Problem 13 - Thomas Calculus

Question

In Exercises \(13-16,\) find the line integrals along the given path \(C .\) $$\int_{C}(x-y) d x, \text { where } C : x=t, y=2 t+1, \text { for } 0 \leq t \leq 3$$

Step-by-Step Solution

Answer

The line integral is \(-\frac{15}{2}.\)

Step 1: Parametrize the Path

The path given is already parametrized with \(x = t\) and \(y = 2t + 1\). We will use parameter \(t\) ranging from 0 to 3 to evaluate the integral.

Step 2: Express Integrand in terms of t

Substitute the expressions for \(x\) and \(y\) into the integrand. Since \(x = t\) and \(y = 2t + 1\), we get:\[(x-y) = t - (2t + 1) = -t - 1.\]

Step 3: Express dx in terms of dt

Since \(x = t\), it follows that \(dx = dt\). We'll substitute this relation into the integral.

Step 4: Set up the Integral

Now replace \(x - y\) and \(dx\) in the integral with their expressions in terms of \(t\):\[\int_{0}^{3}-(t + 1) \cdot dt.\]

Step 5: Evaluate the Integral

Find the integral:\[\int_{0}^{3}-(t + 1) \cdot dt = -\int_{0}^{3}(t + 1) dt.\]This simplifies to:\[-\left[\frac{t^2}{2} + t\right]_{0}^{3}.\]Evaluate the definite integral by substituting the limits:\[-\left(\frac{3^2}{2} + 3\right) + \left(\frac{0^2}{2} + 0\right).\]\[-\left(\frac{9}{2} + 3\right) = -\left(\frac{15}{2}\right) = -\frac{15}{2}.\]

Key Concepts

Parametrized Paths

A parametrized path turns your path or curve into an expression involving a single variable, usually denoted as \( t \). This is particularly handy because it allows you to describe the position along a path using just one parameter instead of working with both \( x \) and \( y \) coordinates separately.
In this exercise, the path \( C \) is already given in a parametrized form where \( x = t \) and \( y = 2t + 1 \). This means as \( t \) changes from 0 to 3, you can see how both \( x \) and \( y \) evolve in a coordinated manner, tracing out the path of the line integral.
Parametrizing is the first step in evaluating line integrals as it simplifies dealing with curves in a plane.

Integrating with Respect to a Parameter

Once we have a parametrized path, we transition to integrating with respect to the parameter \( t \) rather than the coordinate \( x \). This step involves deriving expressions solely in terms of \( t \) to replace any \( x \) or \( y \) terms in your equations.
In this instance, since \( x = t \), \( dx \) translates to \( dt \). This change makes the integration straightforward because it concentrates everything into one variable, \( t \). The original expression becomes manageable and direct, eliminating multiple variables.
By switching our focus to \( t \), we're essentially saying: let's see how changes in \( t \) affect both \( x \) and \( y \), allowing us to carry out the integration in a clean, single-parameter manner.

Definite Integrals

Definite integrals evaluate the accumulation of a quantity, like an area under a curve, between two fixed limits. In the context of line integrals, these limits correspond to values of the parameter \( t \). For this exercise, \( t \) ranges from 0 to 3.
The process for evaluating a definite integral involves evaluating an antiderivative at the upper limit of \( t \) and subtracting the evaluation at the lower limit.
Once we substitute \( t = 3 \) and \( t = 0 \) into our expression derived from the integrand, we get a single numeric value. This final number \(-\frac{15}{2}\) in this case, represents the cumulative impact of the function along the path defined by our limits.

Evaluating Path Integrals

Evaluating path integrals requires following a sequence of steps meticulously to ensure accuracy. First, ensure the path is parametrized, so the line integral can be expressed in relation to \( t \).
Next, substitute \( x \) and \( y \) with their respective parametrized expressions to rewrite the integrand in terms of \( t \).
Then, integrate with respect to \( t \) over the given interval. The path integral sums up all values of the integrand \( (x-y) \) multiplied by \( dx \) as you move along the path from \( t = 0 \) to \( t = 3 \).
Finally, compute the definite integral by evaluating it at the upper and lower bounds to determine the total sum along the path.
These calculations reveal insights into the function's behavior over the specified path, illustrating how path integrals extend traditional integration into more complex geometries.