Question

In Exercises \(7-12,\) a vector field \(\vec{F}\) and a curve \(C\) are given. Evaluate \(\int_{c} \vec{F} \cdot d \vec{r} .\) $$ \begin{aligned} &\vec{F}=\langle x, x+y\rangle ; C \text { is the portion of the parabola } y=x^{2} \text { from }\\\ &(0,0) \text { to }(1,1) \end{aligned} $$

Short answer

The line integral along the curve is \( \frac{13}{6} \).

Step-by-step solution

Parameterize the Curve

The given curve \( C \) is the portion of the parabola \( y = x^2 \) from \( (0,0) \) to \( (1,1) \). We can parameterize this curve by letting \( x = t \) and \( y = t^2 \) where \( t \) ranges from 0 to 1. Thus, the parameterization is \( \vec{r}(t) = \langle t, t^2 \rangle \).

Find the Derivative of Parameterization

We need the derivative of \( \vec{r}(t) \) with respect to \( t \), which is \( \frac{d\vec{r}}{dt} = \langle 1, 2t \rangle \). This represents the tangent vector to the curve at any point \( t \).

Substitute into the Line Integral

The line integral is \( \int_{C} \vec{F} \cdot d\vec{r} = \int_{0}^{1} \vec{F}(\vec{r}(t)) \cdot \frac{d\vec{r}}{dt} \, dt \). Substituting \( \vec{F}(\vec{r}(t)) = \langle t, t + t^2 \rangle \) and \( \frac{d\vec{r}}{dt} = \langle 1, 2t \rangle \), the integral becomes \( \int_{0}^{1} \langle t, t + t^2 \rangle \cdot \langle 1, 2t \rangle \, dt \).

Compute the Dot Product

Compute the dot product \( \langle t, t + t^2 \rangle \cdot \langle 1, 2t \rangle = t(1) + (t + t^2)(2t) = t + 2t^2 + 2t^3 \).

Evaluate the Integral

We now evaluate the integral \( \int_{0}^{1} (t + 2t^2 + 2t^3) \, dt \). Compute each term separately: \( \int_{0}^{1} t \, dt = \frac{1}{2} \), \( \int_{0}^{1} 2t^2 \, dt = \frac{2}{3} \), and \( \int_{0}^{1} 2t^3 \, dt = \frac{1}{2} \). Adding these results gives \( \frac{1}{2} + \frac{2}{3} + \frac{1}{2} = \frac{3}{2} + \frac{2}{3} = \frac{9}{6} + \frac{4}{6} = \frac{13}{6} \).

Final Result

After evaluating the integral, we find that \( \int_{C} \vec{F} \cdot d\vec{r} = \frac{13}{6} \).

Key concepts

Vector Fields

A vector field is a mathematical construct where each point in a given space is assigned a vector. In a two-dimensional plane, this vector field can be visualized using arrows that point in the direction of the vector and whose length represents the vector's magnitude. Vector fields are widely used in physics to model various phenomena such as electromagnetic fields, fluid flow, and gravitational forces.
In the exercise provided, the vector field \( \vec{F} \) is defined as \( \langle x, x+y \rangle \), meaning that at each point \((x, y)\) the field assigns a vector with an \(x\)-component equal to \(x\) and a \(y\)-component equal to \(x+y\). This type of setup is useful for understanding how quantities such as force and velocity vary over a region.

Parameterization

Parameterization is the process of representing a curve by expressing its coordinates as functions of a single variable, often denoted as \(t\). This approach simplifies the study of curves, since analyzing these functions is often easier than dealing with the original equation of the curve.
In our example, the parameterized form of the parabola \(y = x^2\) is \(\vec{r}(t) = \langle t, t^2 \rangle \). Here, \(t\) acts as the parameter, mapping from 0 to 1, thus smoothly describing the curve from point \((0,0)\) to \((1,1)\). By converting the curve into this form, it becomes simpler to manipulate and use in our calculations.

Dot Product

The dot product, also known as the scalar product, is an operation that takes two vectors and returns a single number. The result is obtained by multiplying corresponding components of the vectors and then summing these products. Mathematically, for two vectors \(\vec{A} = \langle a_1, a_2 \rangle\) and \(\vec{B} = \langle b_1, b_2 \rangle\), the dot product is given by \(\vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2\). The dot product has important applications in physics and engineering as it relates to projections and determining angles between vectors.
In our given problem, the dot product \(\langle t, t + t^2 \rangle \cdot \langle 1, 2t \rangle\) results in \(t + 2t^2 + 2t^3\). This step is crucial for simplifying the integrand when evaluating the line integral.

Integral Calculus

Integral calculus is a branch of mathematics that deals with the concept of integration. It allows for the calculation of areas under curves, the accumulation of quantities, and many other phenomena. In the context of vector fields, line integrals—a type of integration—are used to evaluate the interaction of a vector field along a path or curve.
The line integral \(\int_{C} \vec{F} \cdot d\vec{r}\) calculates the total effect (or impact) of the vector field \(\vec{F}\) on the curve \(C\). In our exercise, conducting the line integral led us to evaluate \(\int_{0}^{1} (t + 2t^2 + 2t^3) \, dt\), which was then simplified and calculated as \(\frac{13}{6}\). Line integrals have practical significance in physics for computing work done by a force field or flux across a surface.