91Ó°ÊÓ

Imagine a journey through a path where each position is determined using a third variable, often time, known as a parameter. This is the essence of parametric equations in mathematics. Instead of describing a curve using a single equation in terms of x and y, parametric equations break it down into two separate functions of t (the parameter).
For example, in our exercise, the equations are given as:

• \( x = 4 \sin t \)
• \( y = 2 \cos t \)

Each function describes how x and y change as t varies. This offers flexibility and insight, especially in capturing more complex shapes like circles or ellipses. In essence, parametric equations provide a time-based snapshot of a curve's location, making them a vital tool in mathematical modeling.

A tangent line is like a shadow cast by a curve at a single point, only touching the curve at that point and having the same slope as the curve does at that exact location. To find the equation of a tangent line, we first need its slope and a specific point on the line.
In our problem, the point is defined at \( t = \frac{\pi}{4} \), leading to coordinates \((2\sqrt{2}, \sqrt{2})\) on the curve. We computed the slope \( m \) with the derivative \( \frac{dy}{dx} = -\frac{1}{2} \). Once the slope is known, the point-slope form of the line equation comes into play:

• \( y - y_1 = m(x - x_1) \)

Substituting our specific point and slope, we arrive at:

• \( y = -\frac{1}{2}x + 2\sqrt{2} \)

This formula gives us the tangent line equation, which grants us insights into the curve's behavior at that precise spot.

The first derivative in the context of differential calculus represents the rate of change, or slope, of a function at a particular point. For parametric equations, the process of finding this involves computing both \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) first.
In our exercise:

• \( \frac{dx}{dt} = 4 \cos t \)
• \( \frac{dy}{dt} = -2 \sin t \)

Evaluating these at \( t = \frac{\pi}{4} \) gives us the specific values necessary to determine the tangent's slope:

• \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = -\frac{1}{2} \)

Thus, the first derivative not only helps us find the tangent line slope but also offers insight into how both x and y variables change with respect to time.

The second derivative provides an understanding of how the rate of change itself is changing. It examines the curvature or concavity of a function, revealing whether the graph is bending upwards or downwards.
In cases involving parametric equations, we must derive \( \frac{d^2y}{dx^2} \) using a specific formula:

• \( \frac{d^2y}{dx^2} = \frac{d}{dt}\left( \frac{dy}{dx} \right) \bigg/ \frac{dx}{dt} \)

In our given example, \( \frac{dy}{dx} \) was constant at \(-\frac{1}{2}\), leading to its time-derived value being zero. Thus the second derivative evaluates to zero, hinting that at \( t = \frac{\pi}{4} \), the curve neither hooks up nor down.
This additional level of insight informs us about the stability and shape of the curve at this particular instance.