/*! 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 38 Convert the point from rectangul... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 38

Convert the point from rectangular coordinates into polar coordinates with \(r \geq 0\) and \(0 \leq \theta<2 \pi\). $$ (3, \sqrt{3}) $$

Short Answer

Expert verified
The polar coordinates are \((2\sqrt{3}, \frac{\pi}{6})\).

Step by step solution

01

Determine the radius, r

To convert a point from rectangular coordinates \((x, y)\) to polar coordinates \((r, \theta)\), we first find the radius \(r\) using the formula \(r = \sqrt{x^2 + y^2}\). For the point \((3, \sqrt{3})\), substitute \(x = 3\) and \(y = \sqrt{3}\): \[r = \sqrt{3^2 + (\sqrt{3})^2} = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}\]
02

Determine the angle, 胃

Next, we find the angle \(\theta\) using the formula \(\theta = \tan^{-1} \left( \frac{y}{x} \right)\). For our point, substitute \(x = 3\) and \(y = \sqrt{3}\):\[\theta = \tan^{-1} \left( \frac{\sqrt{3}}{3} \right) = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6}\]
03

Conclusion

Combine the calculated radius \(r = 2\sqrt{3}\) and angle \(\theta = \frac{\pi}{6}\) to find the polar coordinates. Thus, the polar coordinates of the point \((3, \sqrt{3})\) are \(\left( 2\sqrt{3}, \frac{\pi}{6} \right)\).

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.

Rectangular Coordinates
Rectangular coordinates, often referred to as Cartesian coordinates, describe the position of a point in a two-dimensional plane using two values \(x, y\). These represent the horizontal and vertical distances from a reference point known as the origin, which is located at (0, 0). Every point on this grid can be precisely located using these two values.
  • The \(x\) value indicates how far left or right the point is fr氓n the origin.
  • The \(y\) value indicates how far up or down the point is fr氓n the origin.
In our exercise, the point is represented by the coordinates \(3, \sqrt{3}\), where 3 is the horizontal distance and \(\sqrt{3}\) is the vertical distance.
Radius Calculation
The radius \(r\) in polar coordinates is akin to the length of a line drawn from the origin to the point in question. To find this, you utilize the Pythagorean theorem, applied to the right triangle formed by the point with its \(x\) and \(y\) values.For \((3, \sqrt{3})\), the formula is:\[r = \sqrt{x^2 + y^2}\]Substitute \(x = 3\) and \(y = \sqrt{3}\):
  • Calculate \(3^2 = 9\)
  • Calculate \(\sqrt{3}^2 = 3\)
  • Add them to get \(9 + 3 = 12\)
  • Finally, take the square root to get \(r = 2\sqrt{3}\)
Thus, the radius is \(2\sqrt{3}\), the distance from the origin to the point.
Angle Determination
Determining the angle in polar coordinates involves finding the angle \(\theta\) that the line from the origin to the point makes with the positive \(x\)-axis. This angle is generally specified in radians.To calculate \(\theta\), we use:\[\theta = \tan^{-1} \left( \frac{y}{x} \right)\]Insert \(x = 3\) and \(y = \sqrt{3}\):
  • Calculate \(\frac{\sqrt{3}}{3}\)
  • This simplifies to \(\frac{1}{\sqrt{3}}\)
  • The angle \(\theta\) is therefore \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
  • This simplifies to \(\frac{\pi}{6}\)
Consequently, the angle \(\theta\) is \(\frac{\pi}{6}\) radians.
Theta
In polar coordinates, \(\theta\) represents the angle that determines the direction of the point from the origin. Unlike Cartesian coordinates which rely on vertical and horizontal distances, polar coordinates rely heavily on angles to specify direction and orientation.The range for \(\theta\) in polar coordinates is \(0 \leq \theta < 2\pi\). When calculating \(\theta\), make sure it falls within this interval. This ensures the angle correctly represents its position around the circle centered at the origin.In our example:
  • \(\theta\) is initially calculated as \(\frac{\pi}{6}\)
  • This value is already between 0 and \(2\pi\)
Thus, \(\frac{\pi}{6}\) is the correct angle for our polar coordinate.
Trigonometric Functions
Trigonometric functions are essential tools for transitioning between rectangular and polar coordinates. They relate the angles and sides of triangles, which can be extrapolated to the coordinate systems.Key functions include:
  • Sine (sin), which relates the opposite side and hypotenuse.
  • Cosine (cos), relating adjacent side and hypotenuse.
  • Tangent (tan), which uses the formula \(\tan(\theta) = \frac{y}{x}\), essential for finding angles as \(\theta = \tan^{-1}(\frac{y}{x})\).
In the conversion process:
  • Tangent helps in determining \(\theta\)
  • Sin and cos are used in other contexts, such as converting back from polar to rectangular coordinates
These functions allow for efficient and precise translation between different coordinate systems, reinforcing their versatility in geometry and trigonometry.

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

Eliminate the parameter in the equations for projectile motion to show that the path of the projectile follows the curve $$ y=-\frac{g \sec ^{2}(\theta)}{2 v_{0}^{2}} x^{2}+\tan (\theta) x+s_{0} $$ Use the vertex formula (Equation 2.4) to show the maximum height of the projectile is $$ y=\frac{v_{0}^{2} \sin ^{2}(\theta)}{2 g}+s_{0} \quad \text { when } \quad x=\frac{v_{0}^{2} \sin (2 \theta)}{2 g} $$

For the given vector \(\vec{v}\), find the magnitude \(\|\vec{v}\|\) and an angle \(\theta\) with \(0 \leq \theta<360^{\circ}\) so that \(\vec{v}=\|\vec{v}\|\langle\cos (\theta), \sin (\theta)\rangle\) (See Definition 11.8.) Round approximations to two decimal places. $$ \vec{v}=\langle-4,3\rangle $$

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 circle \(r=5\) which lies in Quadrant III.

Two drunken college students have filled an empty beer keg with rocks and tied ropes to it in order to drag it down the street in the middle of the night. The stronger of the two students pulls with a force of 100 pounds at a heading of \(\mathrm{N} 77^{\circ} \mathrm{E}\) and the other pulls at a heading of \(\mathrm{S} 68^{\circ} \mathrm{E}\). What force should the weaker student apply to his rope so that the keg of rocks heads due east? What resultant force is applied to the keg? Round your answer to the nearest pound.

How many petals does the polar rose \(r=\sin (2 \theta)\) have? What about \(r=\sin (3 \theta), r=\sin (4 \theta)\) and \(r=\sin (5 \theta) ?\) With the help of your classmates, make a conjecture as to how many petals the polar rose \(r=\sin (n \theta)\) has for any natural number \(n\). Replace sine with cosine and repeat the investigation. How many petals does \(r=\cos (n \theta)\) have for each natural number \(n\) ?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

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.