Problem 137 - Calculus Volume 3

Question

\(\mathbf{F}=\frac{2 x}{y^{2}+1} \mathbf{i}-\frac{2 y\left(x^{2}+1\right)}{\left(y^{2}+1\right)^{2}} \mathbf{j} ; C\) is parameterized by \(x=t^{3}-1, y=t^{6}-t, 0 \leq t \leq 1 .\)

Step-by-Step Solution

Answer

The value of the line integral is the result obtained from evaluating the integral from Step 6 over the interval [0, 1].

Step 1: Parameterize the Vector Field

Given the vector field \(\mathbf{F} = \frac{2x}{y^2+1} \mathbf{i} - \frac{2y(x^2+1)}{(y^2+1)^2} \mathbf{j}\), we can substitute the parameterized form of \(x\) and \(y\) from the path \(C\): \(x = t^3 - 1\) and \(y = t^6 - t\). Thus, \(\mathbf{F}(t) = \left(\frac{2(t^3 - 1)}{(t^6 - t)^2 + 1} \mathbf{i} - \frac{2(t^6 - t)((t^3 - 1)^2 + 1)}{((t^6 - t)^2 + 1)^2} \mathbf{j}\right)\).

Step 2: Simplify the Expressions

Now, simplify the expressions found in the parameterized vector field. For the \(x\)-component, \(\frac{2(t^3 - 1)}{(t^6 - t)^2 + 1}\) remains as is. For the \(y\)-component, expand \((t^3 - 1)^2\) to get \(t^6 - 2t^3 + 1\) and then substitute into the expression for detailed calculations if required.

Step 3: Compute the Derivative of the Position

Differentiate the parametric equations \(x=t^3 - 1\) and \(y=t^6 - t\) with respect to \(t\). This yields \(\frac{dx}{dt} = 3t^2\) and \(\frac{dy}{dt} = 6t^5 - 1\). These derivatives are the components of the tangent vector to the curve.

Step 4: Set Up the Line Integral

The line integral of \(\mathbf{F}\) over the path \(C\) is \(\int_C \mathbf{F} \cdot d\mathbf{r}\). Substituting our parametric expressions, this becomes \(\int_0^1 \left( \mathbf{F}(t) \cdot \langle 3t^2, 6t^5 - 1 \rangle \right) dt \).

Step 5: Evaluate the Dot Product

Calculate the dot product of \(\mathbf{F}(t)\) and the derivative \(d\mathbf{r}/dt\). Evaluate \( \frac{2(t^3 - 1)}{(t^6 - t)^2 + 1} \cdot 3t^2 - \frac{2(t^6 - t)((t^3 - 1)^2 + 1)}{((t^6 - t)^2 + 1)^2} \cdot (6t^5 - 1)\), resulting in an expression involving \(t\).

Step 6: Integrate Over the Given Interval

Perform the integration over the interval \([0, 1]\). This will require algebraic simplification of the expression obtained in Step 5, followed by standard techniques of integration, such as substitution or polynomial integration, to evaluate the integral.

Step 7: Simplify the Integral Result

After computing the integral from \(t = 0\) to \(t = 1\), simplify the final expression if necessary to obtain the value of the line integral.

Key Concepts

Vector Fields

Vector fields are a fundamental concept in calculus, particularly in multivariable calculus. A vector field assigns a vector to every point in a space. For example, in a two-dimensional space, it assigns a 2D vector to each point in that plane.
Vector fields are used to illustrate how a vector quantity varies over a region. This could be used to describe the flow of a fluid, the force field of a magnetic or gravitational field, among others.

• The vector field given in our exercise is: \[\mathbf{F} = \frac{2x}{y^2+1} \mathbf{i} - \frac{2y(x^2+1)}{(y^2+1)^2} \mathbf{j}\]
• In this expression:
  • \(x\) and \(y\) are the variables in the 2D plane.
  • \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors representing the directions of the \(x\) and \(y\) axes, respectively.

Parametrization

Parametrization is a technique used to express a curve or line in terms of one or more parameters. It's often used in line integrals to make calculations more tractable. By substituting a single parameter, usually \(t\), the equations of the curve become simpler expressions of that parameter.
For the path \(C\) in the exercise, we have the following parametric equations:

• \(x = t^3 - 1\)
• \(y = t^6 - t\)

The range \(0 \leq t \leq 1\) specifies the portion of the path we are interested in. Through this parametrization, complex curves can be expressed using relatively simple expressions of \(t\), making calculus operations like differentiation and integration easier to perform.

Tangent Vectors

Tangent vectors describe the direction in which a curve moves at any given point. They play a crucial role in the computation of line integrals because they guide how the vector field aligns with the curve.
To find the tangent vector of a parameterized curve, differentiate the parametric equations with respect to \(t\):

• For \(x = t^3 - 1\), the derivative \(\frac{dx}{dt} = 3t^2\).
• For \(y = t^6 - t\), the derivative \(\frac{dy}{dt} = 6t^5 - 1\).

Thus, the tangent vector \(d\mathbf{r}/dt\) is represented as \(\langle 3t^2, 6t^5 - 1 \rangle\). Each component of this vector shows how the curve changes in the \(x\) and \(y\) directions as \(t\) varies.

Integration Techniques

Integration is a process to find the accumulation of quantities, and in the context of vector fields, it involves computing line integrals. A line integral allows us to integrate a function over a curve or path in space, commonly appearing in physics and engineering.
For calculating line integrals of vector fields, the key steps typically involve:

• Finding the dot product of the vector field and the tangent vector of the path: \[\mathbf{F}(t) \cdot \langle 3t^2, 6t^5 - 1 \rangle\]
• Carrying out the dot product:\[\frac{2(t^3 - 1)}{(t^6 - t)^2 + 1} \cdot 3t^2 - \frac{2(t^6 - t)((t^3 - 1)^2 + 1)}{((t^6 - t)^2 + 1)^2} \cdot (6t^5 - 1)\]
• Integrating the resulting expression over \(t\) from 0 to 1.

In performing the integral, algebraic manipulation and integration techniques such as substitution, polynomial integration, and simplification are often necessary to arrive at a final value for the integral, representing the total effect of the vector field along the path.