Problem 39 - Thomas Calculus

Question

Find the coordinates of the centroid of the curve $$ x=\cos t, \quad y=t+\sin t, \quad 0 \leq t \leq \pi $$

Step-by-Step Solution

Answer

Centroid is at \( \left( \frac{\pi}{8}, \frac{3\pi^2}{16} \right) \).

Step 1: Understand the Problem

We are given parametric equations for a curve: \( x = \cos t \) and \( y = t + \sin t \), with \( t \) ranging from 0 to \( \pi \). We are asked to find the centroid of this curve over the given interval.

Step 2: Determine the Expressions for Centroid Coordinates

The centroid \((\bar{x}, \bar{y})\) of a curve defined parametrically can be found using: \[ \bar{x} = \frac{1}{L} \int_{t_1}^{t_2} x(t) \frac{dy}{dt} dt \] \[ \bar{y} = \frac{1}{L} \int_{t_1}^{t_2} y(t) \frac{dy}{dt} dt \] where \( L = \int_{t_1}^{t_2} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } dt \) is the curve's arc length.

Step 3: Compute Dy/DT and Arc Length Integral

First, determine \( \frac{dy}{dt} \): Since \( y = t + \sin t \), we have \( \frac{dy}{dt} = 1 + \cos t \). For the arc length \( L \): \( \frac{dx}{dt} = -\sin t \), so the integrand becomes: \[ \sqrt{(-\sin t)^2 + (1+\cos t)^2} = \sqrt{1 + 2\cos t + \cos^2 t + \sin^2 t}\] This simplifies to \[ \sqrt{2 + 2\cos t} = \sqrt{2(1 + \cos t)} = \sqrt{4\cos^2\left( \frac{t}{2} \right)} = 2\left| \cos\left( \frac{t}{2} \right) \right| \]. Since \( 0 \leq t \leq \pi \), \( \cos\left( \frac{t}{2} \right) \) is non-negative, thus: \[ L = \int_0^\pi 2 \cos\left( \frac{t}{2} \right) dt \] To solve, let \( u = \frac{t}{2} \), hence \( du = \frac{1}{2} dt \), and the limits become 0 to \( \frac{\pi}{2} \).

Step 4: Evaluate the Arc Length

Change of variables converts the integral into: \[ L = 4 \int_0^{\frac{\pi}{2}} \cos u \, du \] This evaluates to:\[ L = 4 [\sin u]_0^{\frac{\pi}{2}} = 4(1 - 0) = 4 \].

Step 5: Calculate \( \bar{x} \) and \( \bar{y} \)

Substitute \( \frac{dy}{dt} = 1 + \cos t \) and the expressions for \( x(t) \) and \( y(t) \) into the centroid formulas:\[ \bar{x} = \frac{1}{4} \int_0^\pi \cos t (1+\cos t) dt \] This evaluates to: 1. Split into two integrals: \[ \int_0^\pi \cos t \, dt + \int_0^\pi \cos^2 t \, dt \].2. The first integral is zero and the second, using double angle identity, becomes \( \frac{\pi}{2} \).Thus, \( \bar{x} = \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8} \).Then calculate:\[ \bar{y} = \frac{1}{4} \int_0^\pi (t + \sin t)(1 + \cos t) dt \] Upon evaluation using similar strategies (trigonometric identities and integration by parts), you get \( \bar{y} = \frac{1}{4}(\frac{3\pi^2}{4}) = \frac{3\pi^2}{16} \).

Step 6: Final Step: Combine Results

The coordinates of the centroid are \( \bar{x} = \frac{\pi}{8} \) and \( \bar{y} = \frac{3\pi^2}{16} \).

Key Concepts

Parametric Equations

Parametric equations provide a way to define mathematical functions using a third variable, typically denoted as \( t \). This means you can describe both \( x \) and \( y \) in terms of \( t \), rather than directly relating \( x \) to \( y \). This approach is particularly useful when dealing with curves in a plane. In our exercise, we have the parametric equations \( x = \cos t \) and \( y = t + \sin t \). Here, \( t \) is the parameter that describes the curve's progression from \( 0 \) to \( \pi \). By substituting different values of \( t \) into the equations, you can trace the curve step by step. Parametric equations are advantageous when defining curves that are not functions in the traditional sense, as well as in cases where the curve loops or changes direction.
For example, the equation \( x = \cos t \) produces a wave-like curve, starting at \( x=1 \) when \( t=0 \) and ending at \( x=-1 \) when \( t=\pi \). The equation \( y = t + \sin t \) adds a linear component to the oscillation from \( \sin t \). Together, they create the fascinating path defined by our specific range of \( t \).

Arc Length

The arc length of a curve is a measure of the distance along the curve, from one endpoint to the other. In parametric form, the arc length \( L \) between two points can be calculated using the formula:

• \[ L = \int_{t_1}^{t_2} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]

This formula involves derivatives of the parametric functions to account for changes in both \( x \) and \( y \) as \( t \) progresses. In our problem, we calculated \( \frac{dx}{dt} = -\sin t \) and \( \frac{dy}{dt} = 1 + \cos t \).
The process of finding the arc length begins by substituting these derivatives into the formula and simplifying the expression under the square root.
This procedure is essential in ensuring you are accurately measuring the curve's length rather than just straight-line distance between endpoints. In our solution, through substitution and simplification, we find the arc length \( L \) over the interval \( 0 \) to \( \pi \) is 4. This value helps us determine the normalized centroid coordinates over that portion of the curve.

Integration by Parts

Integration by parts is a method used to solve integrals that are products of two functions, stemming from the product rule of differentiation. Its formula is:

• \[ \int u \, dv = uv - \int v \, du \]

This method is particularly helpful when you have composite functions where one part's derivative easily integrates, such as polynomials, paired with another part that simplifies on differentiation. For our exercise, integration by parts plays a role in solving the integral for \( \bar{y} \), where we integrate \( (t + \sin t)(1 + \cos t) \).
Choosing parts correctly is key: let \( u = g(t) \) and \( dv = h(t)dt \), where \( g(t) \) gets easily differentiated and \( h(t) \) can be integrated without complex additional steps. Carrying through this technique on our problem enabled us to find \( \bar{y} \).
Remember, sometimes multiple integrations by parts may be necessary to fully resolve the expression. Here, applying it strategically alongside trigonometric identities allowed us to handle nested functions appearing in the integrals.

Trigonometric Identities

Trigonometric identities are mathematical expressions involving trigonometric functions that simplify complex trigonometric integrals. In our process, identities such as \( \sin^2 t + \cos^2 t = 1 \) and the double angle formulas are vital tools. These allow transformations that turn complex integrals into more manageable forms.
For instance, to simplify \( \cos^2 t \), we use the identity:

• \[ \cos^2 t = \frac{1 + \cos 2t}{2} \]

This substitution helped us integrate and compute \( \bar{x} \).
Another handy identity is \( 1 + \cos t = 2 \cos^2\left(\frac{t}{2}\right) \), assisting in evaluating the arc length. Trigonometric identities, thus, are not only pivotal in simplifications but also facilitate the application of other calculus techniques.
Having a solid grasp of these identities ensures swift manipulation of trigonometric functions, which is crucial for solving integrals that arise in parametric equations and finding centroids effectively.