Problem 25 - Thomas Calculus

Question

Find the lengths of the curves. $$ x=\cos t, \quad y=t+\sin t, \quad 0 \leq t \leq \pi $$

Step-by-Step Solution

Answer

The length of the curve is 4 units.

Step 1: Parametric Curve Arc Length Formula

To find the length of a curve defined parametrically by \( x=f(t) \) and \( y=g(t) \) for \( a \leq t \leq b \), use the formula: \[ L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \] In this exercise, \( x = \cos t \) and \( y = t + \sin t \), with \( 0 \leq t \leq \pi \).

Step 2: Differentiate x and y with respect to t

Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):- \( x = \cos t \) implies \( \frac{dx}{dt} = -\sin t \).- \( y = t + \sin t \) implies \( \frac{dy}{dt} = 1 + \cos t \).

Step 3: Plug Derivatives into the Arc Length Formula

Substitute \( \frac{dx}{dt} = -\sin t \) and \( \frac{dy}{dt} = 1 + \cos t \) into the arc length formula. The integral becomes: \[ L = \int_{0}^{\pi} \sqrt{(-\sin t)^2 + (1 + \cos t)^2} \, dt \].

Step 4: Simplify the Integrand

The expression inside the square root is: \(( -\sin t )^2 = \sin^2 t\) and \(( 1 + \cos t )^2 = 1 + 2\cos t + \cos^2 t\). Thus: \[ \sin^2 t + 1 + 2\cos t + \cos^2 t = 2 + 2\cos t \].This simplifies the integral to: \[ L = \int_{0}^{\pi} \sqrt{2(1 + \cos t)} \, dt = \int_{0}^{\pi} \sqrt{2(1 + \cos t)} \, dt\].

Step 5: Simplifying Further Using Trigonometric Identity

Use the identity \( 1 + \cos t = 2\cos^2\left(\frac{t}{2}\right) \) to simplify: \[ \sqrt{2(1 + \cos t)} = \sqrt{4\cos^2\left(\frac{t}{2}\right)} = 2\left|\cos\left(\frac{t}{2}\right)\right| \].Since \( 0 \leq t \leq \pi \), \( \cos\left(\frac{t}{2}\right) \) is non-negative, so \( 2\left|\cos\left(\frac{t}{2}\right)\right| = 2\cos\left(\frac{t}{2}\right) \).

Step 6: Compute the Integral

Evaluate the integral: \[ L = \int_{0}^{\pi} 2\cos\left(\frac{t}{2}\right) \, dt \].Perform the substitution \( u = \frac{t}{2} \), \( du = \frac{1}{2} dt \) or \( dt = 2 du \), which changes the limits of integration from \( t=0 \) to \( t=\pi \) into \( u=0 \) to \( u=\frac{\pi}{2} \):\[ L = \int_{0}^{\frac{\pi}{2}} 4\cos(u) \, du = 4\sin(u)\bigg|_{0}^{\frac{\pi}{2}} = 4(1 - 0) = 4 \].

Key Concepts

Parametric Equations

When dealing with parametric equations, we're expressing a curve in terms of a parameter, often labeled as \( t \). In this context, the parameter \( t \) influences both the \( x \) and \( y \) coordinates, shaping the path of the curve.
For instance, in the exercise above, \( x = \cos t \) and \( y = t + \sin t \) describe the curve. Each value of \( t \) between the given interval \( 0 \leq t \leq \pi \) corresponds to a unique point \((x, y)\) on the curve.
Parametric equations are particularly useful when:

• The curve is not a single function like \( y = f(x) \) or \( x = g(y) \).
• The curve forms loops or other complex shapes that are difficult to describe with a single function.

Understanding parametric equations allows us to explore and describe those complex curves, giving us a complete picture of how the curve is drawn across the plane.

Trigonometric Integration

Trigonometric integration is a method used to handle integrals involving trigonometric functions. In our exercise, we use trigonometric identities to simplify the integrand before integrating.
The function within our integral becomes easier by applying the identity \( 1 + \cos t = 2\cos^2\left(\frac{t}{2}\right) \). This allows us to rewrite the square root, transforming it into a more approachable form: \( \sqrt{4\cos^2\left(\frac{t}{2}\right)} = 2\cos\left(\frac{t}{2}\right) \).
Using trigonometric identities:

• Simplifies complex antiderivatives into basic sine and cosine functions.
• Makes integration more straightforward, sidestepping difficult square roots or other complexities.

This reduction in complexity allows us to evaluate the integral effectively, focusing on basic functions like \( \cos \) and \( \sin \), leading to a neat and satisfactory solution to length calculation.

Derivative Calculations

Derivatives are fundamental in determining changes in functions, with crucial applications in finding the arc length of parametric curves. Calculating derivatives with respect to \( t \) shows us how the \( x \) and \( y \) coordinates change as \( t \) changes.
In the given problem:

• We derived \( x = \cos t \), leading to \( \frac{dx}{dt} = -\sin t \).
• For \( y = t + \sin t \), we found \( \frac{dy}{dt} = 1 + \cos t \).

These derivatives help form the components of the arc length formula, providing us with the necessary input to calculate the length of the curve through integration. Using derivatives:

• Facilitates understanding rate changes and behavior along the curve.
• Integrates smoothly into broader calculus concepts, linking derivatives to integrals seamlessly.

This essential calculus tool bridges the gap between geometrical visualization and analytical methods.