Problem 29 - 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 semicircle \(y=\sqrt{1-x^{2}}\) with these two different parametrizations. $$ \begin{array}{l}{\text { a. } x=\cos 2 t, \quad y=\sin 2 t, \quad 0 \leq t \leq \pi / 2} \\ {\text { b. } x=\sin \pi t, \quad y=\cos \pi t, \quad-1 / 2 \leq t \leq 1 / 2}\end{array} $$

Step-by-Step Solution

Answer

The length of the semicircle is \( \pi \) in both parametrizations.

Step 1: Set up the arc length formula for part (a)

To find the length of the curve, use the arc length formula \[ L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]For part (a), we have \( x = \cos 2t \) and \( y = \sin 2t \), with \( 0 \leq t \leq \pi/2 \).

Step 2: Calculate derivatives for part (a)

Compute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):\[ \frac{dx}{dt} = -2\sin 2t \]\[ \frac{dy}{dt} = 2\cos 2t \]

Step 3: Substitute into arc length formula for part (a)

Substitute the derivatives into the arc length formula and simplify:\[ L = \int_{0}^{\pi/2} \sqrt{(-2\sin 2t)^2 + (2\cos 2t)^2} \, dt \]This simplifies to:\[ L = \int_{0}^{\pi/2} \sqrt{4\sin^2 2t + 4\cos^2 2t} \, dt \]Using \( \sin^2 \theta + \cos^2 \theta = 1 \), the expression simplifies:\[ L = \int_{0}^{\pi/2} 2 \, dt \]

Step 4: Evaluate the integral for part (a)

Integrate to find the length:\[ L = 2 \times \left[ t \right]_{0}^{\pi/2} = 2 \times \left( \frac{\pi}{2} - 0 \right) = \pi \]

Step 5: Set up the arc length formula for part (b)

Again use the arc length formula\[ L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]For part (b), we have \( x = \sin \pi t \) and \( y = \cos \pi t \), with \(-1/2 \leq t \leq 1/2 \).

Step 6: Calculate derivatives for part (b)

Compute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):\[ \frac{dx}{dt} = \pi \cos \pi t \]\[ \frac{dy}{dt} = -\pi \sin \pi t \]

Step 7: Substitute into arc length formula for part (b)

Substitute the derivatives into the arc length formula and simplify:\[ L = \int_{-1/2}^{1/2} \sqrt{(\pi \cos \pi t)^2 + (-\pi \sin \pi t)^2} \, dt \]This simplifies to:\[ L = \int_{-1/2}^{1/2} \sqrt{\pi^2 \cos^2 \pi t + \pi^2 \sin^2 \pi t} \, dt \]Using \( \cos^2 \theta + \sin^2 \theta = 1 \), the expression simplifies:\[ L = \int_{-1/2}^{1/2} \pi \, dt \]

Step 8: Evaluate the integral for part (b)

Integrate to find the length:\[ L = \pi \times \left[ t \right]_{-1/2}^{1/2} = \pi \times \left( \frac{1}{2} - \left(-\frac{1}{2}\right) \right) = \pi \]

Key Concepts

Parametrization

Parametrization involves mapping a curve onto a parameter, often denoted as \( t \). It's a way to express the coordinates of points on a curve as functions of a single parameter. For example, the curve defined by \( y = \sqrt{1-x^2} \) can be parametrized using several different methods. This exercise provides two different parametrizations:

• For part (a), \( x = \cos 2t \) and \( y = \sin 2t \), with the interval for \( t \) being \( 0 \leq t \leq \frac{\pi}{2} \).
• For part (b), \( x = \sin \pi t \) and \( y = \cos \pi t \), with \(-\frac{1}{2} \leq t \leq \frac{1}{2} \).

Although each parametrization might look different and use varied intervals for \( t \), they ultimately describe the same path along the curve. This shows the flexibility of parametrization to express the same shape in more than one way. Understanding this concept allows us to choose a parametrization that simplifies calculations or fits specific needs, like bounds of integration.

Semicircle

A semicircle is half of a circle, typically cut along its diameter. In this exercise, we're dealing with the semicircle defined by the equation \( y = \sqrt{1-x^2} \), which corresponds to the upper half of a unit circle. This is because the equation \( y = \sqrt{1-x^2} \) satisfies the equation for a circle \( x^2 + y^2 = 1 \), restricted to positive \( y \) values.
The interesting aspect of this problem is that although the semicircle sits symmetrically on the x-axis from -1 to 1, we see it represented through varying parametrizations that might not immediately suggest this symmetry. Using trigonometric functions for parametrization effectively represents the circular structure, as sine and cosine naturally map out circular paths.

Derivatives

The derivatives of the parametric equations play a pivotal role in determining arc length. For a parametrization given as \( x = f(t) \) and \( y = g(t) \), the derivatives are \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). They depict the rate of change of \( x \) and \( y \) concerning \( t \).
In this problem:

• For part (a), the derivatives are \( \frac{dx}{dt} = -2 \sin 2t \) and \( \frac{dy}{dt} = 2 \cos 2t \), reflecting the changes in \( x \) and \( y \) along the chosen path.
• For part (b), they become \( \frac{dx}{dt} = \pi \cos \pi t \) and \( \frac{dy}{dt} = -\pi \sin \pi t \). These derivatives reflect the rate of change of sine and cosine over the interval, scaled by \( \pi \) to adjust for the parametrization range.

Computing these derivatives is essential, as they eventually form part of the integral that calculates the arc length by defining the motion of the curve.

Integral Calculus

Integral calculus comes into play when calculating the length of a curve, known as arc length. The formula for arc length in parametric form is:
\( L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \ dt \)
In this expression, we integrate over the parameter \( t \) between the limits \( a \) and \( b \). This involves plugging in the derivatives from the previous sections, simplifying under the square root, and evaluating the integral.
Through the examples provided in both parts (a) and (b), the integrals reduced to simpler forms by recognizing identities like \( \sin^2 \theta + \cos^2 \theta = 1 \). This simplification is crucial for evaluating the integral, reducing it to ones that can be directly solved. In both cases, we solved the integrals to find a length of \( \pi \), illustrating that arc length remains consistent regardless of how we parametrize the curve.