Chapter 9: Problem 4
Plot the point whose polar coordinates are given. Then find the Cartesian coordinates of the point. $$(-\sqrt{2}, 5 \pi / 4) \quad \text { (b) }(1,5 \pi / 2) \quad \text { (c) }(2,-7 \pi / 6)$$
Short Answer
Expert verified
(1, 1), (0, 1), (-\sqrt{3}, 1)
Step by step solution
01
Understanding Polar to Cartesian Conversion
To convert from polar coordinates \(r, \theta\) to Cartesian coordinates \(x, y\), use the formulas: \x = r \cos(\theta)\ and \y = r \sin(\theta)\.
02
Conversion for Point (a) \\(-\sqrt{2}, \frac{5 \pi}{4}\\)
Apply the formulas: \x = -\sqrt{2} \cos\left(\frac{5\pi}{4}\right)\ and \y = -\sqrt{2} \sin\left(\frac{5\pi}{4}\right).\Since \frac{5\pi}{4}\ corresponds to 225 degrees, both cosine and sine are negative: \cos\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}}\ \sin\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}}\Thus, \x = -\sqrt{2}(-\frac{1}{\sqrt{2}}) = 1\ and \y = -\sqrt{2}(-\frac{1}{\sqrt{2}}) = 1\.Cartesian coordinates: \(1, 1\).
03
Conversion for Point (b) \\(1, \frac{5\pi}{2}\\)
Since \frac{5\pi}{2} = \frac{3\pi}{2} + 2\pi\ (a full rotation), it lands back at \(\frac{\pi}{2}\) (90 degrees).Thus, \cos(\frac{5\pi}{2}) = 0\ and \sin(\frac{5\pi}{2}) = 1\.Apply the formulas: \x = 1 \cdot 0 = 0\ and \y = 1 \cdot 1 = 1\.Cartesian coordinates: \(0, 1\).
04
Conversion for Point (c) \\(2, -\frac{7\pi}{6}\\)
Add \(2\pi\ (12\pi/6)\) to find the equivalent positive angle: \-\frac{7\pi}{6} + \frac{12\pi}{6} = \frac{5\pi}{6}\, which corresponds to 150 degrees.Thus, \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}\ and \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\.Apply the formulas: \x = 2(-\frac{\sqrt{3}}{2}) = -\sqrt{3}\ and \y = 2(\frac{1}{2}) = 1\.Cartesian coordinates: \(-\sqrt{3}, 1\).
05
Plotting the Points
For plotting:- For point (a) \(1, 1\): Locate \1\ on both the x and y axes and plot.- For point (b) \(0, 1\): Locate x at zero (on the y-axis) and y at 1 and plot.- For point (c) \(-\sqrt{3}, 1\): Locate \(-\sqrt{3}\) on the x-axis extending slightly more than -1.5 and y at 1, and plot.
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar Coordinates
Polar coordinates are a unique way of locating a point in a plane using a combination of a distance and an angle. This is different from the Cartesian system which uses x and y coordinates.
- The polar coordinate system uses the formula \(r, \theta\), where \(r\) is the radial distance from the origin (usually labeled as O) and \(\theta\) is the angle from the positive x-axis.
- The angle \(\theta\) is usually measured in radians, but degrees can also be used.
- Polar coordinates are particularly useful in scenarios where systems are naturally circular or have rotational symmetry, like when dealing with circular motion.
- Start from the origin of your graph.
- Move outwards by a distance equal to \(r\).
- Rotating counterclockwise by the angle \(\theta\).
Cartesian Coordinates
Cartesian coordinates describe the position of a point as specified by distances along perpendicular axes: usually denoted as x (horizontal) and y (vertical).
- To locate a point, trace a line from the origin along the x-axis, to where the x coordinate lands.
- Then trace from there vertically or horizontally depending on the system used. For standard systems, move vertically to the y coordinate.
- This system is intuitive for plotting directly on graph paper, which usually aligns with these axes to form squares.
Trigonometric Functions
Trigonometric functions play a crucial role in connecting polar and Cartesian coordinates. They help us convert coordinates between these two systems using the relationships of a right triangle formed from the point on the plane.The primary functions are:
- Cosine - generates the x-component, \(x = r\cos(\theta)\).
- Sine - generates the y-component, \(y = r\sin(\theta)\).