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# Polar and Cartesian Transformation

## Converting between (r, theta) and (x, y).

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Practice Polar and Cartesian Transformation
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Polar and Cartesian Transformation

You will see during this lesson that points can be converted from rectangular form to polar form with a little algebra and trigonometry.

Can the equation of a shape be converted also? How about a circle, for instance?

### Guidance

Polar Form to Rectangular Form

Sometimes a problem will be given with coordinates in polar form but rectangular form may be needed.

To transform the polar point (4,3π4)\begin{align*}\left (4, \frac{3\pi}{4} \right )\end{align*} into rectangular coordinates: first identify (r, θ).

r = 4 and θ=3π4\begin{align*}\theta = \frac{3\pi}{4}\end{align*}.

Second, draw a vertical line from the point to the polar axis (the horizontal axis). The distance from the pole to where the line you just drew intersects the polar axis is the x value, and the length of the line segment from the point to the polar axis is the y value.

These distances can be calculated using trigonometry:

x = r cos θ and y = r sin θ

x=4 cos 3π4\begin{align*}x = 4 \ \mbox{cos} \ \frac{3\pi}{4}\end{align*} and y=4 sin 3π4\begin{align*}y = 4 \ \mbox{sin} \ \frac{3\pi}{4}\end{align*} or x=22  y=22\begin{align*}x = -2 \sqrt{2} \ \ y = 2\sqrt{2}\end{align*}

(4,3π4)\begin{align*}\left (4, \frac{3\pi}{4} \right )\end{align*} in polar coordinates is equivalent to (22, 22)\begin{align*}(-2\sqrt{2}, \ 2\sqrt{2})\end{align*} in rectangular coordinates.

Rectangular Form to Polar Form

Going from rectangular coordinates to polar coordinates is also possible, but it takes a bit more work. Suppose we want to find the polar coordinates of the rectangular point (2, 2). To begin doing this operation, the distance that the point (2, 2) is from the origin (the radius, r) can be found by

r=x2+y2\begin{align*}r = \sqrt{x^2 + y^2}\end{align*}

r=22+22\begin{align*}r = \sqrt{2^2 + 2^2}\end{align*}

r=8=22\begin{align*}r = \sqrt{8} = 2\sqrt{2}\end{align*}

The angle that the line segment between the point and the origin can be found by

tan θ=yx\begin{align*}\mbox{tan} \ \theta = \frac{y}{x}\end{align*}

tan θ=22\begin{align*}\mbox{tan} \ \theta = \frac{2}{2}\end{align*}

tan θ=1\begin{align*}\mbox{tan} \ \theta = 1\end{align*}

θ=tan11\begin{align*}\theta = \mbox{tan}^{-1} 1\end{align*}

θ=π4\begin{align*}\theta = \frac{\pi}{4}\end{align*}

Since this point is in the first quadrant (both the x and y coordinate are positive) the angle must be 45o or π4\begin{align*}\frac{\pi}{4}\end{align*} radians. It is also possible that when tan θ = 1 the angle can be in the third quadrant, or 5π4\begin{align*}\frac{5\pi}{4}\end{align*} radians. But this angle will not satisfy the conditions of the problem, since a third quadrant angle must have both x and y negative.

Note: when using tan θ=yx\begin{align*}\mbox{tan} \ \theta = \frac{y}{x}\end{align*} to find the measure of θ you should consider, at first, the quotient tan θ=yx\begin{align*}\mbox{tan} \ \theta = \left | \frac{y}{x} \right |\end{align*} and find the first quadrant angle that satisfies this condition. This angle will be called the reference angle, denoted θref. Find the actual angle by analyzing which quadrant the angle must be given the signs of x and y.

#### Example A

Transform the polar coordinates (2,11π6)\begin{align*}\left (2, \frac{11\pi}{6} \right )\end{align*} to rectangular form

Solution:

r=2\begin{align*}r = 2\end{align*} and θ=11π6\begin{align*}\theta = \frac{11\pi}{6}\end{align*}

x=r cos θ\begin{align*}x = r \ \mbox{cos} \ \theta\end{align*} and y=r sin θ\begin{align*}y = r \ \mbox{sin} \ \theta\end{align*}

x=2 cos 11π6\begin{align*}x = 2 \ \mbox{cos} \ \frac{11\pi}{6}\end{align*} and y=2 sin 11π6\begin{align*}y = 2 \ \mbox{sin} \ \frac{11\pi}{6}\end{align*} or x=32  y=1\begin{align*}x = 3\sqrt{2} \ \ y = -1\end{align*}

(2,11π6)\begin{align*}\left (2, \frac{11\pi}{6} \right )\end{align*} is equivalent to (32,1)\begin{align*}(3\sqrt{2}, -1)\end{align*} or in decimal form, approximately (4.342,1)\begin{align*}(4.342, -1)\end{align*}.

#### Example B

Find the polar coordinates for (3,33)\begin{align*}(3, -3\sqrt{3})\end{align*}

x=3\begin{align*}x = 3\end{align*} and y=33\begin{align*}y = -3\sqrt{3}\end{align*}

Solution:

Draw a right triangle in standard form. Find the distance the point is from the origin and the angle the line segment that represents this distance makes with the +x axis:

r=32+(33)2\begin{align*}r = \sqrt{3^2 + (-3\sqrt{3})^2}\end{align*}

=9+27\begin{align*}= \sqrt{9 + 27}\end{align*}

=36\begin{align*}= \sqrt{36}\end{align*}

=6\begin{align*}= 6\end{align*}

And for the angle,

tan θref=(33)3\begin{align*}\mbox{tan} \ \theta_{ref} = \left | \frac{(-3\sqrt{3})}{3} \right |\end{align*}

tan θref=3\begin{align*}\mbox{tan} \ \theta_{ref} = \sqrt{3}\end{align*}

θref=tan13\begin{align*}\theta_{ref} = \mbox{tan}^{-1} \sqrt{3}\end{align*}

θref=π3\begin{align*}\theta_{ref} = \frac{\pi}{3}\end{align*}

So, θref=π3\begin{align*}\theta_{ref} = \frac{\pi}{3}\end{align*} and we can look at the signs of x and y -- (+, -) -- to see that θ=5π3\begin{align*}\theta = \frac{5\pi}{3}\end{align*} since it is a 4th quadrant angle.

The rectangular point (3,33)\begin{align*}(3, -3\sqrt{3})\end{align*} is equivalent to the polar point (6,5π3)\begin{align*}\left (6, \frac{5\pi}{3} \right )\end{align*}.

Recall that when solving for θ, we used

tan θ=(33)3\begin{align*}\mbox{tan} \ \theta = \left | \frac{(-3\sqrt{3})}{3} \right |\end{align*} or tan θ=3\begin{align*}\mbox{tan} \ \theta = \sqrt{3}\end{align*}

We found

θ=5π3\begin{align*}\theta = \frac{5\pi}{3}\end{align*}. BUT, θ could also be θ=2π3\begin{align*}\theta = \frac{2\pi}{3}\end{align*}. You must examine the signs of each coordinate to see that the angle must be in the fourth quadrant in rectangular units or between 3π2\begin{align*}\frac{3\pi}{2}\end{align*} and 2π in polar units. Of the two possible angles for θ, only \begin{align*}\frac{5\pi}{3}\end{align*} is valid. Note that when you use tan-1 on a calculator you will always get an answer in the range \begin{align*}- \frac{\pi}{2} \le \theta \le \frac{\pi}{2}\end{align*}.

#### Example C

Convert the following rectangular coordinates to polar coordinates

a. \begin{align*}(3, 3\sqrt{3})\end{align*}

b. \begin{align*}(-2, 2)\end{align*}

Convert the following polar coordinates to rectangular coordinates:

c. \begin{align*}\left (4, \frac{2\pi}{3} \right )\end{align*}

d. \begin{align*}\left (-1, \frac{5\pi}{6} \right )\end{align*}

Solutions:

a. \begin{align*}(6,66^\circ)\end{align*}

b. \begin{align*}(2\sqrt{2}, 225^\circ)\end{align*}

c. \begin{align*}(-2, 2\sqrt{3})\end{align*}

d. \begin{align*}\left ( \frac{\sqrt{3}}{2}, - \frac{1}{2} \right )\end{align*}

Concept question wrap-up:

Equation of a circle

x2 + y2 = k2 is the equation of a circle with a radius of k in rectangular coordinates.

The equation of a circle is extremely simple in polar form. In fact, a circle on a polar graph is analogous to a horizontal line on a rectangular graph!

You can transform this equation to polar form by substituting the polar values for x, y. Recall x = r cos θ and y = r sin θ.

(r cos θ)2 + (r sin θ)2 = k2,

square the terms: r2 cos2 θ + r2 sin2 θ = k2,

factor the r2 from both terms on the left: r2 (cos2 θ + sin2 θ) = k2

recall the identity: cos2 θ + sin2 θ = 1

r2 = k2

Therefore: \begin{align*}r = \pm k\end{align*} is an equation for a circle in polar units.

When r is equal to a constant, the polar graph is a circle.

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### Guided Practice

1) Change \begin{align*}(3, -45^o)\end{align*} to rectangular coordinates.

2) Change \begin{align*}\left(-3, \frac{\pi}{4}\right)\end{align*} to rectangular coordinates.

3) Express the equation in rectangular form: \begin{align*}r = 6 cos \theta\end{align*}

4) Express the equation in rectangular form: \begin{align*}r = 6\end{align*}

1) To change the description of the point to rectangular, first find the x-value, then the y-value, as follows:

x-coordinate:

\begin{align*}x = r \cdot cos\theta\end{align*} : the \begin{align*}x\end{align*} coordinate is found by multiplying \begin{align*}r\end{align*} by the cosine of \begin{align*}\theta\end{align*}
\begin{align*}x = 3 \cdot cos (-45^o)\end{align*} : substitute the given information for \begin{align*}r\end{align*} and \begin{align*}\theta\end{align*}
\begin{align*}x = 2.121\end{align*}

y-coordinate:

\begin{align*}y = r \cdot sin\theta\end{align*} : the \begin{align*}y\end{align*} coordinate is found by multiplying \begin{align*}r\end{align*} by the sine of \begin{align*}\theta\end{align*}
\begin{align*}y = 3 \cdot sin (-45^o)\end{align*} : substitute the given information for \begin{align*}r\end{align*} and \begin{align*}\theta\end{align*}
\begin{align*}y = -2.121\end{align*}

\begin{align*}\therefore (2.121, -2.121)\end{align*} is the location of \begin{align*}(3, -45^o)\end{align*} in rectangular form.

2) To change \begin{align*}\left( -3, \frac{\pi}{4} \right)\end{align*} to rectangular coordinates, use the same process as Q #1:

x-coordinate:

\begin{align*}-3 cos \left( \frac{\pi}{4} \right)\end{align*} : substituting the values from the problem into \begin{align*}x = r \cdot cos\theta\end{align*}
\begin{align*}-3 cos 45\end{align*} : \begin{align*}\frac{\pi}{4} = 45^o\end{align*}
\begin{align*}x = -2.121\end{align*}

y-coordinate

\begin{align*}-3 sin \left( \frac{\pi}{4} \right)\end{align*} : substituting the values from the problem into \begin{align*}y = r \cdot sin\theta\end{align*}
\begin{align*}-3 sin 45\end{align*} : \begin{align*}\frac{\pi}{4} = 45^o\end{align*}
\begin{align*}y = -2.121\end{align*}

\begin{align*}\therefore (-2.121, -2.121)\end{align*} is the location of \begin{align*}\left( -3, \frac{\pi}{4} \right)\end{align*} in rectangular form.

3) To express \begin{align*}r = 6 cos \theta\end{align*} in rectangular form:

\begin{align*}r^2 = 6 r cos \theta\end{align*} : multiply both sides by \begin{align*}r\end{align*}
\begin{align*}x^2 + y^2 = 6x\end{align*} : Using \begin{align*}x^2 + y^2 = r^2\end{align*} and \begin{align*}x = r cos \theta\end{align*}

\begin{align*}\therefore x^2 + y^2 = 6x\end{align*} is the equation in rectangular form.

4) This one is easy:

\begin{align*}r = 6\end{align*} is the polar form of the equation for a circle
\begin{align*}r^2 = 6^2\end{align*} : square both sides
\begin{align*}x^2 + y^2 = 36\end{align*} : Using \begin{align*}x^2 + y^2 = r^2\end{align*} and simplifying

\begin{align*}\therefore x^2 + y^2 = 36\end{align*} is the equation in rectangular form.

### Explore More

1. How is the point with polar coordinates \begin{align*}(5, \pi)\end{align*} represented in rectangular coordinates?

Plot each point below in polar coordinates (r, θ). Then write the rectangular coordinates (x, y) for the point.

1. \begin{align*}(3, 60^o)\end{align*}
2. \begin{align*}\left( -10, \frac{\pi}{3} \right)\end{align*}
3. \begin{align*}\left( 15, {\pi} \right)\end{align*}

The rectangular coordinates (x, y) are given. For each question: a) Find two pairs of polar coordinates (r, θ), one with r > 0 and the other with r < 0. b) Express θ in radians, and round to the nearest hundredth.

1. \begin{align*}(5, -5)\end{align*}
2. \begin{align*}(0, 10)\end{align*}
3. \begin{align*}(-8, 6)\end{align*}

Transform each polar equation to an equation using rectangular coordinates. Identify the graph, and give a rough sketch or description of the sketch.

1. \begin{align*}\theta =\frac{\pi}{10}\end{align*}
2. \begin{align*}r = 8\end{align*}
3. \begin{align*}r sin \theta = 7\end{align*}
4. \begin{align*}r cos \theta = -3\end{align*}

Transform each rectangular equation to an equation using polar coordinates. Identify the graph, and give a rough sketch or description of the sketch.

1. \begin{align*}x^2 + y^2 - 2x = 0\end{align*}
2. \begin{align*}y = \sqrt{3} x\end{align*}
3. \begin{align*}y = -5\end{align*}
4. \begin{align*}xy = 15\end{align*}

### Vocabulary Language: English Spanish

Tangent

Tangent

The tangent of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the side adjacent to the given angle.
cosine

cosine

The cosine of an angle in a right triangle is a value found by dividing the length of the side adjacent the given angle by the length of the hypotenuse.
polar coordinates

polar coordinates

Polar coordinates describe locations on a grid using the polar coordinate system. The location of each point is determined by its distance from the pole and its angle with respect to the polar axis.
polar form

polar form

The polar form of a point or a curve is given in terms of $r$ and $\theta$ and is graphed on the polar plane.

A quadrant is one-fourth of the coordinate plane. The four quadrants are numbered using Roman Numerals I, II, III, and IV, starting in the top-right, and increasing counter-clockwise.

A quadrant is one-fourth of the coordinate plane. The four quadrants are numbered using Roman Numerals I, II, III, and IV, starting in the top-right, and increasing counter-clockwise.
rectangular coordinates

rectangular coordinates

A point is written using rectangular coordinates if it is written in terms of $x$ and $y$ and can be graphed on the Cartesian plane.
rectangular form

rectangular form

The rectangular form of a point or a curve is given in terms of $x$ and $y$ and is graphed on the Cartesian plane.
sine

sine

The sine of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the hypotenuse.