathbf{F} = -x \mathbf{i} - 2y \mathbf{j}\) interacts with an object moving along a path.

• Vectors are quantities that have both magnitude and direction.
• Force fields are often represented as vector fields, where each point in space has a vector indicating the force at that location.

Vector calculus provides us with tools like the gradient, divergence, and curl, but in this exercise, our focus is on understanding line integrals. Simply put, a line integral of a vector field along a curve accounts for the cumulative effect of the field on a path. This allows us to calculate critical physical properties, such as the work done by a force along a path.

Force Fields

Force fields, in a physical sense, describe the influence that a force has over a certain space. Think of them as invisible webs of force that can change the velocity of objects moving within them. In this exercise, the force field is defined as \(\mathbf{F} = -x \mathbf{i} - 2y \mathbf{j}\).

• The vector \(-x \mathbf{i}\) shows that there is a force pulling towards the negative x-direction.
• The vector \(-2y \mathbf{j}\) indicates a force directed towards the negative y-direction, which increases with y.
• This particular field exerts more influence the further you are from the origin along the y-axis.

Understanding the nature of these vectors within the force field helps us visualize the forces at play and how they accumulate as an object moves through the field along a specific path. This visualization aids in setting up the necessary calculations that form the base for finding the work done by the force field.

Parametric Equations

Parametric equations are a way to express a curve or path using a third variable, often time or another parameter. In our problem, instead of using just x and y, we used the parameter \(x\) and express both x and y coordinates in terms of this parameter.

• The given path \(y = x^3\) is expressed using \(x\) as the parameter.
• The derivative \(dy = 3x^2 \, dx\) indicates how y changes with respect to x along the curve.
• This reformulation enables us to efficiently use calculus operations like integration.

Through parametric equations, complex curves can be broken down into simpler, more manageable parts. In mathematical physics, this encourages us to logically handle parts of the problem and to apply calculus techniques such as integration along a path where the objective is to find accumulative quantities like work done.

Work Done

Work done in physics is a measure of energy transfer when an object is moved by a force over a distance. In the context of force fields and path integrals, it measures how much energy a force field applies to an object as it traverses a specified path.

• The formula for work done involves the line integral \(\int_C \mathbf{F} \cdot d\mathbf{r}\), where \(\mathbf{F}\) is the force vector field and \(d\mathbf{r}\) represents a small movement along the path.
• The dot product \(\mathbf{F} \cdot d\mathbf{r}\) allows us to determine how much of the force affects movement along the path.
• Evaluating this line integral over the curve gives us the total work done, which was \(-34\) in this particular problem, indicating energy expended by the force field.

This concept highlights how integral calculus in combination with vector mathematics provides a powerful method to calculate physical properties, allowing us to predict and understand real-world phenomena.