91Ó°ÊÓ

Parametric equations offer a way to define a set of related quantities as functions of an independent variable, often denoted as \( t \). In the original problem, the curve is defined with parametric equations: \( x = 2 \sin^{-1} t \) and \( y = \ln(1-t^2) \).
These equations describe how each point on the curve is determined by a different value of \( t \). As \( t \) varies within the interval \([0, \frac{1}{2}]\), the corresponding \( x \) and \( y \) values change, tracing out the path of the curve.

When working with parametric equations, it's important to understand that:

• Each value of \( t \) maps to a unique point \((x, y)\).
• The interval over which \( t \) varies determines the portion of the curve being considered.
• By analyzing the behavior of the parametric equations, you can learn about the shape and extent of the curve.

Understanding parametric equations is crucial because they serve as the foundation for computing derivatives and arc lengths in problems like these.

Calculus integration is a powerful tool in handling various mathematical problems. In this context, using integration allows us to find the arc length of a curve defined by parametric equations.
The arc length formula for a parametric curve is:
\[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]
This formula involves integrating over the interval \([a, b]\) which, in the original problem, is \( [0, \frac{1}{2}] \).

Key steps in using integration for arc length include:

• Substitute computed derivatives into the arc length formula.
• Simplify the expression inside the square root to make it easier to integrate.
• Recognize patterns, such as common derivatives or standard integrals, to speed up evaluation.

Through these processes, integration transforms a challenging geometric problem into a manageable analytical task, providing the exact length of the curve efficiently.

Derivative computation involves finding the rate at which one variable changes concerning another. In the case of parametric equations, we need the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) to use the arc length formula.
For \( x = 2 \sin^{-1} t \), compute the derivative:
\[ \frac{dx}{dt} = \frac{2}{\sqrt{1-t^2}} \]
For \( y = \ln(1-t^2) \), compute the derivative:
\[ \frac{dy}{dt} = \frac{-2t}{1-t^2} \]

These computations are essential because:

• They determine the components of the function that describe the curve's direction at any point.
• The resulting expressions are used in the arc length formula, encapsulating the change in \( x \) and \( y \) along the curve as \( t \) varies.
• Accurate derivatives ensure the correct arc length calculation when substituted back into the formula.

Mastering derivative skills enhances problem-solving capabilities and deepens understanding of dynamic systems represented by parametric form.