Problem 44 - Essential Calculus

Question

Write an integral that represents the arc length of the curve on the given interval. Do not evaluate the integral. $$ x=t+\sin t, \quad y=t-\cos t \quad 0 \leq t \leq \pi $$

Step-by-Step Solution

Answer

The integral that represents the arc length of the curve over the interval [0, \(\pi\)] is \(L=\int_{0}^{\pi}{\sqrt{2 + 2\cos t + 2\sin t}\ dt}\).

Step 1: Find the Derivatives

Firstly find the derivatives of x and y with respect to t, \(dx/dt\) and \(dy/dt\). Based on the given parametric equations, \(x=t+\sin t\) and \(y=t-\cos t\), derive \(dx/dt = 1 + \cos t\) and \(dy/dt = 1 + \sin t\).

Step 2: Substitute into the Formula

Substitute \(dx/dt\) and \(dy/dt\) into the arc length formula. This gives: \(L=\int_{0}^{\pi}{\sqrt{(1 + \cos t)^{2}+(1 + \sin t)^{2}}\ dt}\).

Step 3: Simplify the Integral

Now, simplify the integral: \(L=\int_{0}^{\pi}{\sqrt{2 + 2\cos t + 2\sin t}\ dt}\). Remember, that the task is not to evaluate the integral, therefore the final answer is left as it is.

Key Concepts

Parametric Equations

Parametric equations describe a curve by defining both the x and y coordinates in terms of a third variable, usually denoted as t. This variable, t, is often associated with time, providing a dynamic view of how the coordinates change. In the given exercise, the parametric equations are:

• \( x = t + \sin t \)
• \( y = t - \cos t \)

These equations represent a curve that traces out a pathway on a plane as the parameter t changes, specifically from 0 to \( \pi \). This means when t = 0, the initial point of the curve is found, and when t = \( \pi \), it reaches another point on the curve. Parametric equations are particularly useful for curves that cannot be easily expressed as a function of x or y. They allow us to explore complex curves in a straightforward manner, while also revealing relationships between coordinates dependent on time or other variables.

Integral Calculus

Integral calculus is a crucial part of mathematics used to find quantities such as areas, volumes, and in this case, arc lengths. For parametric curves, the arc length from one point to another is determined by integrating a specific function derived from the curve's parametric equations. The general formula for the arc length \( L \) of a parametric curve \( x(t), y(t) \) over an interval \( t \) is:\[ L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \ dt \]In the exercise, \( x = t + \sin t \) and \( y = t - \cos t \), we first take their derivatives. Then substitute these into the arc length integral:\[ L = \int_{0}^{\pi} \sqrt{ (1 + \cos t)^2 + (1 + \sin t)^2 } \ dt \]This integral structure offers a method to compute the physical distance over which the curve stretches from start to end points defined by the parameter t. It integrates the small pieces of length \( \Delta L \), adding them up into a continuous sum, or integral, providing the total arc length.

Derivative of Parametric Functions

Derivatives of parametric functions involve finding the rate of change of the parametric equations with respect to their parameter, typically t. These derivatives are essential for calculating properties like arc length, tangents, and other aspects of the curve.In our exercise, the parametric equations are:

• \( x = t + \sin t \)
• \( y = t - \cos t \)

To find the derivatives, we calculate:

• \( \frac{dx}{dt} = 1 + \cos t \)
• \( \frac{dy}{dt} = 1 + \sin t \)

These represent how the x and y values change as t changes. The derivatives are then used in the arc length formula:\[ L = \int \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \ dt \]Understanding these derivatives is crucial as they offer insight into the curve's behavior, such as its slope and turning points. They provide the foundation for the integral calculus that helps us compute various properties of the parametric curve.