91Ó°ÊÓ

When evaluating line integrals, one significant step involves parameterizing the vector field in question. A vector field assigns a vector to every point in space. In our integral problem, the given vector field is \( \mathbf{F}(x, y) = x y^{2} \mathbf{i} - x^{2} \mathbf{j} \). To express this field along a specific path or curve, such as \( C \) parameterized by \( \mathbf{r}(t) = t^{3} \mathbf{i} + t^{2} \mathbf{j} \), you substitute the path's expressions \( x = t^{3} \) and \( y = t^{2} \) into the vector field. This transforms the continuous spatial vector field into a vector function over the parameter \( t \), simplifying future calculations. The transformed vector field is \( \mathbf{F}(t) = t^7 \mathbf{i} - t^6 \mathbf{j} \), indicating both the magnitude and direction of the vector field as \( t \) varies along the curve.

The dot product is a vital operation in vector calculus used to compute a scalar from two vectors. In contexts like line integrals, the dot product helps in projecting one vector along another. For our exercise, we need to compute \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} \).

• The vector \( \mathbf{F}(t) = t^7 \mathbf{i} - t^6 \mathbf{j} \) is derived from parameterizing the vector field.
• \( \frac{d\mathbf{r}}{dt} = 3t^2 \mathbf{i} + 2t \mathbf{j} \) is the derivative of the path \( \mathbf{r}(t) \).

The dot product, \( (t^7 \cdot 3t^2) + (-t^6 \cdot 2t) \), simplifies to \( 3t^9 - 2t^7 \). It calculates a singular value that gives insight into how the vector field interacts with the curve at each point.

Integral calculus provides the tools to accumulate quantities across a continuous range, which is especially useful in finding areas, volumes, and other quantities in spaces. For line integrals, this involves summing the contributions calculated from the dot product over a specified interval of the parameter \( t \).
To solve our problem, we need to compute the integral: \[\int_{0}^{1} (3t^9 - 2t^7) \, dt\]
This integral accumulates the effects of the vector field along the curve from \( t = 0 \) to \( t = 1 \). By integrating, we discover the accumulated impact of \( \mathbf{F} \) over the curve \( C \), yielding \( \frac{3}{10}t^{10} - \frac{1}{4}t^8 \) evaluated at the boundaries \( 0 \) and \( 1 \), which ultimately gives the result \( \frac{1}{20} \).

Derivatives are a fundamental tool in calculus that measure how a function changes as its input changes. They are crucial in line integrals to describe how the vector path changes. This is encapsulated in the derivative of the parameterized path \( \mathbf{r}(t) \).
In this exercise, the path \( \mathbf{r}(t) = t^{3} \mathbf{i} + t^{2} \mathbf{j} \) has a derivative of \( \frac{d\mathbf{r}}{dt} = 3t^2 \mathbf{i} + 2t \mathbf{j} \). This derivative tells us the velocity vector of the path \( C \) as \( t \) changes. Derivatives allow us to calculate how the path interacts and aligns with the vector field along every distinct segment of \( t \). Knowing \( \frac{d\mathbf{r}}{dt} \) thus enables us to compute the dot product, which contributes directly to evaluating the line integral.