Problem 41 - Thomas Calculus

Question

Length is independent of parametrization To illustrate the fact that the numbers we get for length do not depend on the way we parametrize our curves (except for the mild restrictions preventing doubling back mentioned earlier), calculate the length of the semi-circle \(y=\sqrt{1-x^{2}}\) with these two different parametrizations: $$ \begin{array}{ll}{\text { a. } x=\cos 2 t,} & {y=\sin 2 t, \quad 0 \leq t \leq \pi / 2} \\ {\text { b. } x=\sin \pi t,} & {y=\cos \pi t, \quad-1 / 2 \leq t \leq 1 / 2}\end{array} $$

Step-by-Step Solution

Answer

The arc length is \( \pi \) for both parametrizations, showing it's independent of parametrization.

Step 1: Define the Arc Length Formula

The formula to find the length of a curve defined by parametric equations \( x = f(t), y = g(t) \) is given by \( L = \int_\alpha^\beta \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \). For both parametrizations, we will use this formula to calculate the length of the curve.

Step 2: Calculate Derivatives for Parametrization (a)

Given the parametrization \( x = \cos 2t \) and \( y = \sin 2t \), first find the derivatives: \( \frac{dx}{dt} = -2\sin 2t \) and \( \frac{dy}{dt} = 2\cos 2t \).

Step 3: Set Up the Integral for Parametrization (a)

Substitute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) into the arc length formula: \[ L = \int_0^{\pi/2} \sqrt{(-2\sin 2t)^2 + (2\cos 2t)^2} \, dt \]. Simplify the integrand: \[ \sqrt{4\sin^2 2t + 4\cos^2 2t} = \sqrt{4(\sin^2 2t + \cos^2 2t)} = \sqrt{4} = 2 \].

Step 4: Evaluate the Integral for Parametrization (a)

The integral becomes \[ L = \int_0^{\pi/2} 2 \, dt \]. Evaluating this integral gives \( L = 2t \Big|_0^{\pi/2} = 2(\pi/2) - 2(0) = \pi \).

Step 5: Calculate Derivatives for Parametrization (b)

Given the parametrization \( x = \sin \pi t \) and \( y = \cos \pi t \), find the derivatives: \( \frac{dx}{dt} = \pi \cos \pi t \) and \( \frac{dy}{dt} = -\pi \sin \pi t \).

Step 6: Set Up the Integral for Parametrization (b)

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

Step 7: Evaluate the Integral for Parametrization (b)

The integral becomes \[ L = \int_{-1/2}^{1/2} \pi \, dt \]. Evaluating this integral gives \( L = \pi t \Big|_{-1/2}^{1/2} = \pi(1/2) - \pi(-1/2) = \pi/2 + \pi/2 = \pi \).

Step 8: Conclusion: Compare Results

For both parametrizations, the computed length of the semicircle is \( \pi \), demonstrating that the length is independent of the chosen parametrization.

Key Concepts

Parametrization

In mathematics, parametrization involves expressing the coordinates of a geometric object using a set of variables, often referred to as parameters. By expressing curves using parameters, we can delve into their properties with more freedom and insight. When considering the semicircle given by the equation \(y = \sqrt{1-x^2}\), two parametrizations are provided in the exercise:

• Parametrization (a): \(x = \cos 2t\) and \(y = \sin 2t\) with \(0 \leq t \leq \pi/2\).
• Parametrization (b): \(x = \sin \pi t\) and \(y = \cos \pi t\) with \(-1/2 \leq t \leq 1/2\).

Each parametrization is a different way to describe the path traced by the semicircle. While they look different, both describe the same geometric object. They also help us calculate properties like arc length, as seen in the exercise. When set up correctly, they will always yield the same arc length for the same curve, showing that the arc length is indeed independent of parametrization.

Integral Calculus

Integral calculus is a branch of calculus focused on accumulation and areas under curves. It complements differential calculus, which revolves around rates of change. In the context of arc length, integral calculus allows us to calculate the length of a curve by going through the sum of tiny line segments.Consider the formula used in the exercise:\[ L = \int_\alpha^\beta \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]This formula helps us find the length of a parametric curve from the derivatives of its parametric equations. For both given parametrizations in the semicircle example:

• The derivatives are calculated with respect to time \(t\).
• The expression inside the integral is simplified by recognizing identities such as \(\sin^2 \theta + \cos^2 \theta = 1\).

Integral calculus, therefore, allows us to find an exact measurement of arc length, confirming consistency across different parametrizations.

Semicircle

A semicircle is half of a circle, defined as the set of all points equidistant from a fixed point, known as the center, along with a boundary that forms half a circle. In this problem, we're specifically looking at the top half of the unit circle due to the equation \(y = \sqrt{1-x^2}\), which describes a semicircle centered at the origin with a radius of 1.In tasks involving a semicircle, parametric representations like those provided are often employed:

• Parametrization (a): Directly models a semicircle using trigonometric identities.
• Parametrization (b): Provides an alternative perspective ensuring any consistent computed property, like arc length, is the same.

Whether calculated via standard trigonometry or through different parametric equations, the properties of the semicircle, including the arc length, remain constant. Here, both parametrizations yield an equal arc length of \( \pi \), highlighting the semicircle’s consistent geometric property across different parametric expressions.