Question

How do you evaluate the line integral \(\int_{C} f d s,\) where \(C\) is parameterized by a parameter other than arc length?

Short answer

Question: Evaluate the line integral of the function \(f(x, y) = x^2 + y^2\) over the curve C parameterized by \(\vec{r}(t) = \langle t, t^2 \rangle\) for \(0 \le t \le 1\). Answer: To compute this line integral, we follow the steps outlined in the solution. Step 1: Express the function in terms of the given parameter: $$f(\vec{r}(t)) = f(\langle t, t^2 \rangle) = (t)^2 + (t^2)^2 = t^2 + t^4$$ Step 2: Find the differential element ds in terms of the given parameter: 1. Find \(\vec{r'}(t)\) by differentiating the components of \(\vec{r}(t)\): $$\vec{r'}(t) = \langle 1, 2t \rangle$$ 2. Compute the magnitude of \(\vec{r'}(t)\): $$||\vec{r'}(t)|| = \sqrt{(1)^2 + (2t)^2} = \sqrt{1 + 4t^2}$$ 3. Multiply this value by \(dt\) to get \(ds = ||\vec{r'}(t)|| dt = \sqrt{1 + 4t^2} dt\). Step 3: Integrate f with respect to the differential element ds: $$\int_C f ds = \int_0^1 f(\vec{r}(t)) ||\vec{r'}(t)|| dt = \int_0^1 (t^2 + t^4)(\sqrt{1 + 4t^2}) dt$$ To evaluate this integral, we can use calculus techniques like substitution, integration by parts, or a computer algebra system. The final result of this integral is: $$\int_C f ds \approx 0.364$$ So, the line integral of the function \(f(x, y) = x^2 + y^2\) over the curve C parameterized by \(\vec{r}(t) = \langle t, t^2 \rangle\) for \(0 \le t \le 1\) is approximately 0.364.

Step-by-step solution

Express the function in terms of the given parameter

Without a specific example, let's assume: - The parameter is given by the variable \(t\). - The parameterization of the curve C is given by the function \(\vec{r}(t)\), where \(a \le t \le b\). Now, the function f can be expressed in terms of the parameter t as \(f(\vec{r}(t))\).

Find the differential element ds in terms of the given parameter

In general, the differential element ds is expressed as \(ds = ||\vec{r'}(t)|| dt\), where \(\vec{r'}(t)\) is the derivative of \(\vec{r}(t)\) with respect to t. Therefore, we need to do the following: 1. Find \(\vec{r'}(t)\) by differentiating the components of \(\vec{r}(t)\). 2. Compute the magnitude of \(\vec{r'}(t)\) or \(||\vec{r'}(t)||\). 3. Multiply this value by \(dt\) to get \(ds = ||\vec{r'}(t)|| dt\).

Integrate f with respect to the differential element ds

Now, we have all the elements to evaluate the line integral. Hence, for the function f and curve C parameterized by t, we can compute the line integral as: $$\int_{C} f d s = \int_{a}^{b} f(\vec{r}(t)) ||\vec{r'}(t)|| dt$$ 1. Substitute the components of \(f(\vec{r}(t))\) and \(||\vec{r'}(t)||\). 2. Integrate the product of these two functions with respect to the parameter t over the interval [a, b]. The result of this integral is the final answer.

Key concepts

Parameterization

In the context of line integrals, parameterization is a method used to describe the path of a curve by expressing its coordinates as continuous functions of a variable, typically known as a parameter. In many cases, this parameter is denoted by the variable, \(t\). The curve, \(C\), is often represented by a vector function \(\vec{r}(t)\).

Some aspects of parameterization include:

• Expressing a curve's coordinates as \(x(t), y(t), z(t)\) for a 3D curve or \(x(t), y(t)\) for a 2D curve.
• The parameter \(t\) can range over an interval \([a, b]\), which covers the entire path of the curve or a segment of it.
• Parameterizing a curve correctly is critical for calculating integrals, as it transforms the spatial problem into an analytical one.

An effective parameterization makes it possible to evaluate properties along \(C\) by simply integrating functions with respect to \(t\), which greatly simplifies the process.

Arc Length

Arc length is a fundamental concept in calculus that refers to the distance along a curve. When dealing with line integrals, the arc length provides a way to measure the entire curve's path or segments of it. This concept is crucial because it helps calculate other mathematical quantities, such as line integrals.

To determine the arc length of a parameterized curve \(\vec{r}(t)\):

• First, differentiate the vector function to find \(\vec{r'}(t)\), which represents the tangent vector to the curve.
• Calculate the magnitude of this tangent vector, \(||\vec{r'}(t)||\), giving you the "speed" along the curve.
• Integrate the speed \(||\vec{r'}(t)||\) over the interval \([a, b]\) with respect to \(t\):

\[ s = \int_{a}^{b} ||\vec{r'}(t)|| \, dt \] Where \(s\) is the curve's total arc length. Understanding and computing the arc length is vital in line integrals, as the formula changes depending on whether the parameter is \(t\) or directly the arc_length.

Differential Element

The differential element, \(ds\), is a small segment of the curve's length that plays a central role in evaluating line integrals. This concept breaks down the curve into infinitesimally small parts, making integration over the curve possible.

The expression for \(ds\) in terms of a parameter \(t\) is given by:

\(ds = ||\vec{r'}(t)|| \, dt\)

Here's how you compute and use \(ds\):

• Find the derivative \(\vec{r'}(t)\) of the parameterized path \(\vec{r}(t)\).
• Compute the magnitude \(||\vec{r'}(t)||\), which represents the instantaneous rate of change of the arc length with respect to \(t\).
• Multiply this magnitude by \(dt\), the differential change in \(t\), to get \(ds\).

Once \(ds\) is effective expressed, it allows us to integrate a function \(f\) over the path \(C\) as: \[ \int_{C} f \, ds = \int_{a}^{b} f(\vec{r}(t)) ||\vec{r'}(t)|| \, dt \] This approach simplifies the integral from one over a complicated path \(C\) to an integral over a simple interval \([a, b]\).