Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

, find the length of the parametric curve defined over the given interval. $$ x=t+\frac{1}{t}, y=\ln t^{2} ; 1 \leq t \leq 4 $$

Step-by-Step Solution

Answer

The length of the curve is 3.75.

Step 1: Identify the formula

To find the length of a parametric curve, we use the formula for the arc length from parameter \( a \) to \( b \): \[ L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \ dt \]

Step 2: Find derivatives

Find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Given \( x = t + \frac{1}{t} \), its derivative is \( \frac{dx}{dt} = 1 - \frac{1}{t^2} \). For \( y = \ln t^{2} \), simplifying gives \( y = 2\ln t \), so \( \frac{dy}{dt} = \frac{2}{t} \).

Step 3: Calculate squared derivatives

Square each of the derivatives: \( \left( \frac{dx}{dt} \right)^2 = \left( 1 - \frac{1}{t^2} \right)^2 = 1 - \frac{2}{t^2} + \frac{1}{t^4} \) and \( \left( \frac{dy}{dt} \right)^2 = \left( \frac{2}{t} \right)^2 = \frac{4}{t^2} \).

Step 4: Combine terms and simplify

Combine the squared derivatives inside the square root of the integral:\( \left( 1 - \frac{2}{t^2} + \frac{1}{t^4} \right) + \frac{4}{t^2} \) simplifies to \( 1 + \frac{2}{t^2} + \frac{1}{t^4} \).

Step 5: Set up the integral

Substituting into the arc length formula gives: \[ L = \int_{1}^{4} \sqrt{1 + \frac{2}{t^2} + \frac{1}{t^4}} \ dt \].

Step 6: Simplify under square root

Recognize that \( \sqrt{1 + \frac{2}{t^2} + \frac{1}{t^4}} = \sqrt{\left( 1 + \frac{1}{t^2} \right)^2} = 1 + \frac{1}{t^2} \).

Step 7: Evaluate the integral

Evaluate the integral:\[ \int_{1}^{4} \left( 1 + \frac{1}{t^2} \right) dt \].Split into two integrals:\[ \int_{1}^{4} 1 \, dt + \int_{1}^{4} \frac{1}{t^2} \, dt \].The first integral evaluates to \( [t]_{1}^{4} = 4 - 1 = 3 \).The second integral evaluates as \( \left[-\frac{1}{t}\right]_{1}^{4} = -\frac{1}{4} + \frac{1}{1} = \frac{3}{4} \).

Step 8: Calculate the total length

Add the results of both integrals: \[ L = 3 + \frac{3}{4} = \frac{12}{4} + \frac{3}{4} = \frac{15}{4} = 3.75 \].

Key Concepts

Derivatives of Parametric Equations

To find the length of a parametric curve, the first step involves calculating the derivatives of the parametric equations. This means we need to find how each parameter, typically denoted as \( t \), changes our \( x \) and \( y \) coordinates.

For example, given the parametric equations \( x = t + \frac{1}{t} \) and \( y = \ln t^{2} \), we need to find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Using basic rules of differentiation:

• For \( x = t + \frac{1}{t} \), the derivative is \( \frac{dx}{dt} = 1 - \frac{1}{t^2} \).
• For \( y = \ln t^{2} \), first simplify to \( y = 2\ln t \), then differentiate to get \( \frac{dy}{dt} = \frac{2}{t} \).

Understanding how to differentiate these parametric forms is crucial, as they form the basis for further calculations in finding the arc length of the curve.

Arc Length Formula

The arc length formula is a critical component in determining the length of parametric curves. It is given by the integral from the start to the end of the parameterization. The formula is:

\[ L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \ dt \]

This formula essentially calculates the length by using the Pythagorean theorem in an infinitesimally small interval as you traverse the curve.

• Calculate the square of each derivative computed earlier.
• Add them together, as shown: \( \left( \frac{dx}{dt} \right)^2 = 1 - \frac{2}{t^2} + \frac{1}{t^4} \) and \( \left( \frac{dy}{dt} \right)^2 = \frac{4}{t^2} \).
• These terms are then added together inside the square root of the integral.
• Resulting in: \( \sqrt{1 + \frac{2}{t^2} + \frac{1}{t^4}} \).

This process gives us the integrand function that we will evaluate to find the arc length over the specified interval of our parameter \( t \).

Evaluating Definite Integrals

Once we have set up the integral for the arc length formula, the next step is to evaluate this integral. This involves carrying out integration over the interval \( [a, b] \), where \( a \) and \( b \) are the bounds of our parameter \( t \).

• The integral we need to evaluate in the example is \( \int_{1}^{4} \sqrt{1 + \frac{2}{t^2} + \frac{1}{t^4}} \, dt \).
• Upon simplifying, we recognize \( \sqrt{1 + \frac{2}{t^2} + \frac{1}{t^4}} = 1 + \frac{1}{t^2} \), reducing our computation.
• This leads us to evaluate the new integral: \( \int_{1}^{4} \left( 1 + \frac{1}{t^2} \right) dt \).
• Split this into two separate integrals: \( \int_{1}^{4} 1 \, dt + \int_{1}^{4} \frac{1}{t^2} \, dt \).

Solving these:

• The first integral, \( \int_{1}^{4} 1 \, dt \), evaluates to \( 4 - 1 = 3 \).
• The second integral, \( \int_{1}^{4} \frac{1}{t^2} \, dt \), results in \( \left[-\frac{1}{t}\right]_{1}^{4} = \frac{3}{4} \).
• Add these results: \( L = 3 + \frac{3}{4} = \frac{15}{4} = 3.75 \).

The final step is summing these to find that the arc length is \( 3.75 \). Understanding how to evaluate these integrals cleanly and effectively is key to calculating arc lengths efficiently.