Problem 15 - Calculus: Early Transcendental Functions

Question

Evaluate the line integral. \(\int_{C} 2 x d x,\) where \(C\) is the portion of \(y=x^{2}\) from (2,4) to (0,0)

Step-by-Step Solution

Answer

Therefore, the value of the line integral is 4.

Step 1: Parametrize the Curve

Parametrize the curve C by \(r(t) = \) as \(t\) ranges from 0 to 2. The x-component is \(t\), and the y-component is \(t^2\).

Step 2: Find \(dr(t)\)

Find the derivative \(dr(t)\) of \(r(t)\), this would be the velocity vector of the parameterized curve. This is given by \(<1, 2t>\).

Step 3: Calculate the Line Integral

Evaluate the given line integral \(\int_{C} 2x dx\) over the curve C by using the parameterization. We obtain \(\int_{0}^{2} 2t.dt\).

Step 4: Evaluate the Integral

Evaluate this integral from t=0 to t=2. The answer is \[2\int_{0}^{2} t.dt = [t^2]_{0}^{2} = 4 - 0 = 4\].

Key Concepts

Parametrization of Curves

To start solving problems involving line integrals, we often need to parametrize the curve along which we are integrating. Parametrization allows us to express a curve in terms of a single variable, usually denoted by \(t\). For the curve given by \(y = x^2\), from the point \((2,4)\) to \((0,0)\), a natural choice is to define the curve as a vector function \(r(t)\), where each point on the curve corresponds to a specific value of \(t\).

• The curve can be parameterized by \(r(t) = \langle t, t^2 \rangle\).
• Here, the x-component is \(t\) and the y-component is \(t^2\).
• The parameter \(t\) typically varies over an interval, in this case, from 0 to 2, representing the range of x-values between the start and end point.

Parametrization helps simplify the process of integration over a curve by reducing it to an interval on the real line.

Calculus Problem Solving

In calculus, solving problems like line integrals often involves breaking the problem into smaller, more approachable steps. By parametrizing our curve, we have translated the geometric problem into an algebraic one, allowing us to use calculus tools more effectively. Consider these key steps for problem-solving:

• Find the derivative: Determine \(dr(t)\), which gives us the velocity vector of the curve. It is \(\langle 1, 2t \rangle\) in this example.
• Set up the integral: Use this velocity vector to express the line integral in terms of \(t\). The integral becomes \(\int_{0}^{2} 2t\,dt\).

Each step leads us systematically toward evaluating the integral, ensuring that the problem is manageable even if at first it seems complex. This is the essence of effective calculus problem-solving, transforming seemingly difficult problems through methodical approaches.

Evaluating Integrals

The final step in solving a line integral problem is evaluating the integral itself. Once we've set up the integral using the parameter \(t\), the process becomes straightforward. In our exercise, we've reduced the original integral to \(\int_{0}^{2} 2t\,dt\). Here's how to evaluate it:

• Evaluate the integral: Integrate \(2t\) with respect to \(t\), resulting in \([t^2]_{0}^{2}\).
• Apply the limits: Substitute the limits of integration, starting with \(t = 2\) and then subtracting the result of substituting \(t = 0\).
• Solve: This yields the final value \(4 - 0 = 4\).

Evaluating integrals is often the most rewarding step, as it gives us the final quantitative result of a problem. Practicing these steps will help build a strong foundation in integrals.