Question

Evaluate a line integral to show that the work done in moving an object from point \(A\) to point \(B\) in the presence of a constant force \(\mathbf{F}=\langle a, b, c\rangle\) is \(\mathbf{F} \cdot \overrightarrow{A B}\)

Short answer

Answer: The work done in moving an object from point A to point B in the presence of a constant force \(\mathbf{F}=\langle a, b, c\rangle\) is \(\mathbf{F} \cdot \overrightarrow{A B}\).

Step-by-step solution

1. Path parameterization

Let the path \(\overrightarrow{A B}\) be parameterized by a function \(\mathbf{r}(t)\), where \(t\) ranges from 0 to 1, and \(\mathbf{r}(0)=A\) and \(\mathbf{r}(1)=B\). Then the path can be represented as: $$\mathbf{r}(t) = A + t(\overrightarrow{A B})$$

2. Calculate the derivative of the path

Now we need to find the derivative of \(\mathbf{r}(t)\) with respect to \(t\). This gives us the tangent vector to the path at each point: $$\frac{d\mathbf{r}}{dt} = \frac{d}{dt}(A + t(\overrightarrow{A B})) = \overrightarrow{A B}$$

3. Calculate the dot product

Next, we need to calculate the dot product between the force vector \(\mathbf{F}\) and the tangent vector \(\frac{d\mathbf{r}}{dt}\): $$\mathbf{F} \cdot \frac{d\mathbf{r}}{dt} = \langle a, b, c \rangle \cdot \overrightarrow{A B} = a(A_x - B_x) + b(A_y - B_y) + c(A_z - B_z) = \mathbf{F} \cdot \overrightarrow{A B}$$

4. Integrate the dot product

To find the work done in moving the object from point A to point B, we need to integrate the dot product over the interval from 0 to 1: $$W = \int_{0}^{1} \mathbf{F} \cdot \frac{d\mathbf{r}}{dt} \, dt = \int_{0}^{1} \mathbf{F} \cdot \overrightarrow{A B} \, dt$$

5. Express the work done

Since \(\mathbf{F} \cdot \overrightarrow{A B}\) is a constant with respect to \(t\), we can pull it out of the integral: $$W = \mathbf{F} \cdot \overrightarrow{A B} \int_{0}^{1} dt= \mathbf{F} \cdot \overrightarrow{A B} [t]_{0}^{1} = \mathbf{F} \cdot \overrightarrow{A B} (1-0) = \mathbf{F} \cdot \overrightarrow{A B}$$ Hence, we have shown that the work done in moving an object from point A to point B in the presence of a constant force \(\mathbf{F}=\langle a, b, c\rangle\) is indeed \(\mathbf{F} \cdot \overrightarrow{A B}\).

Key concepts

Vector Calculus

Vector calculus is a fundamental component in the study of physics and engineering, providing a way to describe and analyze the properties of vector fields. At its heart, vector calculus deals with vector quantities that have both magnitude and direction.

In the context of our exercise, we use vector calculus to describe the motion of an object along a path from point A to point B. This path can be represented by a vector function \( \mathbf{r}(t) \), which gives us the position vector of the object at any point in time between the start and the end of its journey. Parameterization is a pivotal concept, representing the path as a smooth curve in space, which is described by continuous and differentiable functions. This allows us to apply the calculus tools of differentiation and integration to paths in three dimensions.

By differentiating the path function, we get a tangent vector to the path, which tells us the direction of the motion along the path at any moment. Then, integrating certain properties of this vector field, such as a force field, over the path, gives us physical quantities of interest, such as the work done on an object as it moves along this path.

Dot Product

The dot product, also known as scalar product, is an operation that takes two equal-length sequences of numbers (usually coordinate vectors), and returns a single number. This operation combines two vectors to produce a scalar, and it’s intimately tied to the concept of the angle between the vectors.

The dot product is calculated as \( \mathbf{A} \cdot \mathbf{B} = A_xB_x + A_yB_y + A_zB_z \), where \( A_x, A_y, A_z \) and \( B_x, B_y, B_z \) are the components of vectors \( \mathbf{A} \) and \( \mathbf{B} \) respectively.

In our exercise, we used the dot product to compute the work done, reflecting the product of the component of the force in the direction of the displacement and the magnitude of that displacement. Intuitively, the dot product measures how much of the force \( \mathbf{F} \) acts in the direction of the path traveled by the object. If the force and the direction are orthogonal, the dot product is zero, implying no work is done by the force in that specific direction.

Work Done by Force

Work done by a force is a key concept in physics, representing the energy transferred to an object via the application of a force through a distance. It is a scalar quantity calculated as the dot product of the force exerted on an object and the displacement of the object.

In mathematical terms, work done W by a constant force \( \mathbf{F} \) on an object that moves from point A to point B is given by \( W = \mathbf{F} \cdot \overrightarrow{A B} \). This statement means that the work done is the product of the magnitude of the force, the magnitude of the displacement, and the cosine of the angle between them.

Our original exercise showcases that if the force is constant and the path of the object is straight, the work done can be simplified to the product of the force and displacement vectors. This simplification occurs because the integral of a constant with respect to a variable is simply the product of that constant and the variable's range of change, in this case, the force vector and the path vector.