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?
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Khan Academy: Polar Coordinates 2
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
r = 4 and
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
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
The angle that the line segment between the point and the origin can be found by
Since this point is in the first quadrant (both the x and y coordinate are positive) the angle must be 45^{o} or
Note: when using
Example A
Transform the polar coordinates
Solution:
Example B
Find the polar coordinates for
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:
And for the angle,
So,
The rectangular point
Recall that when solving for θ, we used
We found
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 wrapup: Equation of a circle x^{2} + y^{2} = k^{2} 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} = k^{2}, square the terms: r^{2} cos^{2} θ + r^{2} sin^{2} θ = k^{2}, factor the r^{2} from both terms on the left: r^{2} (cos^{2} θ + sin^{2} θ) = k^{2} recall the identity: cos^{2} θ + sin^{2} θ = 1 r^{2} = k^{2} 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*}
Answers
1) To change the description of the point to rectangular, first find the xvalue, then the yvalue, as follows:
xcoordinate:
 \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*}
ycoordinate:
 \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:
xcoordinate:
 \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*}
ycoordinate
 \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
 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.
 \begin{align*}(3, 60^o)\end{align*}
 \begin{align*}\left( 10, \frac{\pi}{3} \right)\end{align*}
 \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.
 \begin{align*}(5, 5)\end{align*}
 \begin{align*}(0, 10)\end{align*}
 \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.
 \begin{align*}\theta =\frac{\pi}{10}\end{align*}
 \begin{align*}r = 8\end{align*}
 \begin{align*}r sin \theta = 7\end{align*}
 \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.
 \begin{align*}x^2 + y^2  2x = 0\end{align*}
 \begin{align*}y = \sqrt{3} x\end{align*}
 \begin{align*}y = 5\end{align*}
 \begin{align*}xy = 15\end{align*}