Question

In Exercises 1 and \(2,\) show that the value of \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) is the same for each parametric representation of \(C\). \(\mathbf{F}(x, y)=x^{2} \mathbf{i}+x y \mathbf{j}\) (a) \(\mathbf{r}_{1}(t)=t \mathbf{i}+t^{2} \mathbf{j}, \quad 0 \leq t \leq 1\) (b) \(\mathbf{r}_{2}(\theta)=\sin \theta \mathbf{i}+\sin ^{2} \theta \mathbf{j}, \quad 0 \leq \theta \leq \frac{\pi}{2}\)

Short answer

The value of the line integral over vector field \(\mathbf{F}(x, y)\) should be independent of the parametric representation of the path through the vector field, meaning that \(\int_{C} \mathbf{F} \cdot d\mathbf{r}_{1}\) and \(\int_{C} \mathbf{F} \cdot d\mathbf{r}_{2}\) should yield the same result once computed.

Step-by-step solution

Compute the line integral for \(\mathbf{r}_{1}(t)\)

To compute the line integral over \(\mathbf{F}(x, y)\) using parametric representation \(\mathbf{r}_{1}(t) = t\mathbf{i} + t^{2}\mathbf{j}\), first substitute the parameterized values of \(x = t, y = t^{2}\) into \(\mathbf{F}(x, y)\) for \(0 \leq t \leq 1\) to get \(\mathbf{F}(t\mathbf{i} + t^{2}\mathbf{j}) = t^{2}\mathbf{i} + t^3\mathbf{j}\). Then, compute \(\mathbf{F}(x, y) \cdot d\mathbf{r}_{1}\) = \((t^{2}\mathbf{i} + t^{3}\mathbf{j}) \cdot (dt\mathbf{i} + 2tdt\mathbf{j})\), and integrate over the interval [0,1].

Compute the line integral for \(\mathbf{r}_{2}(\theta)\)

Similarly, for parametric representation \(\mathbf{r}_{2}(\theta) = \sin\theta\mathbf{i} + \sin^{2}\theta\mathbf{j}\), substitute the parameterized values of \(x = \sin\theta, y = \sin^{2}\theta\) into \(\mathbf{F}(x, y)\) for \(0 \leq \theta \leq \frac{\pi}{2}\) to get \(\mathbf{F}(\sin\theta\mathbf{i} + \sin^{2}\theta\mathbf{j}) = \sin^{2}\theta\mathbf{i} + sin^{3}\theta\mathbf{j}\). Then, compute \(\mathbf{F}(x, y) \cdot d\mathbf{r}_{2}\) = \((\sin^{2}\theta\mathbf{i} + \sin^{3}\theta\mathbf{j}) \cdot (cos\theta d\theta\mathbf{i} + 2\sin\theta\cos\theta d\theta\mathbf{j})\), and integrate over [0,\(\frac{π}{2}\)].

Compare the results

After completing step 1 and 2, you should have two results which theoretically should be the same if \(\int_{C} \mathbf{F} \cdot d\mathbf{r}_{1} = \int_{C} \mathbf{F} \cdot d\mathbf{r}_{2}\). This would prove that the value of the line integral over vector field \(\mathbf{F}(x, y)\) is independent of the parametric representation of the path through vector field.

Key concepts

Parametric Representation

Parametric representation is a crucial concept in calculus, particularly when dealing with the geometry of curves. This method articulates each point on a curve as a set of equations, which produce coordinates as functions of one or more parameters.

For example, in the textbook exercise, the parametric representations \( \mathbf{r}_1(t) = t\mathbf{i} + t^2\mathbf{j} \) and \( \mathbf{r}_2(\theta) = \sin \theta\mathbf{i} + \sin^2 \theta\mathbf{j} \) describe the same curve, \(C\), in two different ways using the parameters \(t\) and \(\theta\), respectively. By expressing the curve through parameters, it allows us to compute integrals along paths that may not be easily expressed in terms of just \(x\) and \(y\).

In context to our exercise, employing different parametric equations to describe a pathway should yield the same line integral result over a vector field, indicating the path-independence property of certain line integrals, which is a significant aspect of conservative fields in vector calculus.

Vector Field

Vector fields are pictorial representations of vector functions that map coordinates to vectors. They are essential in physics and engineering because they represent physical quantities which have both magnitude and direction—like velocity, force, or magnetic field—at each point in space.

In the given exercise, the vector field \(\mathbf{F}(x, y)\) is defined by \(x^2\mathbf{i} + xy\mathbf{j}\). The line integral of this vector field along a curve gives us information about the field's behavior over that curve. In intuitive terms, you might think of it as adding up the vector 'force' you feel as you 'walk' along a path within the vector field. If this path leads you through areas with varying 'wind' direction and strength (akin to the vectors in our field), the integral measures the cumulative effect of this 'wind' on your journey.

Thus, the line integral can be seen as a measure of the influence of the vector field along a specific path, in this case, the path defined by \(C\) with its parametric representations.

Calculus

Calculus is the mathematical study of continuous change and is divided primarily into differential calculus and integral calculus. The line integral is a cornerstone concept in calculus, blending these two branches as it involves finding the integral of functions over a curve—the output of a differential process.

Calculus offers the methods needed to compute the line integral over a vector field, as seen in the expansion of the line integral into its components in these solutions. The parametric equations come from differential calculus, where we differentiate the position vectors with respect to the parameters to find the differentials \(d\mathbf{r}\). Integrating these differentials against the vector field using integral calculus yields the line integral that encapsulates the cumulative effect along the path.

The line integral calculations in the exercise steps illustrate how integral calculus gathers and summarizes the incremental contributions of the vector field along the parametrized curve, highlighting the profound connection between calculus and the behavior of dynamic systems.