Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 35-46, find the length of the parametric curve defined over the given interval. $$ x=2 \sin t, y=2 \cos t ; 0 \leq t \leq \pi $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Parametric Equations

The curve is defined by the parametric equations \( x = 2\sin t \) and \( y = 2\cos t \), with the parameter \( t \) varying from \( 0 \) to \( \pi \). These equations describe a circle of radius 2 centered at the origin.

Step 2: Use the Arc Length Formula for Parametric Equations

The formula for the length \( L \) of a parametric curve \( x = f(t) \), \( y = g(t) \) from \( t = a \) to \( t = b \) is given by \[ L = \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt. \]

Step 3: Compute Derivatives

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

Step 4: Substitute and Simplify Under the Square Root

Substitute the derivatives into the arc length formula: \[ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 = (2\cos t)^2 + (-2\sin t)^2 = 4\cos^2 t + 4\sin^2 t. \] Using the identity \( \cos^2 t + \sin^2 t = 1 \), simplify to \[ 4(\cos^2 t + \sin^2 t) = 4. \]

Step 5: Evaluate the Integral

With the expression under the square root simplified to 4, the integral simplifies to \[ L = \int_{0}^{\pi} \sqrt{4} \, dt = \int_{0}^{\pi} 2 \, dt. \] Evaluate this integral: \[ L = 2t \bigg|_{0}^{\pi} = 2\pi - 0 = 2\pi. \]

Step 6: Interpret the Result

The length of the parametric curve over the interval \( 0 \leq t \leq \pi \) is \( 2\pi \), which matches the expected half-circle length for a circle of radius 2.

Key Concepts

Parametric Equations

Parametric equations play a crucial role in calculus, offering a different way to represent curves in the plane. Unlike the typical Cartesian equations that relate two variables, parametric equations use a third variable, the parameter, to define both x and y coordinates independently. In this exercise, the equations are given as \( x = 2\sin t \) and \( y = 2\cos t \), where \( t \) is the parameter.
When \( t \) varies from 0 to \( \pi \), these equations describe a path traced out by a point moving in a circular manner. Here, the curve represents half of a circle with a radius of 2, centered at the origin. This is due to the fact that the sine and cosine functions model circular motion.
Understanding parametric equations extends your ability to describe curves that are difficult or impossible to define using standard functions. With these, you can easily model various types of motion and paths, providing a more dynamic approach to representing relationships between variables.

Derivative of Parametric Equations

The derivative of parametric equations helps us understand how changes in the parameter \( t \) affect the curve. For any parametric curve described by \( x = f(t) \) and \( y = g(t) \), we find their derivatives with respect to \( t \) — these are \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
In this problem, the derivatives are calculated as follows:

• The derivative of \( x = 2\sin t \) with respect to \( t \) is \( \frac{dx}{dt} = 2\cos t \).
• The derivative of \( y = 2\cos t \) with respect to \( t \) is \( \frac{dy}{dt} = -2\sin t \).

These derivatives represent the rate of change of the respective coordinates as the point moves along the curve. They are crucial for calculating the arc length, as seen in later steps, and offer insight into the curve's geometry including tangents and velocity.

Arc Length Formula

The arc length formula is a fundamental tool in calculus used to determine the length of a curve described by parametric equations. For a curve defined by \( x = f(t) \) and \( y = g(t) \) over the interval \( a \leq t \leq b \), the arc length \( L \) is calculated using the integral \[ L = \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt. \]
In this exercise, substituting our derivatives into the formula results in:

• \( \left( \frac{dx}{dt} \right)^2 = (2\cos t)^2 = 4\cos^2 t \)
• \( \left( \frac{dy}{dt} \right)^2 = (-2\sin t)^2 = 4\sin^2 t \)

When combined and simplified using the identity \( \cos^2 t + \sin^2 t = 1 \), we get \( 4(\cos^2 t + \sin^2 t) = 4 \), simplifying the integral to \( L = \int_{0}^{\pi} 2 \, dt = 2\pi \). This result shows us a clean and straightforward approach to measuring the curve's distance covered.

Integral Calculus

Integral calculus is the branch of mathematics concerned with accumulation of quantities, and it plays a critical role in finding the area under curves, among other applications. When computing the arc length of a parametric curve, we'll often integrate to sum up small pieces of the curve to get the total length.
In this exercise, after simplifying the expression under the square root to 4, we find the integral as \[ L = \int_{0}^{\pi} 2 \, dt. \]This integral straightforwardly evaluates to \( 2t \big|_{0}^{\pi} \), giving \( 2\pi \). The use of integral calculus allows us to accurately determine the arc length, reinforcing its importance in tackling problems involving continuous entities.
Understanding how integral calculus operates in this context enhances your ability to solve numerous real-world problems, ranging from physics to engineering and beyond, where finding the "total" or "whole" of some quantity is necessary.