Problem 54 - 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 the Force and Direction

The force \( \mathbf{F} \) has a constant magnitude \( k \) and always points radially outward from the origin. For a point \((x, y)\) on the curve, the vector from the origin to this point is \( \langle x, y \rangle \). Thus, the direction of \( \mathbf{F} \) is along this vector, which means \( \mathbf{F} \) can be expressed as \( k \cdot \frac{\langle x, y \rangle}{\sqrt{x^2 + y^2}} \).

Step 2: Representing the Tangent Vector \( \mathbf{T} \)

The tangent vector \( \mathbf{T} \) at any point on the curve \( y = f(x) \) can be represented as \( \langle 1, f'(x) \rangle \). To use this in our integral, normalize it: \( \mathbf{T} = \frac{\langle 1, f'(x) \rangle}{\sqrt{1 + (f'(x))^2}} \).

Step 3: Dot Product of Force and Tangent Vector

Calculate the dot product \( \mathbf{F} \cdot \mathbf{T} = \left(k \cdot \frac{\langle x, y \rangle}{\sqrt{x^2 + y^2}}\right) \cdot \frac{\langle 1, f'(x) \rangle}{\sqrt{1 + (f'(x))^2}} \). This simplifies to \( k \cdot \frac{x + y \cdot f'(x)}{\sqrt{(x^2 + y^2) \cdot (1 + (f'(x))^2)}} \).

Step 4: Rewriting the Integral for Work Done

The work done \( W \) is given by \( W = \int_{C} \mathbf{F} \cdot \mathbf{T} \, ds \). Substitute \( ds = \sqrt{1 + (f'(x))^2} \, dx \) into the integral: \( W = k \int_{a}^{b} \frac{x + y \cdot f'(x)}{\sqrt{x^2 + y^2}} \, dx \).

Step 5: Simplifying the Integral Expression

Recognize that \( y = f(x) \) and the expression becomes \( W = k \int_{a}^{b} d\sqrt{x^2 + (f(x))^2} \). By the Fundamental Theorem of Calculus, this integrates directly to: \( W = k \left[ \sqrt{b^2 + (f(b))^2} - \sqrt{a^2 + (f(a))^2} \right] \).

Step 6: Conclusion and Result

The work done by the force is \( W = k \left[ \sqrt{b^2 + (f(b))^2} - \sqrt{a^2 + (f(a))^2} \right] \), confirming the problem statement.

Key Concepts

Magnitude of Force

When we talk about force in physics, one of the key characteristics we consider is its magnitude. In this exercise, the force moving the particle has a constant magnitude denoted by \( k \). Think of magnitude as the "size" or "strength" of the force.
For instance, if you are pushing a box with a certain strength, the magnitude is a measure of how strong your push is, without worrying about the direction.
Here, the force's magnitude remains the same (constant) throughout the motion of the particle. This constancy simplifies calculations because we do not need to deal with variable force values. This consistent magnitude also affects how we calculate the work done by the force, as it remains a key factor throughout the process.

Tangent Vector

The tangent vector, usually denoted as \( \mathbf{T} \), is a crucial concept in understanding the direction of movement along a curve. At each point on a smooth curve, there is a tangent vector which tells us the direction in which the curve is heading.
It's like standing on a winding road and looking in the direction the road is going next; that's what the tangent vector represents. In this exercise, for a curve represented by \( y = f(x) \), the tangent vector at any point can initially be expressed as \( \langle 1, f'(x) \rangle \).
However, to ensure consistency across different points, we normalize the vector.

Normalization

Normalization involves scaling a vector so that its "length" is 1, while maintaining its direction. This is done by dividing the vector by its magnitude, which results in the tangent vector \( \mathbf{T} = \frac{\langle 1, f'(x) \rangle}{\sqrt{1 + (f'(x))^2}} \).
By normalizing, we focus purely on direction, which simplifies calculations involving vectors, such as finding the work done by the force along the curve.

Dot Product

The dot product is a mathematical operation that takes two vectors and returns a single number. It is a key tool to find the work done by a force when the force and direction of movement are given as vectors.
In simpler terms, it tells us "how much" of one vector goes in the direction of another. For vectors \( \mathbf{F} \) and \( \mathbf{T} \) in our exercise, the dot product is calculated as:

• \( \mathbf{F} \cdot \mathbf{T} = \left(k \cdot \frac{\langle x, y \rangle}{\sqrt{x^2 + y^2}}\right) \cdot \frac{\langle 1, f'(x) \rangle}{\sqrt{1 + (f'(x))^2}} \)

This product helps in determining how much of the force actually contributes to the work done along the curve.
After simplification, the dot product is \( k \cdot \frac{x + y \cdot f'(x)}{\sqrt{(x^2 + y^2) \cdot (1 + (f'(x))^2)}} \), showing us the effective contribution of the force along the direction of movement at any point.

Integral of Work

In physics, work done by a force is often calculated using integral calculus, especially when dealing with constant forces along curves. The integral combines all the little bits of work over a path, giving us the total work done.
For our problem, we calculate work as \( W = \int_{C} \mathbf{F} \cdot \mathbf{T} \, ds \), where \( C \) is the path of the particle. By substituting \( ds = \sqrt{1 + (f'(x))^2} \, dx \), we rewrite it as:

• \( W = k \int_{a}^{b} \frac{x + y \cdot f'(x)}{\sqrt{x^2 + y^2}} \, dx \)

Here, we integrate over the x-interval \([a, b]\), accumulating all contributions from the force along the curve.
This integral effectively sums up all the infinitesimally small work elements to compute the total work done by the force from the starting to the endpoint.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone concept that links the process of differentiation and integration, allowing us to evaluate definite integrals more easily. In essence, it tells us that integration can be reversed by differentiation.
Applying this theorem makes the integration process straightforward in our exercise. After substituting and simplifying terms in the work integral, the expression becomes:

• \( W = k \int_{a}^{b} d\sqrt{x^2 + (f(x))^2} \)

Because the integral can be directly evaluated using the theorem, the result is:

• \( W = k \left[ \sqrt{b^2 + (f(b))^2} - \sqrt{a^2 + (f(a))^2} \right] \)

This shows how powerful and useful the Fundamental Theorem of Calculus is when dealing with integrals, as it provides a direct route to find the total work done by the force.