Problem 26 - Thomas Calculus

Question

Find a parametrization for the curve. the ray (half line) with initial point \((-1,2)\) that passes through the point \((0,0)\)

Step-by-Step Solution

Answer

The parametrization is \(x(t) = -1 + t\), \(y(t) = 2 - 2t\), for \(t \geq 0\).

Step 1: Understand the Problem

We need to parametrize a ray that starts at the point \((-1, 2)\) and passes through the point \((0, 0)\). This means we need to express the coordinates of the points on the ray as a function of a parameter, commonly denoted as \(t\).

Step 2: Coordinate Differences

Calculate the differences in the x and y coordinates between the point \((-1, 2)\) and the point \((0, 0)\). The differences are \( \Delta x = 0 - (-1) = 1 \) and \( \Delta y = 0 - 2 = -2 \).

Step 3: Direction Vector

The differences calculated give us the direction vector of the ray, which is \( \langle 1, -2 \rangle \). This vector indicates the direction in which the ray extends.

Step 4: Parametric Equations

Using the initial point \((-1, 2)\) and the direction vector \( \langle 1, -2 \rangle \), we write the parametric equations:\[ x(t) = -1 + t \]\[ y(t) = 2 - 2t \]where \(t \geq 0\) as it is a ray extending from the initial point.

Step 5: Verify Parametrization

Substituting \(t = 0\), we get \((x, y) = (-1, 2)\), confirming the starting point. For \(t = 1\), we find \((x, y) = (0, 0)\), confirming the ray passes through this point. Therefore, the parametrization is verified.

Key Concepts

Direction Vector

To understand how to parametrize a ray, one must first grasp the concept of a direction vector. A direction vector is crucial because it gives us the direction in which our ray or line is extending. Imagine it as an arrow pointing from one point to another.

In our problem, we want a ray starting at the point \((-1, 2)\) and passing through the point \((0, 0)\). The direction vector is found by calculating the differences in their coordinates. These differences form the components of our vector.

• For the x-direction: The difference is \(0 - (-1) = 1\).
• For the y-direction: The difference is \(0 - 2 = -2\).

Thus, our direction vector is \(\langle 1, -2 \rangle\). This vector tells us how much to move in the x and y directions to travel from one point to another. It's like the speed and direction of a plane — it helps guide us along the path of the ray.
Understanding and correctly calculating the direction vector is a fundamental step in parametrizing curves.

Coordinate Differences

While tackling parametrization problems, especially those involving rays and lines, calculating the coordinate differences is a fundamental step. These differences are crucial as they're used to derive the direction vector.

Let's see how it applies to our problem. We have two points: the initial point \((-1, 2)\) and another point through which the ray passes, namely \((0, 0)\).

• The difference in the x-coordinates: \(0 - (-1) = 1\).
• The difference in the y-coordinates: \(0 - 2 = -2\).

These differences give us \(\Delta x = 1\) and \(\Delta y = -2\). This means, from the initial point, we would move forward 1 unit in the x-direction and backward 2 units in the y-direction to meet our next point.
Think of these coordinate differences as the stepping stones you need to determine the path the ray takes in its journey. Without understanding this concept, it would be tricky to create an accurate parametric equation.

Parametrization Process

Now, let's dive into the parametrization process. This is where we express the coordinates along the ray in terms of a parameter, typically denoted as \(t\). Imagine \(t\) as a guide that helps us to describe every point along the ray.

We already know our starting point, \((-1, 2)\), and our direction vector \(\langle 1, -2 \rangle\). To parametrize the ray, we use these components to build equations:

• The equation for x becomes \(x(t) = -1 + t\). Initially, when \(t = 0\), \(x = -1\). As \(t\) increases, \(x\) increases positively.
• For y, the equation is \(y(t) = 2 - 2t\). At \(t = 0\), \(y = 2\). As \(t\) increases, \(y\) decreases, moving downwards.

These parametric equations describe the ray starting at \((-1, 2)\) and extending infinitely, with \(t \geq 0\). By substituting different values of \(t\), we can find countless points on the ray. This methodically connects us back to our line, illustrating how algebra and geometry intertwine in the parametrization process.