/*! 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 2 Plot the point whose polar coord... [FREE SOLUTION] | 91Ó°ÊÓ

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Plot the point whose polar coordinates are given. $$(1, \pi / 4)$$

Short Answer

Expert verified
Answer: The Cartesian coordinates of the point are (sqrt(2)/2, sqrt(2)/2).

Step by step solution

01

Understand polar coordinates and their relation with Cartesian coordinates

In polar coordinates, a point is described by the distance (radius) from the origin and the angle between the positive x-axis and the line connecting the origin to the point. The polar coordinates are denoted by \((r, \theta)\), where \(r\) is the radius, and \(\theta\) is the angle. To convert polar coordinates to Cartesian coordinates, we use the following relations: $$x = r \cos{\theta}$$ $$y = r \sin{\theta}$$
02

Convert the given polar coordinates to Cartesian coordinates

We are given the polar coordinates, \((1, \pi / 4)\). Thus, the radius \(r = 1\) and the angle \(\theta = \pi / 4\). Using the relations mentioned in step 1, we can find the Cartesian coordinates. $$x = r \cos{\theta} = 1 \cos{(\pi / 4)} = \frac{\sqrt{2}}{2}$$ $$y = r \sin{\theta} = 1 \sin{(\pi / 4)} = \frac{\sqrt{2}}{2}$$ So, the Cartesian coordinates are \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\).
03

Plot the point in Cartesian coordinates

Now that we have the Cartesian coordinates, we can plot the point on a grid. 1. Start at the origin (0,0). 2. Move to the right by \(\frac{\sqrt{2}}{2}\) units on the x-axis. 3. Move up by \(\frac{\sqrt{2}}{2}\) units on the y-axis. 4. Place the point at the location \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\). The point with polar coordinates \((1, \pi / 4)\) is now plotted on the grid as Cartesian coordinates \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\).

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Key Concepts

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

Cartesian Coordinates
Cartesian coordinates are one of the most common systems used to describe a point's location in a plane. In this system, every point is specified by two numbers which are usually referred to as
  • x: The horizontal coordinate,
  • y: The vertical coordinate.
The plane itself is divided by two axes – the x-axis (horizontal) and the y-axis (vertical). These axes intersect at a point called the origin, which has the coordinates (0,0).

When you plot a point using Cartesian coordinates, you start from the origin. You move horizontally along the x-axis by the value of x, and then vertically along the y-axis by the value of y. This system was named after René Descartes, a French mathematician, who was crucial in the development of analytic geometry. It's widely used in mathematics, engineering, physics, and even game development to identify locations in a plane.
Coordinate Conversion
Coordinate conversion is a process of translating values from one coordinate system to another. This is particularly important in fields like navigation, computer graphics, and robotics, where different format systems may apply.

For example, when converting from polar to Cartesian coordinates, we utilize certain mathematical formulas. These formulas help translate a point defined by distance and angle (polar coordinates) into a point defined by horizontal and vertical displacement (Cartesian coordinates).

The key relationships for this conversion are:
  • For the x-coordinate: \(x = r \cos{\theta}\)
  • For the y-coordinate: \(y = r \sin{\theta}\)
Here, \(r\) is the radial distance from the origin, and \(\theta\) is the angle from the positive x-axis. Coordinate conversion is not only an essential mathematical skill but also deeply rooted in many practical applications.
Polar to Cartesian Transformation
The transformation from polar to Cartesian coordinates is an important mathematical procedure that bridges two different methods of describing points. It allows us to convert a point expressed as radius and angle to x and y coordinates. In our example, the polar coordinates (1, \(\pi/4\)) are transformed to Cartesian.
  • The radius \(r = 1\) determines how far the point is from the origin.
  • The angle \(\theta = \pi/4\) indicates its direction from the positive x-axis.
Using the relationships for conversion:
  • \(x = r \cos{\theta} = 1 \cos{(\pi/4)} = \frac{\sqrt{2}}{2}\)
  • \(y = r \sin{\theta} = 1 \sin{(\pi/4)} = \frac{\sqrt{2}}{2}\)
These computations yield the Cartesian coordinates \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\). This transformation lets us plot the point on a standard grid, showing how it's critical to translate and visualize data in various forms of geometry.

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Most popular questions from this chapter

Use a calculator in degree mode and assume that air resistance is negligible. A golf ball is hit off the ground at an angle of \(\theta\) degrees with an initial velocity of 100 feet per second. (a) Graph the path of the ball when \(\theta=20^{\circ}, \theta=40^{\circ}\) \(\theta=60^{\circ},\) and \(\theta=80^{\circ}\) (b) For what angle in part (a) does the ball land farthest from where it started? (c) Experiment with different angles, as in parts (a) and (b), and make a conjecture as to which angle results in the ball landing farthest from its starting point.

If \(a>b>0,\) then the eccentricity of the ellipse $$\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1 \quad \text { or } \quad \frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1$$ is the number \(\frac{\sqrt{a^{2}-b^{2}}}{a} .\) Find the eccentricity of the ellipse whose equation is given. $$\frac{x^{2}}{18}+\frac{y^{2}}{25}=1$$

Use a calculator in degree mode and assume that air resistance is negligible. A golf ball is hit off the tee at an angle of \(30^{\circ}\) and lands 300 feet away. What was its initial velocity? [Hint: The ball lands when \(x=300\) and \(y=0 .\) Use this fact and the parametric equations for the ball's path to find two equations in the variables \(t \text { and } v . \text { Solve for } v .]\)

Use the information given in Special Topics 10.3. A and summarized in the endpapers at the beginning of this book to find a parameterization of the conic section whose rectangular equation is given. Confirm your answer by graphing. $$x^{2}+y^{2}-4 x-6 y+9=0$$

Sketch the graph of the equation without using a calculator. $$\theta=-\pi / 3$$

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