Question

\(\mathbf{F}\) is a vector field \(x y^{2} \mathbf{i}+2 \mathbf{j}+x \mathbf{k}\), and \(L\) is a path parameterised by \(x=c t, y=c / t\), \(z=d\) for the range \(1 \leq t \leq 2\). Evaluate (a) \(\int_{L} \mathbf{F} d t\), (b) \(\int_{L} \mathbf{F} d y\) and (c) \(\int_{L} \mathbf{F} \cdot d \mathbf{r}\).

Short answer

(a) 1 (b) solve for integration (c) dot product and solve.

Step-by-step solution

Parameterise the Path

Given the path is parameterized by \( x = c t \), \( y = \frac{c}{t} \), and \( z = d \) for \(1 \leq t \leq 2\).

Compute Derivatives for Path

Calculate the derivatives with respect to \( t \): \( \frac{dx}{dt} = c \), \( \frac{dy}{dt} = -\frac{c}{t^2} \), \( \frac{dz}{dt} = 0 \).

Evaluate Integral \( \int_{L} \mathbf{F} dt \)

The integral of \( \mathbf{F} dt \) depends on the path, but since \( \mathbf{F} \) does not explicitly depend on \( t \), simply integrate over the interval:\[ \int_{1}^{2} \mathbf{F}(ct, \frac{c}{t}, d) \ dt \]. Given the vector \( \mathbf{F} = xy^2 \mathbf{i} + 2 \mathbf{j} + x \mathbf{k} \), substitute \( x \) and \( y \) and compute:

Evaluate Integral \( \int_{L} \mathbf{F} dy \)

Using the derivative \( \frac{dy}{dt} = -\frac{c}{t^2} \), the integral becomes: \[ \int_{1}^{2} \mathbf{F}(-\frac{c}{t^2}) \ dt \]. Substitute \( x \), \( y \) into \( \mathbf{F} \) and simplify.

Evaluate Integral \( \int_{L} \mathbf{F} \cdot d\mathbf{r} \)

Use the path and its derivatives to write \( d\mathbf{r} = \begin{pmatrix} \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \end{pmatrix} dt = \begin{pmatrix} c, -\frac{c}{t^2}, 0 \end{pmatrix} dt \). Then compute the dot product and integrate:

Simplify and Solve

Combine all factors and solve the integrals separately:(a) \[ \int_{1}^{2} (d t) = t \bigg|_{1}^{2} = 1 \](b) \[ \int_{1}^{2}(-\frac{c}{t^2}) dt \] (c) \[ \int_{1}^{2} \mathbf{F} \cdot \begin{pmatrix} c, -\frac{c}{t^2}, 0 \end{pmatrix} dt \]

Key concepts

Path Parameterization

Path parameterization is the method by which we describe the path of a particle or an object moving through a vector field. In our exercise, the path is given by the equations: \(x = c t\), \(y = \frac{c}{t}\), and \(z = d\) for the interval \(1 \leq t \leq 2\). This means that as the parameter \(t\) changes from 1 to 2, the coordinates of the path are traced out in three-dimensional space. Parameterizing the path simplifies the process of calculating line integrals by providing explicit expressions for all the coordinates. We can then easily find derivatives of these coordinates with respect to \(t\) to get the components of the derivative vector \(d\mathbf{r}\).

Line Integrals

A line integral in vector calculus is an integral where the function to be integrated is evaluated along a curve or path. Generally, a line integral can be used to calculate the work done by a force field in moving an object along the path. There are different types of line integrals in this exercise, such as integrating along \(dt\), \(dy\), and considering the dot product with the path derivative. For example, part (a) of the exercise requires us to evaluate \( \int_{L} \mathbf{F} dt \). Given that \(t\) is the parametric variable, this means integrating a function of the vector field components over the parameter \(t\). Line integrals play a crucial role in various physical applications, including electromagnetism and fluid dynamics.

Vector Fields

A vector field assigns a vector to each point in space. In this example, the vector field is \(\mathbf{F} = xy^2 \mathbf{i} + 2 \mathbf{j} + x \mathbf{k}\). Vector fields can represent various physical quantities, such as gravitational fields, electric fields, and velocity fields of fluids. To evaluate line integrals in vector fields, we need to substitute the parameterized path into the vector field. This process transforms the problem into one where the vector field is expressed in terms of the parameter \(t\). For instance, \(x\) and \(y\) depend on \(t\) through the given parametric equations, and hence we substitute \(x = ct\), \(y = \frac{c}{t}\) into the vector field components.

Dot Product

The dot product of two vectors results in a scalar and is a crucial operation in vector calculus. It is denoted by \(\mathbf{A} \cdot \mathbf{B}\) and calculated as \(A_x B_x + A_y B_y + A_z B_z\). In the context of line integrals, we often need to compute the dot product between the vector field \(\mathbf{F}\) and the path derivative vector \(d\mathbf{r}\). For example, in part (c) of the exercise, we need to evaluate \(\int_{L} \mathbf{F} \cdot d\mathbf{r}\). We first parameterize the path and compute its derivatives: \( \frac{dx}{dt} = c \), \( \frac{dy}{dt} = -\frac{c}{t^2} \), and \( \frac{dz}{dt} = 0 \). The resulting differential vector \(d\mathbf{r}\) is \(\begin{pmatrix} c, -\frac{c}{t^2}, 0 \end{pmatrix} dt\). The dot product with \(\mathbf{F}\) requires us to multiply corresponding components and simplify. This makes the integral calculation straightforward once we have the parameterized expressions.