To graph the point (r, θ), simply go out r units along the initial ray, then rotate through the angle θ. The point (1, 5π 6

Transcription

1 Polar Coordinates Any point in the plane can be described by the Cartesian coordinates (x, y), where x and y are measured along the corresponding axes. However, this is not the only way to represent points in the plane; in this section we will learn to describe points using the polar coordinate system, which is often a more convenient system for representing points than is the Cartesian coordinate system. To build up the polar coordinate system, we will fix a point O, the origin, and an initial ray (which generally corresponds to the positive part of the x-axis). We describe a point P in the plane using (1) its directed distance r from the origin, and () the directed angle θ from the initial ray to the segment OP : To graph the point (r, θ), simply go out r units along the initial ray, then rotate through the angle θ. The point (1, 5π 6 ) is graphed below: By directed distance, we mean that the sign on r makes a difference in the value of the point; if r > 0, then we go r units to the right of the origin before rotating through θ, whereas if r < 0, we start at r units to the left of the origin. For example, (, π 3 ) and (, π 3 ) are different points: 1

2 Similarly, when we call θ a directed angle, we mean that the sign on θ affects the value of the point; if θ > 0, then we rotate θ radians counterclockwise from the initial ray, and if θ < 0, we rotate θ radians clockwise from the initial ray. The distinct points (1, π 3 ) and (1, π 3 ) are graphed below: Two coordinates that look different might actually describe the same point. For instance, (, 3π) and (, π) are the same point: If (r, θ) are polar coordinates of the point P, then we can find more polar coordinates of P in

3 one of two ways: 1. Rotate through a full circle, i.e. an extra π:. Go to r, then rotate through θ + π: So (r, θ) can be written as (r, θ + nπ) or ( r, θ + (n + 1)π) for any integer n. Example Find all the polar coordinates of P = (4, π 4 ). The point P is graphed below: Since rotating through π 4 is the same as rotating through 7π 4, P = (4, π 4 ) is equivalent to all points of the form (4, 7π 4 + nπ) or ( 4, 7π 4 + (n + 1)π). 3

4 Graphing basic equations in polar coordinates The equation r = a, where θ can be any value, describes a circle of radius a : Alternatively, the equation θ = α, where r can be any number, describes a line that makes an angle of θ with the initial ray: When we apply more restrictions on r and θ, we can get graphs of more complicated regions. Examples The region described by π θ π, r is graphed below in red: 4

5 The region described by θ = π 3, r 1 is graphed below in red: Relating Polar and Cartesian Coordinates If we wish to convert a point s polar coordinates to Cartesian coordinates, or vice-versa, we can use a basic trigonometry to help us out. Recall that, if (x, y) is the Cartesian coordinate of a point with angle θ from the initial ray, and if x + y = r, then sin θ = y r and cos θ = x r : 5

6 So if P has polar coordinates (r, θ), then we can rewrite the coordinates using the conversions x = r cos θ and y = r sin θ. Alternatively, if we have Cartesian coordinates (x, y), then we can determine r and θ using the formulas x + y = r and tan θ = y x. Examples The equation r sin θ = 1 can be rewritten as y = 1, which describes a horizontal line in the plane. The equation r cos θ sin θ = 5 can be rewritten by noting that r cos θ sin θ = r cos θr sin θ = xy. So r cos θ sin θ = 5 is equivalent to xy = 5, or y = 5 x. Let s rewrite r = 3 sin θ using Cartesian coordinates. Since r = x + y, we can solve for r, r = x + y. Since y = r sin θ, we know that sin θ = y r = y x + y. So r = 3 sin θ 6

7 can be rewritten as x + y = 3y x + y, or or x + y = 3y. We can rewrite so that x + y 3y = 0; completing the square, we have x + y 3y = 9 4, (x 0) + (y 3 ) = 9 4. Recall that the equation of a circle centered at (a, b) with radius ρ is (x a) + (y b) = ρ ; so the equation above is that of a circle centered at (0, 3 ) of radius 3. Graphing Polar Equations There are three possibilities for graphing a polar equation: 1. Return the equation to Cartesian coordinates and graph the resulting equation. Make a t-table of θ and r values and graph the equation from the t-table 3. Graph the equation in the rθ plane, then use this as a chart to graph in the xy plane. If the equation can be easily turned into one in Cartesian coordinates whose graph is recognizable, then it may be best to use the first method to graph the equation. For instance, we earlier found that r = 3 sin θ is equivalent to x + (y 3 ) = 9 4 ; since this second equation is simply the graph of a circle of radius 3 centered at (0, 3 ), it is quite simple to draw the curve. However this technique will not always be practicable, so we should be comfortable with the other techniques. Let s use the second technique to graph r = 1 + sin θ. We first note that r = 1 + sin θ is not symmetric about the x-axis: for if (r, θ) is on the curve, then 1 + sin( θ) = 1 sin θ r, so (r, θ) is not necessarily on the curve. However, the curve is symmetric about the y-axis, for if (r, θ) is on the curve, then 1 + sin(π θ) = 1 + sin π cos θ cos π sin θ = 1 + ( sin θ) = 1 + sin θ = r, so (r, π θ) is on the curve. This information means that we will only need to graph the portion of the curve on the right of the y-axis; we can get the rest by simply mirroring the curve across the y-axis. Let s set up a t-table: 7

8 r θ π 6 + π π 3 π 3π 0 3 5π 3 7π π 6 To simplify matters a bit, we will use the approximations 1.4 and Then the table becomes r θ 1 0 π π π 3 π 0 3π.15 5π 3.3 7π π 6 We can now graph the curve based on the table: The other possible graphing technique is to think of both r and θ as spatial coordinates, even though we know that θ is actually related to a coordinate s angle. We graph the curve in the rθ plane (again thinking of θ as a spatial coordinate) then use this graph as a chart to return to the xy plane. Let s use this technique to graph r = cos(θ). We start by drawing a graph of the function in the rθ plane: 8

9 Now we can use the curve to plot the graph in the xy plane. For instance, we read from the graph that as θ increases from 0 to π 4, r decreases from 1 to 0: Then as θ increases from π 4 to π, r decreases from 0 to 1: Continuing in this way, we get the graph of the entire function: 9

10 Rates of Change in Polar Coordinates When an equation for a curve is given in terms of r and θ, the equation may not describe a function; for instance, in the last section we saw that r = 3 sin θ describes a circle, which fails the vertical line test. However, we can still determine the rate at which one variable changes with respect to the other; in particular, if we make suitable restrictions on the values of θ, we can think of r as a function of θ, r = f(θ); then we can calculate f (θ) to determine the rate of change of r with respect to θ. However, if we wished to draw a tangent line to the curve, we would need to know the rate of change of y with respect to x (that is, dy dx ) for the same curve, even though the curve s equation was originally written in polar coordinates. We can determine dy dx using the conversion formulas x = r cos θ and y = r sin θ; in addition, since let s calculate dy and dx : dy dx = dy dx, d y = d (r sin θ) = d (f(θ) sin θ) = f (θ) sin θ + f(θ) cos θ by the product rule. Similarly, d x = d (r cos θ) = d (f(θ) cos θ) = f (θ) cos θ f(θ) sin θ. So dy dx = f (θ) sin θ + f(θ) cos θ f (θ) cos θ f(θ) sin θ. Given r = 3 sin θ, find the slope of the line tangent to the curve at (3, π ). In the previous section, we found that the equation r = 3 sin θ describes the circle centered at (0, 3 ) with radius 3 : 10

11 Based on the graph, we would expect to find that the slope of the tangent line is 0. Let s check using the formulas from the previous section. We can think of r as r = f(θ) = 3 sin θ. Then f (θ) = 3 cos θ; so dy 3 cos θ sin θ + 3 sin θ cos θ = dx 3 cos θ cos θ 3 sin θ sin θ 3 cos θ sin θ + 3 cos θ sin θ = = 3 cos θ 3 sin θ cos θ sin θ cos θ sin θ. Then the value of dy dx at (r, θ) = (3, π ) is cos π sin π cos π sin π = 0. Determining Symmetries In preparation for graphing polar equations, we will learn to determine the symmetries of a polar curve algebraically. A curve can be symmetric about the x-axis, the y-axis, the origin, or a combination of the three. The cases are shown below. The first curve is symmetric about the x-axis; if we folded the paper along that axis, the two halves of the curve would meet exactly: 11

12 If the curve is symmetric about the x-axis, and if (r, θ) is a point on the curve, then (r, θ) must be as well. The curve below is symmetric about the y-axis: If the curve is symmetric about the y-axis, and if (r, θ) is a point on the curve, then (r, π θ) or ( r, θ) must be as well. Finally, a curve is symmetric about the origin if we can twist the graph 180 counterclockwise and have the curve aligned with itself: If the curve is symmetric about the origin and if (r, θ) is a point on the curve, then (r, π + θ) 1

13 or ( r, θ) must be on the curve as well. Examples Determine the symmetries of r = cos(θ). If (r, θ) is on the curve, then cos( θ) = cos(θ) so (r, θ) is also on the curve. This means that the curve is symmetric about the x-axis. To determine if it is symmetric about the y-axis, we check cos((π θ)): = r, cos((π θ)) = cos(π θ) = cos(π) cos(θ) + sin(π) sin(θ) = cos(θ) = r. So if (r, θ) is a point on the graph of r = cos(θ), so is (r, π θ); thus the curve is also symmetric about the y-axis. Finally, we check for symmetry about the origin: cos((π + θ)) = cos(π + θ) = cos(π) cos(θ) sin(π) sin(θ) = cos(θ) = r. Again, if (r, θ) is on the graph, then so is (r, π + θ); so the curve is also symmetric about the origin. 13