91Ó°ÊÓ

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.

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.

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.

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.