/*! 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 7 Convert the Cartesian coordinate... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Convert the Cartesian coordinates to polar coordinates. $$(\sqrt{6},-\sqrt{2})$$

Short Answer

Expert verified
The polar coordinates are \((2\sqrt{2}, 11\frac{\pi}{6})\).

Step by step solution

01

Identify Cartesian Coordinates

We start with the Cartesian coordinates provided: \((x, y) = (\sqrt{6}, -\sqrt{2})\). These will be converted into polar coordinates \((r, \theta)\).
02

Calculate Radius (r)

To find the radius \(r\) in polar coordinates, use the formula \(r = \sqrt{x^2 + y^2}\). Substituting the given values, we get \(r = \sqrt{(\sqrt{6})^2 + (-\sqrt{2})^2} = \sqrt{6 + 2} = \sqrt{8} = 2\sqrt{2}\).
03

Calculate Angle (\theta)

To determine the angle \(\theta\), use the arctangent function: \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\). Here, \(\theta = \tan^{-1}\left(\frac{-\sqrt{2}}{\sqrt{6}}\right)\). Simplifying, \( \frac{-\sqrt{2}}{\sqrt{6}} = -\frac{1}{\sqrt{3}}\), thus \(\theta = \tan^{-1}(-\frac{1}{\sqrt{3}}) = -\frac{\pi}{6}\) or \(\theta = 11\frac{\pi}{6}\) if considering it within the range of \([0, 2\pi)\).
04

Solution Presentation

Thus, the polar coordinates are \((r, \theta) = (2\sqrt{2}, 11\frac{\pi}{6})\). The range for \(\theta\) is chosen as \([0, 2\pi)\) for standard polar coordinate representation.

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.

Cartesian Coordinates
When we talk about Cartesian coordinates, we're diving into the world of a two-dimensional plane most students are familiar with. Cartesian coordinates are expressed as an ordered pair
  • Where the first element represents the position along the horizontal axis (x-axis)
  • And the second element represents the position along the vertical axis (y-axis)
In our given exercise, the Cartesian coordinates are \( (x, y) = (\sqrt{6}, -\sqrt{2}) \).This is essentially a way to describe where a point lies on a flat surface or graph.

Thus each point on this plane is defined by these two numerical values, making it easier to locate it precisely on the grid.
Radius Calculation
To transition from Cartesian to polar coordinates, we first need to calculate the radius. The radius, in the context of polar coordinates, refers to the direct distance from the origin to the point. It gives us a clearer sense of how far away the point is in terms of a circle centered at the origin.

To find this radius, we apply the Pythagorean Theorem in a formula: \[ r = \sqrt{x^2 + y^2} \]Consequently, inputting \( x = \sqrt{6} \) and \( y = -\sqrt{2} \),we find: \[ r = \sqrt{(\sqrt{6})^2 + (-\sqrt{2})^2} = \sqrt{6 + 2} = \sqrt{8} = 2\sqrt{2} \]By calculating this, we essentially gain an understanding of the point’s absolute distance from the origin, crucial for converting into a circular or polar perspective.
Angle Calculation
Calculating the angle \( \theta \) is a key step in converting to polar coordinates. The angle tells us the direction from the origin to the point. To find this angle, we use the arctangent function which is ideal for finding angles based on \( y/x \) ratios.

The formula is:\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]Substituting the values from our problem:\[ \theta = \tan^{-1}\left(\frac{-\sqrt{2}}{\sqrt{6}}\right) \]This simplifies to \(-\frac{1}{\sqrt{3}}\),where\[ \theta = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) \]This results in an angle of :\(-\frac{\pi}{6}\)However, because polar coordinates generally express angles in a range from \([0, 2\pi)\),we convert this angle to a positive measure: \(11\frac{\pi}{6}\).
Coordinate Conversion
The ultimate goal of our task is to convert from Cartesian to polar coordinates. This process involves transitioning from x and y positions on a grid to a circular framework defined by radius and angle.
  • The radius \(r\) is calculated as \( 2\sqrt{2} \)
  • The angle \(\theta\) is found to be \(11\frac{\pi}{6}\).
The resulting polar coordinates are then expressed as \((r, \theta) = (2\sqrt{2}, 11\frac{\pi}{6})\).

These coordinates tell us not just the distance from the origin, but also the specific direction relative to a standard position. Polar coordinates are often favored in contexts where direction and radius are more intuitive or relevant than the Cartesian method. This conversion provides a different perspective, spotlighting the versatility and depth of coordinate systems in mathematics.

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

If we think of an electron as a particle, the function $$P(r)=1-\left(2 r^{2}+2 r+1\right) e^{-2 r}$$ is the cumulative distribution function of the distance, \(r\) of the electron in a hydrogen atom from the center of the atom. The distance is measured in Bohr radii. (1 Bohr radius \(=5.29 \times 10^{-11} \mathrm{m} .\) Niels Bohr \((1885-1962)\) was a Danish physicist.) For example, \(P(1)=1-5 e^{-2} \approx 0.32\) means that the electron is within 1 Bohr radius from the center of the atom \(32 \%\) of the time. (a) Find a formula for the density function of this distribution. Sketch the density function and the cumulative distribution function. (b) Find the median distance and the mean distance. Near what value of \(r\) is an electron most likely to be found? (c) The Bohr radius is sometimes called the "radius of the hydrogen atom." Why?

Give an example of: A rod of length \(2 \mathrm{cm},\) whose density \(\delta(x)\) makes the center of mass not at the center of the rod.

Find the area inside the spiral \(r=\theta\) for \(0 \leq \theta \leq 2 \pi\)

Problems are the statement true or false? Give an explanation for your answer. If the average value of the force \(F(x)\) is 7 on the interval \(1 \leq x \leq 4,\) then the work done by the force in moving from \(x=1\) to \(x=4\) is 21

There are very few elementary functions \(y=f(x)\) for which arc length can be computed in elementary terms using the formula $$\int_{a}^{b} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x$$ You have seen some such functions \(f\) in Problems 18,19 and \(67,\) namely, \(f(x)=\sqrt{x^{3}}, f(x)=\sqrt{4-x^{2}},\) and \(f(x)=\frac{1}{2}\left(e^{x}+e^{-x}\right) .\) Try to find some other function that "works," that is, a function whose are length you can find using this formula and antidifferentiation.

See all solutions

Recommended explanations on Math Textbooks

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.