91Ó°ÊÓ

A vector field is essentially a function that assigns a vector to every point in space. For this problem, the vector field is given by \(\mathbf{F} = x^2 \mathbf{i} - y \mathbf{j}\). Each vector in the field

• Points in a specific direction.
• Has a specific magnitude.

The vector field can be visualized as a flow where each point's vector indicates the direction of the flow and its strength. In our case, the vector at any point \((x, y)\) is determined by the components \(x^2\) and \(-y\), aligning with the coordinate axes \(\mathbf{i}\) and \(\mathbf{j}\). This indicates that:

• The field strength increases with \(x\) squared, enhancing movement in the \(\mathbf{i}\) direction.
• The negative sign on the \(\mathbf{j}\) component suggests downward influence on the y-axis.

Understanding the design and effect of vector fields helps in evaluating the line integral, as it determines how the path interacts with the vector field connected to it.

Parameterization is the technique of expressing the curve's points in terms of a single variable parameter, often \(t\). In this exercise, the curve is defined by \(x = y^2\), with endpoints (4,2) to (1,-1). To parameterize:

• Choose \( y = t \) as the parameter.
• Then \( x = t^2 \), leading to the parameterization: \( \mathbf{r}(t) = (t^2, t) \).

This formula implies that as \( t \) changes from 2 to -1, the curve will be traced backward because:

• At \( t = 2 \), it reaches the starting point (4,2).
• At \( t = -1 \), it reaches the endpoint (1,-1).

Through parameterization, the curve is dissected into infinitesimal segments, simplifying the computation of the line integral later by providing explicit functions for each coordinate based on \(t\).

A tangent vector provides the slope or direction of the curve at a specific point, essentially pointing along the path of the curve. To find the tangent vector for the parameterized curve \(\mathbf{r}(t) = (t^2, t)\),

• Differentiate to get \(\mathbf{r}'(t) = (2t, 1)\).
• The magnitude of this tangent vector, or its norm, is given by \(\|\mathbf{r}'(t)\| = \sqrt{(2t)^2 + 1^2} = \sqrt{4t^2 + 1}\).
• The unit tangent vector is obtained by normalizing \(\mathbf{r}'(t)\): \(\mathbf{T} = \frac{1}{\sqrt{4t^2 + 1}}(2t, 1)\).

Here the unit tangent vector \(\mathbf{T}\) is crucial for the line integral because it defines the direction of each small segment of the curve along which \(\mathbf{F}\)'s work is calculated. By integrating these contributions, the total work along \(C\) can be discerned.

Integration techniques are essential for calculating complex integrals, such as line integrals. In this problem, the line integral of a vector field along the parameterized curve \(\int_C \mathbf{F} \cdot \mathbf{T} \, d s\) necessitates several strategic steps:

• Substitute the parameterized \(\mathbf{F}(\mathbf{r}(t)) = (t^4, -t)\) into the dot product with \(\mathbf{T}\).
• Calculate the dot product: \((t^4, -t) \cdot \left(\frac{1}{\sqrt{4t^2 + 1}}(2t, 1)\right) = \frac{2t^5 - t}{\sqrt{4t^2 + 1}}\).
• The simplified line integral becomes: \(\int_{2}^{-1} (2t^5 - t) \, dt\).
• Integrate each term independently: \(\int 2t^5 \, dt\) and \(\int -t \, dt\).
• Add the results and evaluate the definite integral at limits 2 and -1.

This method streamlines complex multi-variable calculus into simpler stepwise operations, leveraging the derivative and definite properties to yield a result efficiently. The final evaluation ensures the integration captures every nuance from start to end of \(C\), revealing the vector field's net effect.