91Ó°ÊÓ

Parametric equations allow us to express the coordinates of points on a curve as functions of a parameter, usually denoted by \( t \). For the curve \( \mathbf{r}(t)=\langle 6t, 6t-t^2 \rangle \), \( t \) acts as the parameter. It helps us track each point on the simplified path using a single variable.

Understanding parametric equations involves recognizing that:

• The x-coordinate is represented as \( x(t) = 6t \), showing how the x-value changes with \( t \).
• The y-coordinate is represented as \( y(t) = 6t - t^2 \), showing how the y-value changes as \( t \) varies.

This makes the study of curves more tangible as it breaks down complex expressions into manageable parts. The parametric form is particularly useful in optimization problems where direct solutions might be challenging. By using parametric equations, we can easily isolate the behavior of one coordinate, optimize it, and find specific points of interest like a maximum or minimum on the curve.

A quadratic function is characterized by its general form \( y = ax^2 + bx + c \) and describes a symmetrical curve known as a parabola. In our scenario, the expression \( y(t) = 6t - t^2 \) is a quadratic function in terms of \( t \).

Key properties include:

• The direction in which the parabola opens is determined by the sign of \( a \). If \( a < 0 \), like here where \( a = -1 \), it opens downwards. Otherwise, it opens upwards.
• The vertex of the parabola represents either the highest or lowest point, depending on the direction it opens.

By focusing on the quadratic form, mathematicians and students can predict the behavior and solution of various optimization problems. In many real-world applications, such as maximizing height or minimizing cost, quadratic functions come in handy.

The vertex of a parabola is a crucial point indicating the parabola's maximum or minimum value. For the downward-opening parabola given by \( y(t) = 6t - t^2 \), it informs us the apex of its trajectory. The vertex formula \( t = -\frac{b}{2a} \) can help efficiently find this point.

Let's break down the process:

• Identify coefficients \( a = -1 \) and \( b = 6 \) from \( y(t) = -t^2 + 6t \).
• Substitute these into the formula to find \( t = -\frac{6}{2(-1)} = 3 \), which identifies the parameter value at the vertex.
• The vertex gives a maximum y-value for the function at \( t = 3 \). Calculating \( y(3) \) yields \( 9 \).
  

Finding the vertex allows us to understand not only the nature of quadratic functions but gives concrete solutions for locating optimal points in a given scenario. In practical terms, students and professionals frequently use this method to identify ideal conditions in various fields, from physics to economics.