Question

Using the Fundamental Theorem for line integrals Verify that the Fundamental Theorem for line integrals can be used to evaluate the given integral, and then evaluate the integral. \(\int_{C} \nabla\left(1+x^{2} y z\right) \cdot d \mathbf{r},\) where \(C\) is the helix \(\mathbf{r}(t)=\langle\cos 2 t, \sin 2 t, t\rangle\) for \(0 \leq t \leq 4 \pi\)

Short answer

Question: Verify the Fundamental Theorem for line integrals for the given scalar function and curve, and compute the evaluation. Scalar function: \(f(x, y, z) = 1 + x^2yz\) Curve C: \(\mathbf{r}(t) = \langle \cos{2t}, \sin{2t}, t \rangle\) for \(0 \leq t \leq 4\pi\) Answer: The line integral \(\int_C \nabla f \cdot d\mathbf{r}\) evaluates to \(0\).

Step-by-step solution

Compute the gradient of the scalar function

We have the scalar function \(f(x, y, z) = 1 + x^2yz\). We need to compute the gradient of f: \(\nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\rangle\). Taking the partial derivatives, we get: \(\frac{\partial f}{\partial x} = 2xyz\) \(\frac{\partial f}{\partial y} = x^2z\) \(\frac{\partial f}{\partial z} = x^2y\) So, \(\nabla f = \langle 2xyz, x^2z, x^2y \rangle\).

Parameterize the curve C

The curve C is parameterized by \(\mathbf{r}(t) = \langle \cos{2t}, \sin{2t}, t \rangle\) for \(0 \leq t \leq 4\pi\).

Compute the derivative of the parameterization

We need the derivative of the parameterization \(\mathbf{r}(t)\), which is \(\mathbf{r}'(t) = \langle \frac{d(\cos{2t})}{dt}, \frac{d(\sin{2t})}{dt}, \frac{dt}{dt} \rangle\). Compute the derivatives to get: \(\mathbf{r}'(t) = \langle -2\sin{2t}, 2\cos{2t}, 1 \rangle\)

Use the Fundamental Theorem for line integrals

Now, apply the Fundamental Theorem for line integrals: \(\int_C \nabla f \cdot d\mathbf{r} = f(\mathbf{r}(b)) - f(\mathbf{r}(a))\), where \(a = 0\) and \(b = 4\pi\). First, we plug \(\mathbf{r}(0)\) and \(\mathbf{r}(4\pi)\) into the scalar function f: \(f(\mathbf{r}(0)) = f(\langle 1, 0, 0 \rangle) = 1 + (1)^2(0)(0) = 1\) \(f(\mathbf{r}(4\pi)) = f(\langle 1, 0, 4\pi \rangle) = 1 + (1)^2(0)(4\pi) = 1\) Now, compute the difference \(f(\mathbf{r}(4\pi)) - f(\mathbf{r}(0))\): \(1 - 1 = 0\) Thus, the line integral \(\int_C \nabla f \cdot d\mathbf{r}\) equals \(0\).

Key concepts

Gradient of Scalar Field

Imagine standing on a hilly terrain, and every point on the ground has its height recorded as a number. A scalar field in mathematics is somewhat analogous to this terrain, where instead of height, each point in space has a scalar (a single number) associated with it. When we talk about the gradient of this field, we’re looking at the direction and rate at which these numbers increase the most rapidly.

The gradient is represented by a vector that points in the direction of the greatest increase of the scalar field and whose length corresponds to the rate of increase. In our problem, the scalar field given by the function f(x, y, z) = 1 + x^2yz describes a three-dimensional space where each point has a scalar value assigned according to this formula.

Computing the gradient involves taking the partial derivatives of the scalar function with respect to each variable—the result is a vector function. It’s like figuring out the slope of the terrain in every possible direction. The gradient vector for our function is \( abla f = \langle 2xyz, x^2z, x^2y \rangle \), where each component tells us about changes in the terrain (our field) with respect to x, y, and z directions.

Line Integral

If the gradient tells us how steep our terrain is, the line integral takes us on a specific path (or curve) across it and sums up the influence of the field along this journey. Picture walking along a path on the terrain and measuring the slope under your feet with respect to the direction you’re moving. The line integral does exactly this—but mathematically.

In this context, we are dealing with a line integral of a gradient vector field, which simplifies our work significantly. According to the Fundamental Theorem for line integrals, if we have a path C and a gradient vector field derived from a scalar field, the result of the line integral over C will be the difference between the values of the scalar field at the endpoints of the path. This means we don't have to do the walk in small steps summing up slopes; we just look at where we started and where we ended.

Parameterization of Curves

So, how do we describe the path which we'll take on our terrain? Through parameterization. It’s a way to express our path as a single equation where a third variable, usually 't', acts as a guide from the beginning to the end of the path. This variable can often represent time, which helps visualize the curve’s shape and direction as time progresses.

In our helix example, the curve C is given by \( \mathbf{r}(t) = \langle \cos 2t, \sin 2t, t \rangle \). The trigonometric functions describe a circle when projected onto the x-y plane, while 't' ensures that as time goes by, the path moves upward in a spiral—just like a helix. Parameterization can be particularly powerful because it allows us to convert a potentially complicated path into a single equation that's much easier to work with mathematically.