Problem 46 - Thomas Calculus

Question

Work done by a radial force with constant magnitude \(A\) particle moves along the smooth curve \(y=f(x)\) from \((a, f(a))\) to \((b, f(b)) .\) The force moving the particle has constant magnitude \(k\) and always points away from the origin. Show that the work done by the force is $$ \int_{C} \mathbf{F} \cdot \mathbf{T} d s=k\left[\left(b^{2}+(f(b))^{2}\right)^{1 / 2}-\left(a^{2}+(f(a))^{2}\right)^{1 / 2}\right] $$

Step-by-Step Solution

Answer

The work done is \( k \left[\sqrt{b^2 + (f(b))^2} - \sqrt{a^2 + (f(a))^2} \right] \).

Step 1: Understand Force and Path

The force \( \mathbf{F} \) has a constant magnitude \( k \) and is radial, meaning it points in the direction of the vector from the origin \((0, 0)\) to a point \((x, y)\). The path, \( C \), is defined by the curve \( y = f(x) \), and we're moving from \((a, f(a))\) to \((b, f(b))\).

Step 2: Define the Tangent Vector

The unit tangent vector \( \mathbf{T} \) to the curve at any point is defined by parameterizing \( y = f(x) \) as \( \mathbf{r}(t) = (x(t), f(x(t))) \). The tangent vector \( \mathbf{T} \) is \( \frac{d\mathbf{r}}{dt} / \left| \frac{d\mathbf{r}}{dt} \right| \).

Step 3: Express the Force in Terms of Coordinates

The force \( \mathbf{F} \) with magnitude \( k \) is in the direction of \((x, y)\), thus \( \mathbf{F} = \frac{k}{\sqrt{x^2 + y^2}} (x, y) \).

Step 4: Compute the Dot Product

The work done is the integral along the path \( C \) of the dot product \( \mathbf{F} \cdot \mathbf{T} \). Calculate \( \mathbf{F} \cdot \mathbf{T} = \frac{k}{\sqrt{x^2 + y^2}} (x \cos \theta + y \sin \theta) \), where \( \theta \) defines the angle of the curve's tangent vector.

Step 5: Integrate Along the Path

The integration step is \( \int_{C} \mathbf{F} \cdot \mathbf{T} \, ds = k \int_{a}^{b} \frac{x \, dx + f(x) \, f'(x) \, dx}{\sqrt{x^2 + (f(x))^2}} = k \left[ \sqrt{x^2 + (f(x))^2} \right]_a^b \).

Step 6: Simplify the Result

After integrating, plug in the end points to find the total work: \[ k \left[ \sqrt{b^2 + (f(b))^2} - \sqrt{a^2 + (f(a))^2} \right] \]. This matches exactly with the given expression for work done by force.

Key Concepts

Vector Calculus

Vector calculus can be thought of as an extension of the calculus you're familiar with, but it operates in multi-dimensional spaces. It's all about vectors, which include directions and magnitudes. In vector calculus, we often focus on fields, which are functions defined over space with vector outputs.

In the exercise on work done by a radial force, we're dealing with vector fields. The radial force acting on a particle is not a single value; instead, it varies with direction and position along the path between two points. Understanding how these vector fields operate and interact with paths is central to applying vector calculus concepts.

• Vector calculus helps us analyze and visualize the behavior of vectors in space.
• It is particularly useful for physical applications, such as understanding forces and motion.

This exercise also uses these concepts to derive the work done by a force, which involves looking at the path and the force together.

Line Integrals

Line integrals are a crucial concept in vector calculus. They allow you to calculate quantities along a curve or path, which is important when dealing with physical phenomena like work, where force is applied across a distance.

In simple terms, a line integral is an accumulation of values along a line or a curve. In this context, you're summing up the contributions of force along a path defined by the curve \(y = f(x)\).

• The idea is to multiply the force by an infinitesimally small segment of the path.
• Mathematically, this involves integration, combining the concepts of vectors and paths.

The process here involves computing the dot product of the force vector \(\mathbf{F}\) and the tangent vector \(\mathbf{T}\), followed by integrating along the entire path. This gives the total work done by the radial force on the particle as it moves between the two points.

Radial Force Analysis

When analyzing radial forces, which originate from or point towards a common point (often the origin), a distinct approach is needed. These forces have a geometry that affects their interaction with paths and objects.

In the problem context, the radial force is constant but always points away from the origin. Analyzing this requires understanding the direction and magnitude of the force along the curve \(y = f(x)\). The force's radial nature means it aligns with the position vector, affecting calculations of work done along the path.

• Radial forces are always directed along the line from the origin to the point of interest.
• This makes their magnitude dependent on the position of the point in the field.

The exercise highlights how such forces can cause work, calculated by integrating the dot product of the radial force and the tangent vector over a path. It reveals that the work done is proportional only to specific spatial properties of the path endpoints, showing the unique characteristics of radial forces.