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Convert the point from polar coordinates into rectangular coordinates. $$ (6, \arctan (2)) $$

Short Answer

Expert verified
The rectangular coordinates are \(\left(\frac{6\sqrt{5}}{5}, \frac{12\sqrt{5}}{5}\right)\).

Step by step solution

01

Understand Polar Coordinates

Polar coordinates are given in the form \((r, \theta)\), where \(r\) is the radius (distance from the origin) and \(\theta\) is the angle (direction from the origin). Here, \(r = 6\) and \(\theta = \arctan(2)\).
02

Use the Rectangular Coordinate Formulas

Rectangular coordinates \((x, y)\) can be found using the formulas \(x = r \cos \theta\) and \(y = r \sin \theta\). We will apply these formulas to convert the given polar coordinates into rectangular coordinates.
03

Calculate the \(x\)-coordinate

To find \(x\), use \(x = 6 \cos(\arctan(2))\). Since \(\arctan(2)\) means that in a right triangle, the opposite side to the angle is 2 and the adjacent side is 1, the hypotenuse will be \(\sqrt{2^2 + 1^2} = \sqrt{5}\). Thus, \(\cos(\theta) = \frac{1}{\sqrt{5}}\). Substitute to get \(x = 6 \times \frac{1}{\sqrt{5}} = \frac{6}{\sqrt{5}}\).
04

Calculate the \(y\)-coordinate

To find \(y\), use \(y = 6 \sin(\arctan(2))\). With \(\arctan(2)\), where opposite = 2 and adjacent = 1, we have \(\sin(\theta) = \frac{2}{\sqrt{5}}\). Thus, \(y = 6 \times \frac{2}{\sqrt{5}} = \frac{12}{\sqrt{5}}\).
05

Simplify the Rectangular Coordinates

Now we have \(x = \frac{6}{\sqrt{5}}\) and \(y = \frac{12}{\sqrt{5}}\). For simplicity in calculations or presentation, these can be multiplied by \(\sqrt{5}/\sqrt{5}\) to rationalize the denominator if preferred. So, \(x = \frac{6\sqrt{5}}{5}\) and \(y = \frac{12\sqrt{5}}{5}\).

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Key Concepts

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

Polar Coordinates
Polar coordinates are a way of representing points in a plane using two components: the radial distance from the origin and the angular direction from the positive x-axis. These two components are typically denoted as \((r, \theta)\), where \(r\) is the distance from the origin to the point, often called the radius, and \(\theta\) is the angle measured in radians or degrees between the positive x-axis and the line segment connecting the origin to the point.

Unlike rectangular (or Cartesian) coordinates, polar coordinates offer an efficient way to describe locations that are naturally circular or rotational in nature. This makes them particularly useful in fields like engineering and physics, where rotational symmetry occurs regularly. For our problem, the point is represented as \((6, \arctan(2))\), indicating a radius of 6 and an angle whose tangent is 2.
Rectangular Coordinates
Rectangular coordinates, also known as Cartesian coordinates, express a point by its distances from two perpendicular axes intersecting at a point called the origin. These coordinates are given in the form \((x, y)\), where \(x\) and \(y\) represent the horizontal and vertical distances from the origin, respectively.

This coordinate system is extremely intuitive for most everyday applications, as it mirrors familiar dimensions of everyday space. Rectangular coordinates make graphing and calculations straightforward and are the basis for many tools and technologies in both mathematics and physical sciences. In our exercise, the conversion from polar coordinates using methods like trigonometry allows us to translate the point into this familiar \((x, y)\) form.
Coordinate Conversion
To convert from polar to rectangular coordinates, we utilize the relationship between the radius \(r\), the angle \(\theta\), and the coordinate points \((x, y)\). This is done through the formulas:
  • \(x = r \cos \theta\)
  • \(y = r \sin \theta\)
These formulas derive from trigonometric definitions and rest on the geometric fact that any point described in polar coordinates can be projected onto the x and y axes.

In the given exercise, with \(r = 6\) and \(\theta = \arctan(2)\), we determined that \(\cos(\theta)\) corresponds to \(\frac{1}{\sqrt{5}}\) and \(\sin(\theta)\) to \(\frac{2}{\sqrt{5}}\) based on the right triangle known from the tangent function.
Trigonometric Functions
Trigonometric functions are essential in converting between polar and rectangular coordinates because they define the relationships between the angles and sides of right triangles. Some of the primary trigonometric functions include sine, cosine, and tangent.

For the angle \(\theta = \arctan(2)\), the tangent of \(\theta\) is the ratio of the opposite side to the adjacent side in a right triangle, which equals 2. Knowing this, we set up a right triangle where the opposite side is 2 and the adjacent side is 1, leading us to calculate the hypotenuse as \(\sqrt{5}\). This triangle configuration allows us to find:
  • \(\cos(\theta) = \frac{1}{\sqrt{5}}\)
  • \(\sin(\theta) = \frac{2}{\sqrt{5}}\)
These ratios then allow us to use the sine and cosine values to find precise rectangular coordinates, showing how trigonometry aids in coordinate conversion.

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Most popular questions from this chapter

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. $$ \vec{v}=\langle 3,4\rangle $$

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 \(r=5\).

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.

Carl's friend Jason competes in Highland Games Competitions across the country. In one event, the 'hammer throw', he throws a 56 pound weight for distance. If the weight is released 6 feet above the ground at an angle of \(42^{\circ}\) with respect to the horizontal with an initial speed of 33 feet per second, find the parametric equations for the flight of the hammer. (Here, use \(\left.g=32 \frac{f t}{s^{2}} .\right)\) When will the hammer hit the ground? How far away will it hit the ground? Check your answer using a graphing utility.

We know that \(|x+y| \leq|x|+|y|\) for all real numbers \(x\) and \(y\) by the Triangle Inequality established in Exercise 36 in Section 2.2. We can now establish a Triangle Inequality for vectors. In this exercise, we prove that \(\|\vec{u}+\vec{v}\| \leq\|\vec{u}\|+\|\vec{v}\|\) for all pairs of vectors \(\vec{u}\) and \(\vec{v}\). (a) (Step 1) Show that \(\|\vec{u}+\vec{v}\|^{2}=\|\vec{u}\|^{2}+2 \vec{u} \cdot \vec{v}+\|\vec{v}\|^{2}\). 6 (b) (Step 2) Show that \(|\vec{u} \cdot \vec{v}| \leq\|\vec{u}\|\|\vec{v}\| .\) This is the celebrated Cauchy-Schwarz Inequality. (Hint: To show this inequality, start with the fact that \(|\vec{u} \cdot \vec{v}|=|\|\vec{u}\|\|\vec{v}\| \cos (\theta)|\) and use the fact that \(|\cos (\theta)| \leq 1\) for all \(\theta\).) (c) (Step 3) Show that \(\|\vec{u}+\vec{v}\|^{2}=\|\vec{u}\|^{2}+2 \vec{u} \cdot \vec{v}+\|\vec{v}\|^{2} \leq\|\vec{u}\|^{2}+2|\vec{u} \cdot \vec{v}|+\|\vec{v}\|^{2} \leq\|\vec{u}\|^{2}+\) \(2\|\vec{u}\|\|\vec{v}\|+\|\vec{v}\|^{2}=(\|\vec{u}\|+\|\vec{v}\|)^{2}\) (d) (Step 4) Use Step 3 to show that \(\|\vec{u}+\vec{v}\| \leq\|\vec{u}\|+\|\vec{v}\|\) for all pairs of vectors \(\vec{u}\) and \(\vec{v}\). (e) As an added bonus, we can now show that the Triangle Inequality \(|z+w| \leq|z|+|w|\) holds for all complex numbers \(z\) and \(w\) as well. Identify the complex number \(z=a+b i\) with the vector \(u=\langle a, b\rangle\) and identify the complex number \(w=c+d i\) with the vector \(v=\langle c, d\rangle\) and just follow your nose!

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