91Ó°ÊÓ

A vector field is a mathematical construct that assigns a vector to every point in space. In simple terms, you can imagine it as a way of showing how a vector (which has both magnitude and direction) changes across different locations. In this exercise, the vector field given is \( \vec{F} = x \vec{j} \). Here, the vector field is dependent on the \( x \) coordinate and points in the direction of the \( y \)-axis because of the \( \vec{j} \) unit vector.

• \( \vec{j} \) is a unit vector along the \( y \)-axis.
• The vector's magnitude depends linearly on \( x \).

In practical terms, if you were to visualize this field, you'd see vectors placed along the \( y \)-axis, with a length that grows as \( x \) increases. This insight is crucial when understanding how the field behaves as we compute line integrals.

The dot product is a way to multiply two vectors, resulting in a scalar, which means it is a number rather than a vector. In the context of line integrals, the dot product helps measure how much one vector goes in the direction of another.
For two vectors \( \vec{A} = (a_1, a_2) \) and \( \vec{B} = (b_1, b_2) \), the dot product is calculated as:\[ \vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2 \] In this exercise, the dot product \( \vec{F} \cdot d\vec{r} \) becomes \( t \cdot (0) = 0 \) because the direction of \( d\vec{r} = (1, 0) \) has no component along \( \vec{j} \).

• The dot product is zero when vectors are perpendicular.
• It helps determine the portion of one vector along another.

The key takeaway here is that a zero dot product implies no work done in the direction of the path, leading to a result of zero for the integral.

Parametrization is the process of defining a curve using a parameter, often denoted as \( t \). It's crucial for describing lines in terms of a single variable rather than coordinates or vectors.
In our problem, the path along the \( x \)-axis from \((1,0)\) to \((3,0)\) is parametrized by \( \vec{r}(t) = (t, 0) \). The parameter \( t \) varies from 1 to 3, representing points along the path.

• \( \vec{r}(t) \) helps shift focus from fixed coordinates to a dynamic path.
• It allows easy integration by replacing \( x \) and \( y \) with a single \( t \).

Thus, parametrization simplifies the integration process of line integrals by providing a clear path definition.

The \( x \)-axis in a coordinate plane is a key reference line, where \( y = 0 \) for any point on it. This exercise involves a path entirely on the \( x \)-axis, which simplifies the vector field integration immensely.

• Along the \( x \)-axis, a point is described solely by its \( x \) value.
• Movement occurs only in the \( x \)-direction, making \( y \) irrelevant in parametrization.

For our line integral, positioning the path entirely on the \( x \)-axis (\((1,0)\) to \((3,0)\)) ensures the vector field component \( \vec{F} = x \vec{j} \) is perpendicular to the path, as its direction strictly points along \( \vec{j} \). This clarifies why the line integral results in zero since there is no \( y \)-component in the path for the vector field to interact with.