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Polar Coordinates & Polar Functions (College Board AP® Precalculus): Flashcards

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FrontThe Polar Coordinate System
Define polar coordinates.

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Cards in this collection (30)

Define polar coordinates.
Polar coordinates are an ordered pair $(r, θ)$ locating a point, where $θ$ is the angle in standard position whose terminal ray includes the point and $r$ is a signed radius value related to its distance from the origin.
The positive $x$ -axis, from which $θ$ is measured, is called the polar axis.
Where does the point with polar coordinates $(- 3, \frac{π}{12})$ lie?
Draw the terminal ray for $\frac{π}{12}$ , then, because $r$ is negative, measure along the opposite ray instead, a distance of $3$ from the origin.
A negative signed radius reverses the direction, so this point ends up down and to the left rather than up and to the right.
True or False?
If $r = 0$ then the point is at the origin, whatever the value of $θ$ .
True.
A signed radius of $0$ means no distance is measured along the ray at all, so every angle produces the same point.
The origin is therefore the one point whose angle is completely unconstrained.
Why can one point have many different polar coordinate representations?
For two separate reasons:
- adding a whole number of full revolutions leaves the point where it is, so $(r, θ)$ and $(r, θ + 2 π k)$ give the same point
- reversing the ray and the sign of the radius together also returns it, so $(r, θ)$ and $(- r, θ + π)$ give the same point
Fill in the two gaps in the formulas for converting polar coordinates to rectangular ones.
$x =______, y =______$
The completed formulas are:
$x = r \cos θ, y = r \sin θ$
They work for every value of $r$ and $θ$ , a negative radius included, which is why no separate rule is needed for points on the opposite ray.
How do you convert rectangular coordinates to polar ones?
Find the radius from $r = \sqrt{x^{2} + y^{2}}$ .
For the angle use $θ = \arctan \frac{y}{x}$ when $x > 0$ , and $θ = \arctan \frac{y}{x} + π$ when $x < 0$ .
Why does that conversion need $π$ adding when $x$ is negative?
Because arctangent only ever returns an angle in $(- \frac{π}{2}, \frac{π}{2})$ , and every one of those terminal rays points into the right half-plane.
A point with a negative $x$ -coordinate lies in the left half-plane, so the ray has to be turned through half a revolution to reach it.
A point has rectangular coordinates $(- 2 \sqrt{3}, 2)$ . Find polar coordinates for it with $r > 0$ and $0 \leq θ < 2 π$ .
The radius is $r = \sqrt{12 + 4} = 4$ .
Since $x < 0$ , the angle is $\arctan (- \frac{1}{\sqrt{3}}) + π = - \frac{π}{6} + π = \frac{5 π}{6}$ , so the polar coordinates are $(4, \frac{5 π}{6})$ .
Define complex number.
A complex number is a number of the form $a + b i$ , where $a$ and $b$ are real numbers and $i$ is the imaginary unit, defined by $i^{2} = - 1$ .
The real number $a$ is called its real part and the real number $b$ its imaginary part.
True or False?
Every real number is also a complex number.
True.
Any real number $a$ can be written as $a + 0 i$ , which is of the required form with an imaginary part of $0$ .
The real numbers are therefore a subset of the complex numbers rather than a separate kind of thing.
How is a complex number represented as a point?
In the complex plane, whose horizontal axis is the real axis and whose vertical axis is the imaginary axis.
The number $a + b i$ is the point with rectangular coordinates $(a, b)$ , so $4 + 2 i$ sits at $(4, 2)$ .
Where in the complex plane do the purely real and purely imaginary numbers lie?
A purely real number such as $2 = 2 + 0 i$ has imaginary part $0$ , so it lies on the real axis.
A purely imaginary number such as $i = 0 + i$ has real part $0$ , so it lies on the imaginary axis.
How do you convert $a + b i$ to polar form?
Treat $(a, b)$ exactly as a pair of rectangular coordinates and convert them to polar coordinates $(r, θ)$ in the usual way.
The only difference is how the answer is written, as $(r \cos θ) + i (r \sin θ)$ rather than as an ordered pair.
Fill in the two gaps in the polar form of the complex number with polar coordinates $(4, \frac{π}{3})$ .
$(______\cos \frac{π}{3}) + i (______\sin \frac{π}{3})$
The completed polar form is:
$(4 \cos \frac{π}{3}) + i (4 \sin \frac{π}{3})$
The radius multiplies both the cosine and the sine, and leaving it out of one of them is the commonest error here.
Evaluating gives $2 + 2 \sqrt{3} i$ back in $a + b i$ form.
Convert $- \sqrt{3} + i$ to polar form.
Here $a = - \sqrt{3}$ and $b = 1$ , so $r = \sqrt{3 + 1} = 2$ , and since $a < 0$ the angle is $\arctan (- \frac{1}{\sqrt{3}}) + π = \frac{5 π}{6}$ .
The polar form is therefore $(2 \cos \frac{5 π}{6}) + i (2 \sin \frac{5 π}{6})$ .
What is a polar function, and what does its graph consist of?
A polar function has the form $r = f (θ)$ , taking an angle as its input and returning a signed radius as its output.
Its graph is every point whose polar coordinates $(f (θ), θ)$ satisfy the equation, traced out as $θ$ runs across the domain.
How do you sketch a polar graph from its equation?
Make a small table of values, working out $r$ at several angles spread across the domain, then plot each point and join them with a smooth curve.
For $r = 3 + 2 \cos θ$ that gives a radius of $5$ at $θ = 0$ and of $1$ at $θ = π$ .
A polar function is given by $r = 2 + 2 \cos θ$ . Fill in the two gaps in its table of values.
At $θ = \frac{π}{2}$ the radius is $______$ and at $θ = π$ the radius is $______$ instead.
The completed values are:
At $θ = \frac{π}{2}$ the radius is $2$ and at $θ = π$ the radius is $0$ instead.
They come from $\cos \frac{π}{2} = 0$ and $\cos π = - 1$ substituted into the equation.
At an angle where $f (θ) = 0$ what happens to the graph of a polar function?
The curve passes through the origin, because a signed radius of zero places the point there whatever the angle.
Those angles are worth finding early, since they are where the curve can change which side of the origin it is drawn on.
What does a negative output value do to a polar graph?
It puts the point on the ray opposite the terminal ray, at a distance of $|r|$ from the origin.
So that part of the curve is drawn in the opposite direction from the angle that produced it, as happens with $r = 2 + 3 \cos θ$ for angles pointing left.
True or False?
A polar graph with an inner loop must come from a function that takes negative values.
True.
An inner loop appears exactly where $f (θ) < 0$ , since those points are drawn on the opposite ray and fold back in towards the origin.
Spotting a loop therefore tells you straight away that the output changes sign somewhere in the domain.
Can the output of a polar function change while the distance from the origin stays the same?
Yes, because that distance is $|f (θ)|$ rather than $f (θ)$ itself.
An output changing from $- 2$ to $2$ leaves the distance at $2$ throughout.
It is much like a change in $y$ from $- 2$ to $2$ leaving the distance from the horizontal axis unchanged.
How is a polar graph affected by restricting its domain?
Only the portion traced out between the endpoints of the restricted interval remains, found by following the points as $θ$ increases across it.
That portion can consist of more than one visible piece, if the curve passes through the origin somewhere inside the interval.
Fill in the two gaps in the rule for when the distance from the origin is increasing.
The distance between the point with polar coordinates $(f (θ), θ)$ and the origin is increasing when $f$ is positive and $______$ , and also when $f$ is negative and $______$ instead.
The completed rule is:
The distance between the point with polar coordinates $(f (θ), θ)$ and the origin is increasing when $f$ is positive and increasing, and also when $f$ is negative and decreasing instead.
In both cases the value of $f$ is moving away from zero, which is what makes the distance grow.
When is the distance from the origin to a point on a polar graph decreasing?
When $f$ is positive and decreasing, and also when $f$ is negative and increasing.
In both of those cases the value of $f$ is moving towards zero, so the point is closing in on the origin.
True or False?
If a polar function $f$ is decreasing, then the point on its graph is moving closer to the origin.
False.
It depends on the sign of $f$ as well as on its direction of change.
If $f$ goes from $- 1$ to $- 3$ then $f$ is decreasing, yet the distance from the origin has grown from $1$ to $3$ .
What does a relative extremum of a polar function correspond to on the graph?
A point that is relatively closest to or farthest from the origin, compared with the points near it.
A relative extremum occurs where $f$ changes from increasing to decreasing, or from decreasing to increasing.
A polar function has a relative maximum at an angle where $f$ is negative. Is that point closest to or farthest from the origin?
Closest.
A relative maximum of a negative $f$ is where $f$ is least negative, so its value sits nearest to zero and the distance $|f (θ)|$ is at its smallest there.
The convenient shortcut, that maxima give farthest points, only holds while $f$ is positive.
The average rate of change of $r$ with respect to $θ$ is found over an interval. What does it measure, and what are its units?
It is the ratio of the change in the radius values to the change in $θ$ , so graphically it gives the rate at which the radius is changing per radian.
Its units are therefore units of $r$ per radian.
A polar function has $f (\frac{π}{4}) = 6.0$ and an average rate of change of $- \frac{6}{π}$ on $[\frac{π}{4}, \frac{π}{2}]$ . Estimate $f (\frac{5 π}{12})$ .
Treat $f$ as changing at that constant rate across the interval, and add the rate multiplied by the change in angle to the known value.
Here $\frac{5 π}{12} - \frac{π}{4} = \frac{π}{6}$ , so the estimate is $6.0 + (- \frac{6}{π}) (\frac{π}{6}) = 5.0$ .

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