Question

Given the force field \(\mathbf{F}\), find the work required to move an object on the given oriented curve. $$\begin{aligned}&\mathbf{F}=\frac{\langle x, y, z\rangle}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} \text { on the line segment from }(1,1,1) \text { to }\\\&(10,10,10)\end{aligned}$$

Short answer

Question: Calculate the work done by the force field \(\mathbf{F}(\mathbf{r})=\frac{\mathbf{r}}{|\mathbf{r}|^3}\) in moving an object along the oriented curve from point A(1,1,1) to point B(10,10,10). Answer: The work done by the force field \(\mathbf{F}\) in moving the object along the given oriented curve is \(\frac{13}{90}\) units.

Step-by-step solution

Represent the curve as a vector function

To represent the line segment from point A(1,1,1) to point B(10,10,10), we use a vector function \(\mathbf{r}(t)\): $$\mathbf{r}(t) = \mathbf{A} + t(\mathbf{B} - \mathbf{A})$$ where \(\mathbf{A} = \langle 1,1,1 \rangle\) and \(\mathbf{B} = \langle 10,10,10 \rangle\), and \(t\) is a parameter ranging from 0 to 1. So, the vector function becomes: $$\mathbf{r}(t) = \langle 1,1,1 \rangle + t(\langle 10,10,10 \rangle - \langle 1,1,1 \rangle) = \langle 1,1,1 \rangle + t\langle 9,9,9\rangle$$

Compute the derivative of the vector function

Next, we need to find the derivative of \(\mathbf{r}(t)\) with respect to the parameter \(t\). This is given by: $$\frac{d\mathbf{r}}{dt} = \frac{d}{dt}\langle 1+9t, 1+9t, 1+9t \rangle = \langle 9, 9, 9 \rangle$$

Determine the limits of integration

The parameter \(t\) ranges from 0 to 1. Therefore, the limits of integration are 0 and 1.

Substitute the vector function and its derivative into the line integral formula

The line integral formula for the work done by the force field \(\mathbf{F}\) along the curve is given by: $$W=\int_{0}^{1} \mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt} dt$$ First, we substitute \(\mathbf{r}(t)\) into the force field \(\mathbf{F}\), which is given by: $$\mathbf{F}(\mathbf{r}(t)) = \frac{\langle 1+9t, 1+9t, 1+9t\rangle}{\left((1+9t)^{2}+(1+9t)^{2}+(1+9t)^{2}\right)^{3 / 2}}$$ Now, we need to find the dot product between \(\mathbf{F}(\mathbf{r}(t))\) and \(\frac{d\mathbf{r}}{dt}\).

Calculate the line integral

With the components of the line integral in place, we can compute the integral: $$W=\int_{0}^{1}\frac{\langle 1+9t, 1+9t, 1+9t\rangle}{\left((1+9t)^{2}+(1+9t)^{2}+(1+9t)^{2}\right)^{3 / 2}} \cdot \langle 9, 9, 9 \rangle dt = 9\int_{0}^{1} \frac{(1+9t)}{((3+27t)^{2})^{3 / 2}} dt$$ We can make a substitution here: Let \(u = 3+27t\), so \(du = 27 dt\), and when \(t=0\), \(u=3\) and when \(t=1\), \(u=30\). Thus, our integral becomes: $$W = \int_{3}^{30} \frac{(u/9 - 2/9) du}{u^{3}}$$ Solve the integral: $$W =\left[ -\left(\frac{13}{18}\right)\frac{1}{u^2}\right]_{3}^{30} =-\left(\frac{13}{18}\right)\left(\frac{1}{900} - \frac{1}{9}\right) = \frac{13}{90}$$ So, the work required to move the object along the given oriented curve is \(\frac{13}{90}\) units.

Key concepts

Force Fields

A force field is a vector field that represents a force acting at every point in space. Imagine \( \mathbf{F} = \frac{\langle x, y, z\rangle}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} \), as a field of arrows hovering in the air. Each arrow in the vector field indicates the direction and magnitude of the force exerted at that specific position. This concept is crucial in physics, where force fields are used to describe gravitational, electric, and magnetic forces.

In this context, the force field provides a mathematical representation of a force that varies with position. This helps determine the work done by the force when an object traverses a path within that region. Work is essentially the energy transferred by the force, which can be calculated using a line integral method. Understanding force fields enables students to apply mathematical concepts to real-world physical problems, making vector calculus meaningful and applicable outside of the classroom.

Vector Calculus

Vector calculus is a branch of mathematics focused on vector fields and differentiable functions of multiple variables. It is a key tool for solving physical problems involving motion and forces.

• It focuses on differentiation and integration of vector fields.
• Calculates quantities such as gradients, divergences, and curls.
• Essential for analyzing physical phenomena like fluid dynamics and electromagnetism.

In our example, we use vector calculus to describe and analyze the motion of an object through a force field along a specified curve. First, we represent the curve using a parameterized vector function: \( \mathbf{r}(t) = \langle 1+9t, 1+9t, 1+9t \rangle \).

Then, we compute its derivative \( \frac{d\mathbf{r}}{dt} = \langle 9, 9, 9 \rangle \). This derivative highlights how the vector function changes with respect to the parameter \( t \). Utilizing vector calculus allows us to perform complex calculations that are necessary for understanding how different vectors, such as those in force fields, interact with objects moving along paths in space.

Integration Techniques

Integration techniques are methods used to compute integrals, especially line integrals, which are integrals evaluated over a path or curve. They're a bit different from ordinary integrals, because they operate over vector fields and are often parameterized in terms of curves.

In the example, the work performed by the force field along a path is calculated using a line integral. The steps involve drilling into the force vector dotted with the derivative of the position vector, all integrated over the parameter \( t \) from 0 to 1:

• Substitute the vector function and its derivative into the line integral formula.
• Perform substitution: set \( u = 3 + 27t \).
• Change the limits of the integration from \( [0, 1] \) to \( [3, 30] \).

This process requires a clear grasp of how substitutions and limits work in integration. Students should recognize how this technique simplifies solving complicated integrals involved in calculating work done by a force along a curve. Mastering these methods makes problem-solving involving line integrals and vector fields more approachable.