/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Find a parametrization for the c... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 36

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

Short Answer

Expert verified
Parametrization: \( x = -1 + t \), \( y = 2 - 2t \) for \( t \geq 0 \).

Step by step solution

01

Identify the Direction Vector

The ray starts at the point \((-1, 2)\) and passes through the point \((0, 0)\). The direction vector is found by subtracting the coordinates of the initial point from the point it passes through: \( (0 - (-1), 0 - 2) = (1, -2) \). This gives us the direction vector \( \vec{d} = (1, -2) \).
02

Formulate the Parametric Equations

Using the initial point \((-1, 2)\) and the direction vector \((1, -2)\), the parametric equations for the ray can be written. The general formula for a parametric line is \( (x, y) = (x_0, y_0) + t \cdot \vec{d} \), where \( t \) is the parameter. Therefore, \( x = -1 + t(1) \) and \( y = 2 + t(-2) \).
03

Simplify the Parametric Equations

Substitute and simplify the equations: \( x = -1 + t \) and \( y = 2 - 2t \). These are the parametric equations for the ray starting at \((-1, 2)\) and passing through \((0, 0)\).
04

Domain of the Parameter

Since the problem specifies a ray (half-line), the parameter \( t \) must be non-negative (\( t \geq 0 \)) because the ray starts at \((-1, 2)\) and extends indefinitely in the direction of the vector \((1, -2)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Direction Vector
A direction vector is a fundamental concept in geometry, particularly when working with lines or rays. It essentially determines the direction in which a line is oriented. For any given line or ray in a Cartesian plane, a direction vector can be found by taking two distinct points on the line.
Here's how you find it:
  • Pick two points on the ray or line, let's call them \((x_1, y_1)\) and \((x_2, y_2)\).
  • Subtract the coordinates of the first point from the second point: \((x_2 - x_1, y_2 - y_1)\).
This gives you the direction vector \(\vec{d} = (x_2 - x_1, y_2 - y_1)\). In our scenario, with the initial point \((-1, 2)\) and the point \((0, 0)\), the direction vector is calculated as \((1, -2)\).
Direction vectors are vital because they standardize the way we express the orientation of a line or ray.
Parametrization
Parametrization is a method used to describe the points on a line, ray, or curve using a parameter, typically denoted as \(t\). It translates complex geometric concepts into an algebraic form.
For a line or ray:\( (x, y) = (x_0, y_0) + t \cdot \vec{d} \), where:
  • \((x_0, y_0)\) is the initial point.
  • \(\vec{d}\) is the direction vector.
  • \(t\) is the parameter, which takes on different values to yield different points on the line or ray.
The parameter \(t\) often represents time, moving from one point forward on the line. For the ray starting at \((-1, 2)\) with direction \((1, -2)\), we formed the parametric equations \(x = -1 + t\) and \(y = 2 - 2t\).
By adjusting \(t\), you can find any point along that ray, providing a flexible and comprehensive description of the ray's geometry.
Rays in Geometry
A ray is a part of a line that begins at a particular point, called the initial point, and extends infinitely in one direction. It is essential in various fields of mathematics and physics. Unlike a full line, a ray has one endpoint.
Understanding Rays:
  • The initial point is where the ray starts.
  • The ray passes through another point, which provides its direction.
  • It can be described using a direction vector and a parameterization.
  • In our exercise, the ray starts at \((-1, 2)\) and continues through \((0, 0)\).
Rays are described using a parameter \(t\), usually with \(t \geq 0\). This non-negative restriction on the parameter aligns with the ray extending infinitely in the direction noted, without extending backward.
This characteristic makes rays incredibly useful in geometry, serving as a simplified linear representation in directions for models and computations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

a. Find the length of one arch of the cycloid $$x=a(t-\sin t), \quad y=a(1-\cos t)$$ b. Find the area of the surface generated by revolving one arch of the cycloid in part (a) about the \(x\) -axis for \(a=1\)

Graph the sets of points whose polar coordinates satisfy the equations and inequalities in Exercises \(11-26\). $$r \geq 1$$

Graph the sets of points whose polar coordinates satisfy the equations and inequalities in Exercises \(11-26\). $$0 \leq \theta \leq \pi / 2, \quad 1 \leq|r| \leq 2$$

Replace the polar equations in Exercises \(27-52\) with equivalent Cartesian equations. Then describe or identify the graph. $$r \sin \theta=-1$$

Find the point on the parabola \(x=t, y=t^{2},-\infty
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.