/*! 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 13 Plot the point given in polar co... [FREE SOLUTION] | 91Ó°ÊÓ

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Plot the point given in polar coordinates and then give three different expressions for the point such that (a) \(r<0\) and \(0 \leq \theta \leq 2 \pi\) (b) \(r>0\) and \(\theta \leq 0\) (c) \(r>0\) and \(\theta \geq 2 \pi\) $$ \left(-3,-\frac{11 \pi}{6}\right) $$

Short Answer

Expert verified
(-3, \(\frac{\pi}{6}\)), (3, -\(\frac{5\pi}{6}\)), (3, \(\frac{7\pi}{6}\)) are valid representations.

Step by step solution

01

Understand the Given Polar Coordinates

The given point is \((-3, -\frac{11\pi}{6})\). In polar coordinates, a point \((r, \theta)\) can be expressed with various values of \(r\) and \(\theta\). The angle \(-\frac{11\pi}{6}\) lies in the fourth quadrant.
02

Plot the Point in Polar Coordinates

Since plotting involves visualizing the location based on radius and angle, \(r = -3\) means you move 3 units in the opposite direction of \(-\frac{11\pi}{6}\). This angle is equivalent to \(\frac{\pi}{6}\) measured clockwise in standard position, or \(11\pi/6\) counterclockwise.
03

Find Equivalent Expression (Condition a)

For condition (a), where \(r < 0\) and \(0 \leq \theta \leq 2\pi\), maintain the negative radius and adjust \(-\frac{11\pi}{6}\) by adding \(2\pi\) to set \(\theta\) within the desired range. Thus, \(\theta = \frac{\pi}{6}\): \((-3, \frac{\pi}{6})\).
04

Find Equivalent Expression (Condition b)

For condition (b), where \(r > 0\) and \(\theta \leq 0\), convert \(r\) to positive by adjusting \(\theta\) by adding \(\pi\) (or 80°) to the original, so it makes a half-turn:\(\theta = -\frac{11\pi}{6} + \pi = -\frac{5\pi}{6}\): \((3, -\frac{5\pi}{6})\).
05

Find Equivalent Expression (Condition c)

For condition (c), where \(r > 0\) and \(\theta \geq 2\pi\), make \(r\) positive and convert \(\theta\) to a value greater than \(2\pi\). Use \(-\frac{11\pi}{6} + 3\pi = \frac{7\pi}{6}\) adjusted again for a full circle:\((3, \frac{7\pi}{6})\).

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Key Concepts

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

Understanding Radians in Polar Coordinates
Radians are a way of measuring angles based on the radius of a circle. A full circle in radians is equal to \(2\pi\), which corresponds to 360 degrees.
This method is integral in calculus and trigonometry, especially when dealing with polar coordinates.
In polar coordinates, angles are often easier to understand in radians because they directly relate to the circle's arc length. The basic structure is:
  • \(\pi\) radians = 180 degrees
  • \(\frac{\pi}{2}\) radians = 90 degrees
  • \(2\pi\) radians = 360 degrees
When converting between degrees and radians, a simple formula can be used:
Multiply the degree measure by \(\frac{\pi}{180}\) to convert to radians, and for the reverse, multiply radians by \(\frac{180}{\pi}\). For polar coordinates, knowing the quadrant in which your angle lies (e.g., through negative or positive \(r\) and \(\theta\)) is also important, as it helps in correctly plotting the point relative to the origin.
Mastering Angle Conversion
Converting angles is crucial in polar coordinates because polar functions often require adjustments to conform to specific rules or quadrants. Sometimes this involves adding or subtracting \(2\pi\), allowing the angle to fit within a single circular rotation.
To convert angles, it helps to visualize them:
  • Positive angles measure counterclockwise from the positive x-axis.
  • Negative angles measure clockwise from the positive x-axis.
For example, an angle of \(-\frac{11\pi}{6}\) radians can be modified by adding \(2\pi\) to give \(\frac{\pi}{6}\), which lies in the first quadrant, making the computation easier to visualize.
Similarly, if a negative radius is involved, flipping it to positive by adjusting the angle by \(\pi\) brings the point a half-circle around the origin, which is often necessary for calculations set within a certain angular framework. These conversions make handling and interpreting polar coordinates much more manageable and precise.
Converting Polar to Cartesian Coordinates
Polar coordinates represent a point's location based on its distance from the origin (radius) and the angle from the positive x-axis.
They can be converted to Cartesian coordinates (x, y) for easier plotting on a standard grid. The conversion formulas are:
  • \(x = r \cos(\theta)\)
  • \(y = r \sin(\theta)\)
For example, given a polar point \((-3, -\frac{11\pi}{6})\), substituting into the formulas becomes:
  • \(x = -3 \cos\left(-\frac{11\pi}{6}\right)\)
  • \(y = -3 \sin\left(-\frac{11\pi}{6}\right)\)
These calculations translate the polar point to its Cartesian counterpart, providing a graphic x-y plot. It's an essential tool for bridging polar and Cartesian systems, enhancing the ability to work across different mathematical frameworks seamlessly.
This is particularly useful in fields like physics and engineering, where understanding these transformations aids in comprehending various phenomena on a broader conceptual level.

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Most popular questions from this chapter

Every polar curve \(r=f(\theta)\) can be translated to a system of parametric equations with parameter \(\theta\) by \(\\{x=r \cos (\theta)=f(\theta) \cos (\theta), y=r \sin (\theta)=f(\theta) \sin (\theta) .\) Convert \(r=6 \cos (2 \theta)\) to a system of parametric equations. Check your answer by graphing \(r=6 \cos (2 \theta)\) by hand using the techniques presented in Section \(11.5\) and then graphing the parametric equations you found using a graphing utility.

We say that two non-zero vectors \(\vec{v}\) and \(\vec{w}\) are parallel if they have same or opposite directions. That is, \(\vec{v} \neq \overrightarrow{0}\) and \(\vec{w} \neq \overrightarrow{0}\) are parallel if either \(\hat{v}=\hat{w}\) or \(\hat{v}=-\hat{w}\). Show that this means \(\vec{v}=k \vec{w}\) for some non- zero scalar \(k\) and that \(k>0\) if the vectors have the same direction and \(k<0\) if they point in opposite directions.

In Exercises \(41-50\), use set-builder notation to describe the polar region. Assume that the region contains its bounding curves. The region inside the cardioid \(r=2-2 \sin (\theta)\) which lies in Quadrants I and IV.

In Exercises \(25-39\), find a parametric description for the given oriented curve. the circle \((x-3)^{2}+(y+1)^{2}=117\), oriented counter-clockwise

The goal of this exercise is to use vectors to describe non-vertical lines in the plane. To that end, consider the line \(y=2 x-4 .\) Let \(\vec{v}_{0}=\langle 0,-4\rangle\) and let \(\vec{s}=\langle 1,2\rangle .\) Let \(t\) be any real number. Show that the vector defined by \(\vec{v}=\vec{v}_{0}+t \vec{s}\), when drawn in standard position, has its terminal point on the line \(y=2 x-4\). (Hint: Show that \(\vec{v}_{0}+t \vec{s}=\langle t, 2 t-4\rangle\) for any real number \(t\).) Now consider the non-vertical line \(y=m x+b\). Repeat the previous analysis with \(\vec{v}_{0}=\langle 0, b\rangle\) and let \(\vec{s}=\langle 1, m\rangle .\) Thus any non-vertical line can be thought of as a collection of terminal points of the vector sum of \(\langle 0, b\rangle\) (the position vector of the \(y\) -intercept) and a scalar multiple of the slope vector \(\vec{s}=\langle 1, m\rangle\).

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