Polar coordinates locate a point by its distance from the origin and its angle from the positive -axis. Converting between polar and rectangular form also gives a formula for rotating a point about the origin, which the last problem uses on a tilted ellipse.
Polar coordinates
Definition 16.2.1. In polar coordinates a point is written , where , the radial coordinate, is the distance from the origin, and , the angular coordinate, is the angle the segment from the origin makes with the positive -axis.
Problem 16.2.2. Convert to polar:
(a) (b) (c)
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(a) .
(b) .
(c) . The legs of the triangle are and and the hypotenuse is . Since , it is a –– triangle with the angle at the origin, and the point is in the third quadrant: .
Problem 16.2.3. Convert to rectangular:
(a) (b) (c)
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(a) Drop a perpendicular: a right triangle with hypotenuse , angles ,,, leg along the -axis and leg opposite the angle. By the law of sines,
so and . Thus .
(b) points the same way as . With we get ,, so .
(c) , and since the triangle is ––, it’s the same for both legs, so
Problem 16.2.4. (a) What are the rectangular coordinates of the point with polar coordinates ?
(b) What about , assuming ?
(c) Suppose point is in rectangular coordinates and in polar coordinates, . Find in terms of and . How must we modify our answer if we remove the restriction ? How is related to and ?
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(a) .
(b) The point is times as far out along the same ray, so
Since , we have . Check with from part (c) of the previous problem: and , giving .
(c) If we have and , then , so we can calculate:
As for how is related to ? Well,
Problem 16.2.5. What is the graph of ? Of ? (In polar.)
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is the circle of radius about the origin, and is the line through the origin at angle .
Problem 16.2.6. (a) Graph in polar and use the graph to guess the rectangular form of this equation.
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(a)
A circle with radius , so .
(b) So and is what we have. Recall ,,,. So
Problem 16.2.7. In rectangular form, the general form of an equation whose graph is the circle with center and radius is . Find the polar form of this equation. That is, find the general form of the equation whose graph is the circle with center (in polar) and radius .
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First try: , which expands to
wait, no: (and ). So
So as an equation we have: given centered at , the polar equation centered at that corresponds to our rectangular form is
Rotating points about the origin
Problem 16.2.8. (a) What are the rectangular coordinates of the point that results when is rotated counterclockwise about the origin?
(b) Show that if the point is rotated counterclockwise about the origin, the result has rectangular coordinates .
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(a) , so . The triangle with legs and has a angle at the origin: indeed and . Rotating adds the angles, , so
So rotated counterclockwise is .
(b) We know that and , so rotation (let’s call it to avoid confusion) by about the origin can be given by . It is, however, known that
so we have new coordinates of
But wait! We know , and likewise for , so substituting in we have
as desired.
Remark (Scratch work for the rotation problem). Let be polar coordinates for ; then the rotation result is , i.e. . OHH.
Checking the formula against (a), with and :
Ellipses
The standard form of an ellipse centered at the origin is
where is the -radius and is the -radius. With (as drawn), is the major radius or semi-major axis and is the minor radius or semi-minor axis.
Minor axis: the short side.
Major axis: the long side.
The semi-minor axis is since the minor axis is (the whole axis in the ellipse).
Example 16.2.9.
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So the major axis is , with radius , and is vertical; the minor axis is , with radius , for the horizontal.
A circle fits the same pattern: . Shifting: moves the graph right , and moves it down .
Foci. Say for , so the major axis is horizontal. For any point on the ellipse, the sum of its distances to the two foci is : in the picture, ,, and so on for every point.
The gaps (from the left end to ) and (from to the right end) are equal, since the ellipse is symmetric about its center. Taking the point at the right end of the major axis: its distance to is , and its distance to is the stretch from the left end to (since ), so , which is the whole major axis.
Now take the point at the top, . There , so . With the distance from the center to each focus, the right triangle with legs and and hypotenuse gives
Example 16.2.10.,,, so and .
Say you have as center and ; then ,.
Now a tilted ellipse.
Problem 16.2.11. Let be constants and positive constants with . The graph of an equation of the form
is an ellipse whose major axis has length , and every ellipse in the Cartesian plane with a horizontal major axis is of this form. Find the length of the major axis of the following ellipse:
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Try to factor it like a square: , with and … yikes.
The term means the ellipse is tilted, so rotate it to make the major axis horizontal. Dividing by :
In polar:
so
✗ (Dead end.)
Rotate instead:
Then : the major axis has length .