Problem 53 - Thomas Calculus

Question

Work and area Suppose that \(f(t)\) is differentiable and positive for \(a \leq t \leq b .\) Let \(C\) be the path \(\mathbf{r}(t)=t \mathbf{i}+f(t) \mathbf{j}, a \leq t \leq b\) and \(\mathbf{F}=y \mathbf{i} .\) Is there any relation between the value of the work integral $$\int_{C} \mathbf{F} \cdot d \mathbf{r}$$ and the area of the region bounded by the \(t\) -axis, the graph of \(f\) and the lines \(t=a\) and \(t=b ?\) Give reasons for your answer.

Step-by-Step Solution

Answer

The work integral equals the area under the curve of \( f(t) \).

Step 1: Understanding the Work Integral

The work integral \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \) calculates the work done by a force field \( \mathbf{F} \) along a path \( C \). Here, \( \mathbf{F} = y \mathbf{i} \), which contains only a horizontal component dependent on \( y \). The path \( C \) is parameterized by \( \mathbf{r}(t) = t \mathbf{i} + f(t) \mathbf{j} \). Thus, \( d\mathbf{r} = (1 \mathbf{i} + f'(t) \mathbf{j}) dt \).

Step 2: Calculating the Dot Product

For the work integral, we need to calculate the dot product \( \mathbf{F} \cdot d\mathbf{r} \). With \( \mathbf{F} = f(t) \mathbf{i} \), we find \( \mathbf{F} \cdot d \mathbf{r} = f(t) \cdot 1 dt = f(t) dt \). This simplifies the integral to just depend on \( f(t) \).

Step 3: Evaluating the Work Integral

Now, the work integral becomes \( \int_{a}^{b} f(t) \, dt \), which represents the area under the curve of \( f(t) \) from \( t = a \) to \( t = b \). This matches the definition of the definite integral to compute the area between \( f(t) \) and the \( t \)-axis, bounded by \( t=a \) and \( t=b \).

Step 4: Relating Work to Area

Since the work integral \( \int_{a}^{b} f(t) \, dt \) directly equals the integral representing the area under the graph of \( f \), it shows that the work done by \( \mathbf{F} \) along the path \( C \) is indeed equivalent to the area between the graph of \( f(t) \), the \( t \)-axis, and the vertical lines \( t = a \) and \( t = b \).

Key Concepts

Work Integral

In physics, the concept of a work integral is essential in understanding how forces act over distances. A work integral \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \) is used to compute the work done by a force field \( \mathbf{F} \) along a specific path \( C \). This path is usually described by a parametric equation. In our context, the force \( \mathbf{F} = y \mathbf{i} \) means the work is calculated considering only horizontal movement, as the force has no vertical component.
The expression \( d\mathbf{r} \) represents the differential path element, or the small steps taken along the path. Calculating \( d\mathbf{r} \) gives us the full description of movement along the path, which was found to be \( (1 \mathbf{i} + f'(t) \mathbf{j}) dt \).
The dot product \( \mathbf{F} \cdot d\mathbf{r} \) signifies how much of the force is aligned with the movement along \( C \). Here, it boils down to \( f(t) \cdot 1 \ dt = f(t) dt \) because \( \mathbf{F} \) aligns entirely with the \( i \) component, leading directly to our integral of \( f(t) \) over the domain \( [a, b] \).

Path Integral

The path integral concept is closely tied to the essence of work done across a path. When performing a path integral, one evaluates a function along a curved line or a path. The path \( C \) here is defined by \( \mathbf{r}(t) = t \mathbf{i} + f(t) \mathbf{j} \), illustrating a trajectory through space based on the variable \( t \).
This integral sums up the effect of the vector field \( \mathbf{F} \) across each infinitesimally small segment of the path, measuring how much work is exerted in the direction of travel. As computed, it showed that the work along the path \( C \) relies solely on the changes and the integral of \( f(t) \), thus, converting into a calculation of area as well.
Path integrals are powerful in applications involving electromagnetism and fluid dynamics, where fields impact moving objects.

Definite Integral

A definite integral is a fundamental concept in calculus that allows us to calculate the accumulated quantity of a function between two limits, \( a \) and \( b \). It is represented by \( \int_{a}^{b} f(t) \, dt \) and provides the net area between the curve \( f(t) \) and the t-axis between the points \( t=a \) and \( t=b \).
In our context, the definite integral has a dual role: it equates to the work integral \( \int_{a}^{b} f(t) \, dt \), demonstrating how the physical concept of work correlates to mathematical area. The definite integral's value represents this work done, or area under the curve, making it a versatile tool for connecting geometry to physical phenomena.
Integrals can help solve various problems, from finding displacement from velocity to computing total produced by a varying process.

Area under a Curve

The area under a curve is a geometric interpretation of the definite integral. It represents how much space lies beneath the curve of a function \( f(t) \) above the t-axis, within a specific interval \( [a,b] \). This area is crucial for many real-world applications.
The calculation of this area as shown in the solution is directly connected to work done, as they are found using the same integral \( \int_{a}^{b} f(t) \, dt \). This reveals the intimate relationship between the work done by a force and the geometrical area interpretation.
The idea of area under a curve is not limited to physics; it also finds use in probability (Cumulative Distribution Functions), economics (Consumer Surplus), and many other fields, making it a key link between algebraic and graphical analysis of functions.