91Ó°ÊÓ

Parametric equations are a way to represent curves by specifying coordinates both as functions of an independent parameter, often denoted as \( t \). This kind of representation is particularly useful for describing complex movements and shapes. In the context of the exercise, the curve is defined by the equations \( y = 8t^3 - 1 \) and \( x = 4t^2 + 3 \). Here, both \( x \) and \( y \) depend on \( t \), making \( t \) a parameter that traces the curve as it changes value.

**Why Use Parametric Equations?**

• They can describe curves that cannot be expressed as a single function \( y = f(x) \) or \( x = g(y) \).
• Useful for modeling real-world scenarios, like the path of a moving object.
• Simplifies calculations, especially when involving calculus.

In our problem, parametric equations allow us to treat \( t \) as a variable to find specific points on the curve. As \( t \) varies, it generates different \( x \) and \( y \) coordinates, which can be used to determine properties such as tangents and normals.

Derivatives represent the rate at which one quantity changes with respect to another. When working with parametric equations, instead of directly differentiating \( y \) with respect to \( x \), we find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) first. This method allows us to determine the rate of change of \( y \) and \( x \) as the parameter \( t \) changes.

For our given curve, we have:
- \( \frac{dy}{dt} = 24t^2 \)
- \( \frac{dx}{dt} = 8t \)

To find the derivative \( \frac{dy}{dx} \) (i.e., the slope of the tangent to the curve), we divide these two rates:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{24t^2}{8t} = 3t \]

This result provides the slope of the tangent line at any point \( t \) on the curve. Understanding derivatives is essential, as it allows us to analyze motion dynamics along the curve and identify points where the line is tangent or normal.

The slope of a line is a measure of how steep the line is, denoted by \( m \). In calculus, the slope of a line tangent to a curve at a given point is the derivative of the curve at that point. In our problem, the slope of the tangent line is \( 3t \), derived from the parametric equations.

**Key Points about Slope:**

• A positive slope indicates that the line rises as it moves from left to right.
• A negative slope indicates that the line falls.
• A zero slope means the line is horizontal.
• An undefined slope (division by zero) suggests a vertical line.

For the tangent line at a given parameter \( t_1 \), the slope is \( m_1 = 3t_1 \). The normal line, which is perpendicular to the tangent, has a slope that is the negative reciprocal, so if the tangent's slope is \( 3t_1 \), the normal's slope would be \( -\frac{1}{3t_2} \). This relationship helps in determining points on the curve where specific conditions, like tangency and normalcy, occur at the same line but different locations.