Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems 25-32, find the arc length of the given curve. } $$ $$ x=2 \cos t, y=2 \sin t, z=3 t ;-\pi \leq t \leq \pi $$

Step-by-Step Solution

Answer

The arc length of the curve is \( 2\pi\sqrt{13} \).

Step 1: Find the Derivatives

Calculate the derivatives of the parametric equations with respect to the parameter \( t \): - For \( x = 2 \cos t \), the derivative is \( x'(t) = -2 \sin t \). - For \( y = 2 \sin t \), the derivative is \( y'(t) = 2 \cos t \). - For \( z = 3t \), the derivative is \( z'(t) = 3 \).

Step 2: Set Up the Arc Length Integral

The formula for the arc length \( L \) of a parametric curve \((x(t), y(t), z(t))\) is given by \[ L = \int_{a}^{b} \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} \, dt \]Substitute the derivatives we found: \[ L = \int_{-\pi}^{\pi} \sqrt{(-2 \sin t)^2 + (2 \cos t)^2 + 3^2} \, dt \]

Step 3: Simplify the Expression Inside the Integral

Simplify the expression under the integral's square root: \((-2 \sin t)^2 + (2 \cos t)^2 + 3^2 = 4 \sin^2 t + 4 \cos^2 t + 9\). Use the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \) to further simplify:\[ 4(\sin^2 t + \cos^2 t) + 9 = 4 \cdot 1 + 9 = 13 \]Thus, the integral becomes:\[ L = \int_{-\pi}^{\pi} \sqrt{13} \, dt \]

Step 4: Evaluate the Integral

Since \( \sqrt{13} \) is a constant, the integral simplifies to:\[ L = \sqrt{13} \int_{-\pi}^{\pi} 1 \, dt \]. The integral of 1 over \([-\pi, \pi]\) is the length of the interval:\[ L = \sqrt{13} (\pi - (-\pi)) = \sqrt{13} \cdot 2\pi \].

Key Concepts

Parametric Equations

Parametric equations are a way of defining a curve by using parameters. Typically, we use a parameter like \( t \) to describe the coordinates of a curve in terms of equations for \( x \), \( y \), and sometimes \( z \) in three-dimensional space. This allows us to express complex curves easily and efficiently. Let's consider the parametric equations from our example:

• \( x = 2 \cos t \)
• \( y = 2 \sin t \)
• \( z = 3t \)

These equations describe a curve in 3D space as \( t \) varies. In this specific case, as \( t \) moves from \(-\pi\) to \( \pi \), the \( x \) and \( y \) components describe a circular motion, which, in combination with the linear \( z \) component, forms a helical shape.By plotting the values of \( x \), \( y \), and \( z \), you can visualize the parametric curve. Every point on the curve corresponds to a specific value of the parameter \( t \). Understanding parametric equations is essential for describing motion and trajectories in physics and engineering.

Derivatives

Derivatives measure how a function changes as its input changes. When dealing with parametric equations, derivatives help us understand how quickly the curve is changing with respect to the parameter \( t \). Calculating derivatives for each parametric equation provides the rates of change of the positions \( x \), \( y \), and \( z \).In the given problem, we derived the following:

• The derivative of \( x = 2 \cos t \) is \( x'(t) = -2 \sin t \).
• The derivative of \( y = 2 \sin t \) is \( y'(t) = 2 \cos t \).
• The derivative of \( z = 3t \) is \( z'(t) = 3 \).

These derivatives are critical for calculating the arc length of our curve. They provide the necessary components to plug into the arc length formula, which requires the square root of the sum of the squares of these derivatives. Mastering calculus operations like differentiation is crucial for dealing with any changes on curves represented by parametric equations.

Integral Calculus

Integral calculus deals with accumulation, such as finding areas under curves or, in our case, finding the arc length of a parametric curve. The arc length formula for a parametric curve \( (x(t), y(t), z(t)) \) uses integral calculus to sum the infinitesimally small lengths along the curve.The formula is given by:\[ L = \int_{a}^{b} \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} \, dt \]In our exercise, after substituting derivatives of our parametric equations, we simplify the expression under the square root and solve:

• First, compute \((-2 \sin t)^2 + (2 \cos t)^2 + 3^2\).
• Apply the identity \( \sin^2 t + \cos^2 t = 1 \) to simplify further.
• This yields a constant of \( \sqrt{13} \).
• The integral evaluates simply over the specified interval, resulting in \( 2\pi \).

Thus, the integral is \( \sqrt{13} \cdot 2\pi \), which gives the total arc length.Through integral calculus, complex geometric properties become tangible. You not only find arc lengths but also understand the effects of rates of change over specific intervals.