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:

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: 1

2 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 the (distinct) points graphed below: 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:

3 Identifying Polar Coordinates Two coordinates that look different might actually describe the same point. For instance, it is clear that (, π/3) = (, 4π/3): If (r, θ) are the polar coordinates of the point P, then we can find more polar coordinates of P in one of two ways: 3

4 1. Rotate through a full circle, i.e. an extra π:. Go to r, then rotate through θ + π: Thus a point with polar coordinates (r, θ) also has polar coordinates for any integer n. (r, θ + nπ) and ( r, θ + (n + 1)π) 4

5 Example. Find all the polar coordinates of P = (4, π 4 ). 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). Converting Between 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. Consider the graph below, where the point in question has Cartesian coordinates (x, y) and polar coordinates (r, θ): Using the triangle above, we see that, if we know the polar coordinates (r, θ) for the point, then we can find Cartesian coordinates (x, y) for it via the formulas x = r cos θ, y = r sin θ. Alternatively, if we already have Cartesian coordinates (x, y), then we can determine polar coordinates (r, θ) for the same point using the formulas x + y = r, tan θ = y x. Example. The equation r sin θ = 1 can be rewritten as y = 1, which describes a horizontal line in the plane. 5

6 Example. 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. Example. Rewrite the equation r = 3 sin θ in Cartesian coordinates. Using the conversion r = x + y, we can solve for r, r = x + y. 6

7 On the other hand, since y = r sin θ, we know that sin θ = y r = y x + y. Putting all of this together, we see that the equation r = 3 sin θ can be rewritten as x + y = 3y x + y, or x + y = 3y. We can rewrite so that x + y 3y = 0; completing the square for y, we have x + y 3y = 9 4, or (x 0) + (y 3 ) = 9 4. Recall that the equation of a circle centered at (h, k) with radius ρ is (x h) + (y k) = ρ ; so the equation above is that of a circle centered at (0, 3 ) of radius 3. Graphs of Basic Equations in Polar Coordinates For some very simple polar equations, it is quite easy to construct a graph. We look at two simple polar curves here, and will consider more complicated graphs below. The equation r = a, (where there are no restrictions on θ), describes a circle of radius a : 7

8 Alternatively, the equation θ = α, (where there are no restrictions on r) describes a line that makes an angle of θ with the initial ray: 8

9 Graphing Polar Equations Most of the polar equations we run into will not be as simple as the two above, so in the rest of this section, we will consider techniques for graphing these equations. Below are three possible techniques for drawing the graph of 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 ), thus it is quite simple to draw the curve: However, this technique will not always be practical, 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. 9

10 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: r θ π 6 π 4 π 3 π 3π π 3 7π 4 11π 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: 10

11 The other possible graphing technique is to think of both r and θ as Cartesian coordinates (even though we know that θ is actually related to a coordinate s angle). We graph the curve in the rθ plane (as we would graph a Cartesian equation in the xy plane) then use this graph as a chart to return to the xy plane. Example. Draw a graph pf r = cos(θ). We start by drawing a graph of the function in the rθ plane: 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: 11

12 Then as θ increases from π 4 to π, r decreases from 0 to 1: 1

13 As θ increases from π to 3π 4, r increases from 1 to 0: 13

14 Continuing in this way, we get the graph of the entire function: 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 of x; 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 14

15 function of θ, r = f(θ); then we can calculate dr/ = 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, using the differentiation rule for parametric equations, we can write Let s calculate the derivatives dy and dx : dy dx = dy dx d y = d (r sin θ) = dr sin θ + r cos θ, using the product rule to find the derivative in the second line. Similarly,. So d x = d (r cos θ) dy dx = = dr cos θ r sin θ. dy dx = dr dr sin θ + r cos θ. cos θ r sin θ Example. Given r = 3 sin θ, find the slope of the line tangent to the curve at (r, θ) = (3, π ). We saw above that the equation r = 3 sin θ describes the circle centered at (0, 3 ) with radius 3 : 15

16 Based on the graph, we expect the slope of the tangent line to be 0. Let s check using the formulas above: with r = f(θ) = 3 sin θ, we have so dr = 3 cos θ; dr dy dx = sin θ + r cos θ cos θ r sin θ = = = dr 3 cos θ sin θ + 3 sin θ cos θ 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 for dy dx at (r, θ) = (3, π ) is cos π sin π cos π sin π thus the formula confirms that the tangent line to the curve r = 3 cos θ at (3, π/) is horizontal. = 0; 16