Problem 110 - Calculus Volume 3

Question

Find the arc length of the curve on the indicated interval of the parameter. $$ x=\cos (2 t), \quad y=\sin (2 t), \quad 0 \leq t \leq \frac{\pi}{2} $$

Step-by-Step Solution

Answer

The arc length is \( \pi \).

Step 1: Parameters introduction

The given parametric equations are \( x = \cos(2t) \) and \( y = \sin(2t) \). We need to find the arc length of the curve for \( 0 \leq t \leq \frac{\pi}{2} \).

Step 2: Formula for arc length

The arc length \( L \) of a parametrically defined curve given by \( x = f(t) \) and \( y = g(t) \) over the interval \([a, b]\) is calculated using:\[L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \ dt\].

Step 3: Compute derivatives

Compute the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). For \( x = \cos(2t) \), \( \frac{dx}{dt} = -2\sin(2t) \). For \( y = \sin(2t) \), \( \frac{dy}{dt} = 2\cos(2t) \).

Step 4: Substitute into arc length formula

Substitute the derivatives into the arc length formula:\[L = \int_{0}^{\frac{\pi}{2}} \sqrt{(-2\sin(2t))^2 + (2\cos(2t))^2} \ dt\]

Step 5: Simplify the integral's expression

Simplify the expression under the square root:\[(-2\sin(2t))^2 + (2\cos(2t))^2 = 4\sin^2(2t) + 4\cos^2(2t)\]. Use the Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \) to simplify it to:\[4(\sin^2(2t) + \cos^2(2t)) = 4\].

Step 6: Evaluate the integral

The integral becomes:\[L = \int_{0}^{\frac{\pi}{2}} \sqrt{4} \ dt = \int_{0}^{\frac{\pi}{2}} 2 \ dt\]. This evaluates to:\[L = 2\left[t\right]_{0}^{\frac{\pi}{2}} = 2\left(\frac{\pi}{2}\right) - 2(0) = \pi\].

Step 7: Conclusion

The arc length of the curve from \( t = 0 \) to \( t = \frac{\pi}{2} \) is \( \pi \).

Key Concepts

Understanding Parametric Equations

Parametric equations are a way of expressing geometric shapes using parameters. In our exercise, we have two equations:

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

Each equation involves the same parameter, \( t \), allowing us to describe the coordinates \((x, y)\) on a curve. This method particularly helps in tracing paths that are not easily described using standard Cartesian equations. For example, describing a circle or a spiral becomes much more straightforward with parametric forms than with traditional x-y equations. By varying the parameter, you create a path or a curve on the plane. This is particularly useful in physics and engineering where paths of moving objects are calculated in real-time.

Diving into Derivatives in Calculus

In calculus, derivatives help us understand how functions change. They are essential for finding tangents, optimizing values, and solving real-world problems involving rates of change. In this scenario, we use derivatives to find the rate of change of \( x \) and \( y \) with respect to \( t \).
For our parametric equations:

• \( \frac{dx}{dt} = -2\sin(2t) \)
• \( \frac{dy}{dt} = 2\cos(2t) \)

These derivatives signify how fast \( x \) and \( y \) change as the parameter \( t \) varies. This information is crucial when calculating the arc length as it reflects the curve's rate and direction of movement. Calculating these derivatives correctly allows us to accurately determine the length of the path.

Exploring the Pythagorean Identity

The Pythagorean identity is a fundamental concept in trigonometry. It states that: \[\sin^2(\theta) + \cos^2(\theta) = 1\] This identity is pivotal in simplifying complex expressions involving trigonometric functions. In our problem, it allows us to convert the expression under the square root in the arc length calculation into a much simpler form.
When we have \[ (-2\sin(2t))^2 + (2\cos(2t))^2 = 4\sin^2(2t) + 4\cos^2(2t)\] Using the identity, this simplifies to: \[ 4(\sin^2(2t) + \cos^2(2t)) = 4\] It effectively reduces our problem, making the subsequent integration process straightforward, since the constant 4 can be treated outside the integral.

Integral Calculus Made Easy

Integral calculus is the study of accumulation and everything related to areas under curves. It's essentially the reverse process of differentiation. In this exercise, we use integral calculus to determine the arc length of our parametric curve. The formula for arc length involves an integral of the functions describing the path, specifically: \[ L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \] The integral calculates the sum of tiny segments of the curve, which add up to the total arc length. After simplifying the expression under the square root using the Pythagorean identity, we integrate over the interval from 0 to \( \frac{\pi}{2} \): \[ L = \int_{0}^{\frac{\pi}{2}} 2 \, dt\] This becomes a straightforward calculation resulting in the arc length of \( \pi \), showing the elegance and utility of integral calculus in solving real-life geometric problems.