Problem 19 - Calculus Single Variable

Question

Calculate the length of the given parametric curve. $$ x=4 \cos (2 t) \quad y=4 \sin (2 t) \quad-\pi / 6 \leq t \leq \pi / 6 $$

Step-by-Step Solution

Answer

The length of the curve is \( \frac{8\pi}{3} \).

Step 1: Parametric Formulas for Derivatives

To find the length of a parametric curve, we first need the derivatives of the parametric equations. Given:\[ x(t) = 4 \cos(2t) \]\[ y(t) = 4 \sin(2t) \]The derivatives are:\[ \frac{dx}{dt} = \frac{d}{dt}[4 \cos(2t)] = -8 \sin(2t) \]\[ \frac{dy}{dt} = \frac{d}{dt}[4 \sin(2t)] = 8 \cos(2t) \]

Step 2: Length of Parametric Curve Formula

The length of a parametric curve from \( t = a \) to \( t = b \) is given by the formula:\[ L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]For our problem, \( a = -\frac{\pi}{6} \) and \( b = \frac{\pi}{6} \).

Step 3: Substitute and Simplify Under the Integral

Substitute the derivatives into the integral formula:\[ L = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \sqrt{(-8 \sin(2t))^2 + (8 \cos(2t))^2} \, dt \]\[ = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \sqrt{64 \sin^2(2t) + 64 \cos^2(2t)} \, dt \]Using the Pythagorean identity \(\sin^2(2t) + \cos^2(2t) = 1\), we simplify further:\[ = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \sqrt{64} \, dt = \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} 8 \, dt \]

Step 4: Evaluate the Integral

Evaluate the integral:\[ L = 8 \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} dt \]This simplifies to:\[ L = 8 \left[ t \right]_{-\frac{\pi}{6}}^{\frac{\pi}{6}} = 8 \left( \frac{\pi}{6} - \left(-\frac{\pi}{6}\right) \right) = 8 \times \frac{\pi}{3} = \frac{8\pi}{3} \]

Step 5: Conclusion

The length of the parametric curve is \( \frac{8\pi}{3} \).

Key Concepts

Parametric Curve Length

Calculating the length of a parametric curve might seem daunting at first, but it simplifies with the right understanding. In parametric equations, we express both the x-coordinate and y-coordinate of points on a curve as functions of a third variable, often 't'. This allows us to create equations like:

• \( x(t) = f(t) \)
• \( y(t) = g(t) \)

To find the length of such a curve, we use a specific integral formula. This formula integrates over both the rates of change of x and y:\[ L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]Here, \(L\) is the length of the curve from \(t = a\) to \(t = b\). It accounts for every little segment of the curve, calculating a cumulative total.

Derivatives of Parametric Equations

Understanding derivatives is crucial when dealing with parametric equations. The derivatives tell us the rate at which each coordinate changes with respect to 't'. For example, given our parametric equations:

• \( x(t) = 4 \cos(2t) \)
• \( y(t) = 4 \sin(2t) \)

The derivative for \(x(t)\) is \( \frac{dx}{dt} = -8 \sin(2t) \), and for \(y(t)\), we have \( \frac{dy}{dt} = 8 \cos(2t) \). These derivatives help us form the integral needed to find the curve length. In the formula \( L = \int \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \), these derivatives plug in to create a clearer path to compute the arc length.

Integral Calculus

The field of integral calculus is vital in solving many problems involving areas, volumes, and in this case, the length of curves. When computing the length of the parametric curve, the process involves evaluating an integral:\[ L = \int_{a}^{b} \sqrt{(-8 \sin(2t))^2 + (8 \cos(2t))^2} \, dt \]This integral allows us to sum up all infinitesimally small sections of the curve within the specified range of 't'. Each segment contributes to the total length, leading to a comprehensive calculation of the curve's full length. Integral calculus thus provides a powerful tool to bridge discrete segments of calculations into continuous totals.

Pythagorean Identity

In many trigonometric integrals, the Pythagorean identity is a celebrated savior. This identity states that \( \sin^2(\theta) + \cos^2(\theta) = 1 \). In our parametric curve problem, we see its importance when simplifying the integral. Our expression inside the integral simplifies:\[ 64 \sin^2(2t) + 64 \cos^2(2t) \]Utilizing \( \sin^2(2t) + \cos^2(2t) = 1 \), it simplifies to\[ 64 \times 1 = 64 \]This reduction simplifies the integral to solving a straightforward constant integral. Thus, the Pythagorean identity makes trigonometric integrals more approachable and manageable, lending itself to simplifying the calculation of complex parametric curves.