91Ó°ÊÓ

Parametric equations are a way of expressing a mathematical curve by defining two separate equations. Instead of relating two variables directly, each variable is defined independently in terms of a third variable, usually denoted as \( t \). This third variable is known as the parameter.

In this exercise, the parametric equations are given:

• \( x = \sqrt{t} \)
• \( y = 3t - 1 \)

The parameter \( t \) varies from 0 to 1. This means, as \( t \) changes within this range, the values of \( x \) and \( y \) follow accordingly to form a curve.

Such parametric representations are essential because they allow for the analysis of curves that might be difficult to handle with traditional \(y = f(x) \) or \( x = g(y) \) forms. Thus, offering a more flexible framework to model curves and motion in physics and engineering.

Differentiation is the mathematical process used to find the rate at which a function is changing at any given point. In the context of parametric equations, it involves finding the derivatives of the functions \( x(t) \) and \( y(t) \) with respect to the parameter \( t \).

For the functions given:

• Derive \( x(t) = \sqrt{t} \) to get \( \frac{dx}{dt} = \frac{1}{2\sqrt{t}} \).
• Derive \( y(t) = 3t - 1 \) to get \( \frac{dy}{dt} = 3 \).

Differentiating these equations helps determine the slope of the curve described as \( t \) changes. These derivatives are foundational in computing the arc length of the curve, as they represent the "tiny little parts" (or infinitesimal changes) along the curve. Understanding differentiation here is crucial since it allows us to find how the curve stretches or bends as the parameter \( t \) varies.

The arc length of a curve described by parametric equations is calculated using a specific formula. This formula is an adaptation of the Pythagorean theorem, taking into account both the horizontal and vertical changes as the parameter varies.

The formula for arc length \( L \) in terms of parameter \( t \) is:\[ L = \int_a^b \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 } dt \]

Here, \( a \) and \( b \) are the limits of the parameter \( t \). For our exercise, plugging in the derivatives \( \frac{dx}{dt} = \frac{1}{2\sqrt{t}} \) and \( \frac{dy}{dt} = 3 \) gives the integral:\[ \int_0^1 \sqrt{ \left(\frac{1}{2\sqrt{t}}\right)^2 + (3)^2 } dt \]

This transforms the problem of finding the length of a curve into evaluating an integral over its defined interval for \( t \).

Evaluating the integral derived from the arc length formula requires combining algebraic manipulation and integration techniques. Given the integral expression:

• \( \int_0^1 \sqrt{ \frac{1}{4t} + 9 } dt \)

It can look intimidating. However, understanding and rearranging it can help simplify the task. In this exercise, it was resolved through methods beyond basic calculus, possibly involving numerical integration or advanced substitution, yielding the result: \(2\left(3+t\right)^{\frac{1}{2}}\bigg|_0^1\).

Upon calculating, this results in:

• \( (4 + 2\sqrt{2} - 2\sqrt{3}) \)

This process shows that calculus is about more than rote computation—it's understanding how to manipulate expressions cleverly and appropriately to uncover solutions. Integral evaluation is a powerful tool for solving real-world problems involving continuous changes and measurements.