Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems 25-32, find the arc length of the given curve. } $$ $$ x=2 \cos t, y=2 \sin t, z=t / 20 ; 0 \leq t \leq 8 \pi $$

Step-by-Step Solution

Answer

The arc length of the curve is \(\frac{2\pi\sqrt{1601}}{5}\).

Step 1: Understand the Arc Length Formula

To find the arc length of a parametric curve defined by functions \(x(t)\), \(y(t)\), and \(z(t)\), we use the formula \( L = \int_{a}^{b} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2} \, dt \). Here, \(x = 2\cos t\), \(y = 2\sin t\), and \(z = t/20\), with \(t\) from 0 to \(8\pi\).

Step 2: Calculate Derivatives

Find \(\frac{dx}{dt}\), \(\frac{dy}{dt}\), and \(\frac{dz}{dt}\). - \(\frac{dx}{dt} = -2\sin t\)- \(\frac{dy}{dt} = 2\cos t\)- \(\frac{dz}{dt} = 1/20\)

Step 3: Compute the Integrand

Substitute these derivatives into the arc length formula's integrand: \( \sqrt{(-2\sin t)^2 + (2\cos t)^2 + (1/20)^2} \). Calculate:\[ = \sqrt{4\sin^2 t + 4\cos^2 t + \frac{1}{400}} \]Since \(\sin^2 t + \cos^2 t = 1\), simplify to:\[ \sqrt{4 + \frac{1}{400}} \]

Step 4: Simplify the Integrand

Simplify \(\sqrt{4 + \frac{1}{400}}\):\[= \sqrt{\frac{1600}{400} + \frac{1}{400}} = \sqrt{\frac{1601}{400}} \]\[= \frac{\sqrt{1601}}{20} \]

Step 5: Evaluate the Integral

Now, integrate the constant: \( L = \int_{0}^{8\pi} \frac{\sqrt{1601}}{20} \, dt \).\[= \frac{\sqrt{1601}}{20}(8\pi - 0) \]Simplify:\[ = \frac{8\pi\sqrt{1601}}{20} \]\[ = \frac{2\pi\sqrt{1601}}{5} \]

Step 6: Finalize the Result

The arc length of the curve from \(t = 0\) to \(t = 8\pi\) is \(\frac{2\pi\sqrt{1601}}{5}\).

Key Concepts

Parametric Equations

Parametric equations are essential in expressing the coordinates of the points that make up a geometric object as functions of a variable, typically known as a parameter. Unlike standard equations which directly relate x and y, parametric equations use one or more parameters, often denoted as "t", to express the positions of objects in space.
For example, in the given problem, the parametric equations are:

• \(x = 2\cos t\)
• \(y = 2\sin t\)
• \(z = \frac{t}{20}\)

These equations describe the x, y, and z coordinates of the curve as t varies from 0 to \(8\pi\). This form allows for a comprehensive three-dimensional representation of the curve, which can be particularly useful in fields such as physics and engineering. By understanding these equations, one can analyze various properties of the curve.

Implementation of Arc Length Formula

Calculating the arc length of a curve represented by parametric equations involves implementing a specific formula. This formula is integral-based, reflecting how integration plays a key role in measuring lengths and areas geometrically. The arc length formula for a three-dimensional parametric curve is given as:\[L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt\]To apply this, we first determine the derivatives \(\frac{dx}{dt}\), \(\frac{dy}{dt}\), and \(\frac{dz}{dt}\). In this example, they are calculated with respect to "t", yielding:

• \(\frac{dx}{dt} = -2\sin t\)
• \(\frac{dy}{dt} = 2\cos t\)
• \(\frac{dz}{dt} = \frac{1}{20}\)

Substituting these into the arc length formula provides the integral to solve. This process illustrates the step-by-step implementation of using calculus to find arc lengths, an essential task in higher mathematics.

Derivatives of Parametric Equations

Derivatives of parametric equations play a fundamental role in analyzing the motion and characteristics of a curve. When dealing with parametric forms, each parameter function needs to be differentiated to find the rate of change in each spatial dimension.
For the equations:

• \(x = 2\cos t\)
• \(y = 2\sin t\)
• \(z = \frac{t}{20}\)

The derivatives are derived as follows:

• \(\frac{dx}{dt} = -2\sin t\) captures the change in the x-direction.
• \(\frac{dy}{dt} = 2\cos t\) reflects the change in the y-direction.
• \(\frac{dz}{dt} = \frac{1}{20}\) shows the change in the z-direction.

Understanding these derivatives not only aids in calculating the arc length but also provides insights into the geometry and movement of the curve in space. This knowledge is crucial, especially in fields such as kinematics, where describing object motion is vital.

Integral Calculus

Integral calculus is the mathematical language used to find areas, volumes, central points, and more. In this context, integration helps determine the length of a curve by accumulating infinitesimal lengths over the parameter interval. The integral used here:\[L = \int_{0}^{8\pi} \frac{\sqrt{1601}}{20} \, dt\]evaluates to give the total arc length of the curve from \(t = 0\) to \(t = 8\pi\). In this instance, the integration is simplified to a constant term, emphasizing how the process can vary in complexity.Integration converts local rates of change from derivatives into the total change over an interval. This approach is critical in mathematics and applied sciences for calculating trajectory lengths, evaluating dynamic systems, and other essential tasks where dimensions come into play.