Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 35-46, find the length of the parametric curve defined over the given interval. $$ x=t+\frac{1}{t}, y=\ln t^{2} ; 1 \leq t \leq 4 $$

Step-by-Step Solution

Answer

The length of the curve is \( \frac{15}{4} \).

Step 1: Define the Parametric Equations

The problem provides us with the parametric equations: \( x(t) = t + \frac{1}{t} \) and \( y(t) = \ln(t^2) \). The parameter \( t \) varies from 1 to 4. Our task is to find the length of this parametric curve over the given interval.

Step 2: Find Derivatives

The first step to finding the arc length of a parametric curve is to compute the derivatives of \( x(t) \) and \( y(t) \). For \( x(t) = t + \frac{1}{t} \), the derivative is \( x'(t) = 1 - \frac{1}{t^2} \). For \( y(t) = \ln(t^2) \), the derivative is \( y'(t) = \frac{2}{t} \) based on the chain rule.

Step 3: Apply the Arc Length Formula

The length \( L \) of a parametric curve is given by the integral: \( L = \int_{a}^{b} \sqrt{\left( x'(t) \right)^2 + \left( y'(t) \right)^2} \, dt \). With \( x'(t) = 1 - \frac{1}{t^2} \) and \( y'(t) = \frac{2}{t} \), we plug these into the formula: \( L = \int_{1}^{4} \sqrt{\left(1 - \frac{1}{t^2}\right)^2 + \left(\frac{2}{t}\right)^2} \, dt \).

Step 4: Simplify the Integrand

Now we need to simplify the expression under the square root: \( \left(1 - \frac{1}{t^2}\right)^2 + \left(\frac{2}{t}\right)^2 \). This simplifies to \( 1 - \frac{2}{t^2} + \frac{1}{t^4} + \frac{4}{t^2} \), or \( 1 + \frac{2}{t^2} + \frac{1}{t^4} \).

Step 5: Simplify Further and Evaluate Integral

The expression \( \sqrt{1 + \frac{2}{t^2} + \frac{1}{t^4}} \) simplifies to \( \sqrt{\left(1 + \frac{1}{t^2}\right)^2} \), which is \(1 + \frac{1}{t^2}\) since the square root of a perfect square is its base (considering the value as non-negative). Hence, \( L = \int_{1}^{4} \left(1 + \frac{1}{t^2}\right) \, dt \).

Step 6: Compute the Integral

Compute the integral: \( L = \int_{1}^{4} 1 \, dt + \int_{1}^{4} \frac{1}{t^2} \, dt \). This is equivalent to \( [t]_{1}^{4} + [-\frac{1}{t}]_{1}^{4} \). Evaluating these integrals, we get \( (4 - 1) \) for the first integral and \( (-\frac{1}{4} - (-1)) \) for the second integral. Therefore, \( L = 3 + \frac{3}{4} = \frac{15}{4} \).

Key Concepts

Parametric Equations

A parametric equation involves one or more equations that express a set of quantities as explicit functions of a number of independent variables, known as parameters. In our given exercise, we have the equations:

• \( x(t) = t + \frac{1}{t} \)
• \( y(t) = \ln(t^2) \)

The variable \( t \) serves as the parameter and dictates the position on the curve. These types of equations are often used to model the motion of objects or to trace complex shapes because they provide a way to describe curves distinctly in terms of the parameter \( t \). The interval \( 1 \leq t \leq 4 \) specifies the range over which we'll examine the curve, determining our starting and ending points.

Derivatives of Parametric Equations

To find the arc length of a parametric curve, we first need the derivatives of the parameterized functions with respect to the parameter \( t \). In our problem, we find:

• The derivative of \( x(t) = t + \frac{1}{t} \) is \( x'(t) = 1 - \frac{1}{t^2} \).
• The derivative of \( y(t) = \ln(t^2) \) is \( y'(t) = \frac{2}{t} \), which is found using the chain rule — an essential rule for differentiating composite functions.

These derivatives are crucial because they describe the rate of change of \( x \) and \( y \) concerning \( t \). Understanding these rates is key to calculating the distance along the curve, as it helps us grasp how the curve bends and moves within the given interval.

Arc Length Formula

To determine the arc length of a curve represented by parametric equations, we use the arc length formula: \[ L = \int_{a}^{b} \sqrt{(x'(t))^2 + (y'(t))^2} \, dt \].This formula essentially sums up small pieces of the curve — each piece having a length calculated using the Pythagorean theorem. By integrating these segments over the interval \([a, b]\), we can calculate the total length of the curve.
In our exercise, using \( x'(t) = 1 - \frac{1}{t^2} \) and \( y'(t) = \frac{2}{t} \), we substitute into the formula to get:\[ L = \int_{1}^{4} \sqrt{\left(1 - \frac{1}{t^2}\right)^2 + \left(\frac{2}{t}\right)^2} \, dt \].The goal here is to simplify the expression under the square root to make the integration manageable.

Definite Integral Calculation

After deriving the expression for the arc length \( L \), we need to evaluate the integral over the interval \( [1, 4] \). Simplifying the expression under the square root, we find:

• The integrand simplifies to \( \sqrt{\left(1 + \frac{1}{t^2}\right)^2} = 1 + \frac{1}{t^2} \).

Having a simpler form allows us to easily integrate:\[ L = \int_{1}^{4} \left(1 + \frac{1}{t^2}\right) \, dt \].We split this into two simpler integrals:

• \( \int_{1}^{4} 1 \, dt = [t]_{1}^{4} = 4 - 1 = 3 \)
• \( \int_{1}^{4} \frac{1}{t^2} \, dt = [-\frac{1}{t}]_{1}^{4} = -\frac{1}{4} + 1 = \frac{3}{4} \)

Adding these results gives the total arc length: \( \frac{15}{4} \). This process shows how the integral, an aggregate of infinitesimal contributions, computes the curve’s length.