Problem 45 - 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 \(f(t)\) from \(t = a\) to \(t = b\).

Step 1: Understanding the Problem

The problem asks about the relationship between a work integral and a specific area. Here, the curve \(C\) is defined by the position vector \(\mathbf{r}(t) = t\mathbf{i} + f(t)\mathbf{j}\) and the force field is \(\mathbf{F} = y\mathbf{i}\). We need to compute the line integral \(\int_{C} \mathbf{F} \cdot d\mathbf{r}\) and find if it relates to the area under the curve \(f(t)\) from \(t = a\) to \(t = b\).

Step 2: Define the Differential Path Element

We need to find \(d\mathbf{r}\), the differential path element. Since \(\mathbf{r}(t) = t\mathbf{i} + f(t)\mathbf{j}\), its derivative with respect to \(t\) is \(\mathbf{r}'(t) = \mathbf{i} + f'(t)\mathbf{j}\). Therefore, \(d\mathbf{r} = (1\mathbf{i} + f'(t)\mathbf{j}) dt\).

Step 3: Find the Dot Product

Find the dot product \(\mathbf{F} \cdot d\mathbf{r}\). Since \(\mathbf{F} = y\mathbf{i}\), substitute \(y = f(t)\) into \(\mathbf{F}\), giving \(\mathbf{F} = f(t)\mathbf{i}\). Thus, \(\mathbf{F} \cdot d\mathbf{r} = f(t)\mathbf{i} \cdot (\mathbf{i} + f'(t)\mathbf{j}) dt = f(t) dt\).

Step 4: Evaluate the Line Integral

The work integral becomes \[ \int_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{a}^{b} f(t) \, dt \]. This integral calculates the area under the curve \(f(t)\) from \(t = a\) to \(t = b\), essentially representing the area of the region defined by the curve \(f\), the \(t\)-axis, and the lines \(t=a\) and \(t=b\).

Step 5: Conclusion

The line integral \(\int_{C} \mathbf{F} \cdot d\mathbf{r}\) is equal to the area under the curve \(f(t)\) between \(t=a\) and \(t=b\). Therefore, there is a direct relationship between the work integral and the area of the specified region.

Key Concepts

Work Integral

The concept of a work integral is key in vector calculus and physics as it relates to the calculation of work done by a force over a certain path. In simple terms, the work integral involves calculating how much work a force vector does as an object moves along a path. A line integral represents this calculation when the force field is described by vectors.
In this specific exercise, the force field is given as \( \mathbf{F} = y \mathbf{i} \). This implies the force works in the direction of the x-axis with a magnitude equivalent to the y-component at any point on the curve. Work integral is computed using

• the force vector \( \mathbf{F} \)
• the differential path element \( d\mathbf{r} \)

Their dot product \( \mathbf{F} \cdot d\mathbf{r} \) simplifies the work done calculation over the path \( C \). Here, the work integral equals the integral of \( f(t) \) with respect to \( t \) from \( t = a \) to \( t = b \). This line integral, surprisingly, mirrors the area under the curve \( f(t) \), illustrating an elegant connection between work in vector calculus and geometry.

Area Under the Curve

The area under a curve is a fundamental concept in calculus. It represents the accumulation of values (or the integral) of a function over a given interval. This exercise smartly links the area under the curve of \( f(t) \) to a concept from vector calculus known as the work integral.
When we calculate the integral \( \int_{a}^{b} f(t) \, dt \), it gives us a straight measure of the area between the curve, the \( t \)-axis, and the vertical lines at \( t = a \) and \( t = b \). Instead of dealing with the complicated geometry of the curve \( f(t) \), this integral approach boils the calculation down to a simple sum over a continuous span.
Because \( f(t) \) is positive and differentiable, every infinitesimal part of this curve contributes positively to the total. The exercise shows that this integral, by design, mirrors the physical work done by a vector force field, emphasizing the intrinsic relationship between these mathematical values.

Vector Calculus

Vector calculus is a branch of mathematics that deals with vectors and the differentiation and integration of vector fields. It is crucial to physics and engineering to describe and compute the behavior of various vector quantities in space.
In our exercise, vector calculus shows its power through the ability to calculate both a work integral and determine the area under a curve using vectors. The vector representation \( \mathbf{r}(t) = t\mathbf{i} + f(t)\mathbf{j} \) describes a path in the plane, combining parametric descriptions with vector notations for clarity and accuracy. The integral of \( \mathbf{F} \cdot d\mathbf{r} \) employs vector calculus methods, clearly elucidating the relationship between physical phenomena like force and geometric representations like area.

• Visualizing paths via vectors allows us to conduct operations over curves.
• Use of dot product helps convert vector integrals to scalar values.

This exercise exemplifies how vector calculus elegantly ties physical concepts to mathematical analysis, providing profound insights into the geometry behind many physical problems.