Calculating the Length of a Curve Defined by Parametric Equations Using Integration

In this article, we will explore the process of finding the length of a curve defined by parametric equations. We will use the formula for arc length and apply it to a specific example to demonstrate the steps involved in the calculation.

Introduction to Parametric Equations and Curve Length

Parametric equations are typically used to describe the position of a moving object in a plane. Instead of defining a curve y f(x), a set of parametric equations defines the curve as x f(t) and y g(t), where t is a parameter.

Formula for Arc Length

The length of a curve described by parametric equations can be found using the arc length formula:

L ∫a^b √[ (dx/dt)^2 (dy/dt)^2 ] dt

Where a and b are the limits of the parameter t that define the interval of interest.

Example Problem: Curve Length Calculation

Consider the parametric equations given below:

x 2θ sin(2θ) y cos(2θ)

We are tasked with finding the length of the curve defined by these equations for π/6 ≤ θ ≤ π/3.

Step 1: Calculate Derivatives

First, we need to find the derivatives of x and y with respect to θ:

dx/dθ 2(2 cos(2θ)) 4 cos(2θ) dy/dθ -2 sin(2θ)

Step 2: Plug into the Arc Length Formula

Now, we substitute these derivatives into the arc length formula:

L ∫π/6^π/3 √[ (4 cos(2θ))^2 (-2 sin(2θ))^2 ] dθ

After simplifying the expression inside the square root:

256 cos^2(2θ) 4 sin^2(2θ) 16 (16 cos^2(2θ) sin^2(2θ))

This can be further simplified using the trigonometric identity cos^2(2θ) sin^2(2θ) 1:

16 cos^2(2θ) 4 sin^2(2θ) 16 cos^2(θ) 4 sin^2(θ) 16 cos^2(θ)

Hence, the equation becomes:

L 4 ∫π/6^π/3 cos(θ) dθ

Step 3: Evaluate the Integral

Evaluating the integral:

L 4 [sin(θ)]π/6^π/3

Substituting the limits:

L 4 [sin(π/3) - sin(π/6)]

Using the values of sine for special angles:

sin(π/3) √3/2 and sin(π/6) 1/2

Hence:

L 4(√3/2 - 1/2) 2√3 - 2

Conclusion

The length of the curve described by the parametric equations ( x 2θ sin(2θ) ) and ( y cos(2θ) ) for ( frac{π}{6} ≤ θ ≤ frac{π}{3} ) is ( 2√3 - 2 ).

Further Exploration

To further explore this topic, you might consider calculating the length of curves defined by different parametric equations or exploring applications of parametric equations in various fields such as physics and engineering.

Now, let's look at a similar problem using a different set of parametric equations: ( x θ sin(θ) ) and ( y cos(θ) ) where ( θ in [frac{π}{3}, 2π/3] ).

Comparison with Another Example

Given these parametric equations, the curve length will be calculated as follows:

L ∫π/3^2π/3 √[ (1 cos(θ))^2 (sin(θ))^2 ] dθ

Further simplifying:

L ∫π/3^2π/3 √[ 1 cos^2(θ) sin^2(θ) ] dθ ∫π/3^2π/3 √[ 2 2 cos(θ) ] dθ 2 ∫π/3^2π/3 cos(θ/2) dθ

Evaluating the integral:

L 2 [2 sin(θ)]π/3^2π/3 4 [sin(2π/3) - sin(π/3)] 4 [√3/2 - √3/2] 2√3 - 1

Thus, the length of the curve in this case is approximately 1.4641.

Key Takeaways

Understanding the concept and formula for finding the length of a curve defined by parametric equations. Mastering the integration techniques and trigonometric identities to simplify expressions. Practicing with various parametric equations to develop problem-solving skills.

By following these steps, you can calculate the length of any curve defined by parametric equations, expanding your knowledge in calculus and its applications.