Question

, find the length of the parametric curve defined over the given interval. $$ x=2 \sin t, y=2 \cos t ; 0 \leq t \leq \pi $$

Short answer

The length of the curve is \(2\pi\).

Step-by-step solution

Understand the Problem

We need to find the length of the parametric curve defined by the equations \(x=2\sin t\) and \(y=2\cos t\) over the interval \(0 \leq t \leq \pi\). This is the equation for a circle with a radius of 2 that completes a half-circle as \(t\) goes from 0 to \(\pi\).

Recall the Formula for Arc Length

The length of a parametric curve given by \((x(t), y(t))\) for \(a \leq t \leq b\) is \( L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\).

Calculate the Derivatives

Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). \(x = 2 \sin t\) gives \(\frac{dx}{dt} = 2 \cos t\). \(y = 2 \cos t\) gives \(\frac{dy}{dt} = -2 \sin t\).

Substitute Derivatives into Arc Length Formula

Substitute \(\frac{dx}{dt} = 2 \cos t\) and \(\frac{dy}{dt} = -2 \sin t\) into the formula: \[ L = \int_0^\pi \sqrt{(2\cos t)^2 + (-2\sin t)^2} \, dt \] which simplifies to \[ L = \int_0^\pi \sqrt{4\cos^2 t + 4\sin^2 t} \, dt \]

Simplify the Integral Expression

Use the trigonometric identity \(\cos^2 t + \sin^2 t = 1\) to further simplify: \[ L = \int_0^\pi \sqrt{4(\cos^2 t + \sin^2 t)} \, dt = \int_0^\pi \sqrt{4} \, dt = \int_0^\pi 2 \, dt \]

Integrate and Solve for Length

Integrate the simplified expression: \[ L = \int_0^\pi 2 \, dt = 2 \left[t\right]_0^\pi = 2(\pi - 0) = 2\pi \]

Verify the Result

Since the curve is part of a circle of radius 2, the circumference of the full circle is \(2\pi \times 2 = 4\pi\). The calculated length \(2\pi\) is correct, as it represents half of the circle.

Key concepts

Arc Length of Curves

When trying to find the length of a curve described by parametric equations, we use a specific formula tailored for parametric forms. The arc length of a curve can be understood as the distance you would travel if you were to walk from the start to the end of the curve. It's similar to measuring how much string you would need to lay along the curve. To calculate this in a parametric setting, we use: \[ L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \ dt \] Where \( x(t) \) and \( y(t) \) define the curve, and the interval \([a, b]\) is the parameter range over which you want to calculate the length. This approach helps to accurately assess the distance along curves that aren’t straightforward, using calculus to break it down into smaller, manageable parts.

Trigonometric Identities

Fundamental to simplifying the arc length formula are trigonometric identities. In this exercise, one crucial identity was used: \[ \cos^2 t + \sin^2 t = 1 \] This identity simplifies the expression under the square root when substituting derivatives into the arc length formula. Trigonometric identities like this one make calculations cleaner and easier by reducing complexity. Always keep in mind that identities can transform what seems like a complicated problem into a simple one—for instance, in our problem by transforming \(4\cos^2 t + 4\sin^2 t\) into \(4\). Becoming comfortable with these identities is a powerful tool in calculus and beyond.

Parametric Equations

Parametric equations provide a way of defining a curve through parameters, often with time \(t\) as a variable. In this particular problem, the parametric equations given are:

• \(x = 2\sin t\)
• \(y = 2\cos t\)

These effectively describe a curve by specifying both the \(x\) and \(y\) coordinates in terms of \(t\), allowing for a more versatile description of shapes and motions. Parametric equations are useful in many areas of mathematics and physics for modeling paths and trajectories and are especially helpful for curves that don't fit neatly into standard x-y function forms. Understanding this allows one to transform complex curves into manageable equations.

Calculus

Calculus is the mathematical study of change and motion, and it provides the tools to find the length of complex curves, among many other applications. Key ideas within calculus, such as differentiation and integration, help in determining how quantities change and accumulate. Differentiation yielded \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), critical for the arc length formula. Then, integration allowed these pieces to come together, computing the total change from the derivatives over the interval. It’s the essence of calculus: breaking down something complex to understand how all the small changes fit together. Whether finding the area under a curve, the volume of a shape, or the path traveled, calculus provides a framework to find precise solutions to these interesting problems.