Polar to Rectangular Form

5.1 (CONTINUED) Transforming an equation from Polar to Rectangular Form. Example 8 Transform r = 4sinθ to rectangular coordinates. Since x2 + y2 = r2 multiply both sides by r to get r2 on the left. r2 = 4rsinθ Substitute x2 + y2 for r2 and y for rsinθ x2 + y2 = 4y Put everything on the left to see this is an equation of a circle. x2 + y2 - 4y = 0 Recall that we must complete the square to find out what the radius and center are. We do nothing to the x terms, since there is no “ax” term, but since “by” = -4y, add (-4/2)2 , which is 4, to both sides (x-0)2 + (y2 - 4y +4) = 0 + 4 (x-0)2 + (y - 2)2 = 4 = 22 So on the rectangular coordinate system, this is a circle with center (0,2) and radius = 2. On the polar grid, rectangular coordinate (0,2) translates to (2, π/2). Notice that the original polar form of the equation was r = 4sinθ and the center of the circle in rectangular coordinates is (0,2). Another way to write this equation is r = 2asinθ where a=2 You’ll find that any polar equation in this form will be a circle with center at (0,a). (We’ll talk more about this later.) Example 9 Transforming an equation from Rectangular to Polar Form 4xy = 9 Remember, x = rcosθ and y = rsinθ 4(rcosθ)(rsinθ) = 9 4r2cosθsinθ = 9 When you see sinθ and cosθ multiplied together this should remind you that sin(2θ) = 2sinθcosθ, so rearranging our equation gives: 2r2(2cosθsinθ) = 2r2(2sinθcosθ) = 2r2 sin(2θ) = 9 5.2 INDENTIFYING AND GRAPHING POLAR EQUATIONS Example 1 Identify and graph r = 3. There is no θ in this equation, which means θ can be anything. This is the graph of all the points at a distance r = 3 from the pole. The set of all points equidistant from a center is a circle. Recall r2 = x2 + y2, which is a circle if r is constant. In this case, x2 + y2=9 π/2 = -3π/2 90°=-270° 2π/3 =- 4π/3 120°=-240° 3π/4= -5π/4 135°=- 225° π/3 =- 5π/3 60°= -300° π/4 =- 7π/4 45°=-315° 5π/6= - 7π/6 150°=-210 ° π= - π 5 180°=- 180° π/6 = -11π/6 30°=-330° 4 3 2 1 1 2 4 3 5 0 =2π = -2π 0° = 360° 7π/6=- 5π/6 210°=- 150° 11π/6= - π/6 330°= - 30° 7π/4= - π/4 315°= - 45° 5π/4= -3π/4 225°=- 135° 4π/3=- 2π/3 240°= - 120° 5π/3 = -π/3 300°=- 60° 3π/2=- π/2 270°=- 90° By the way, if you graph a circle using your calculator in rectangular points, you have to break it into 2 functions: Y1 = r 2 − x 2 and Y2 = − r 2 − x 2 (Upper semicircle) (Lower semicircle) since x2 + y2 = r2 is not a function, but in polar coordinates, it r=3 is a function, because for each choice of θ there is only one corresponding value or r, namely, r =3, so your calculator will have no problem. Example 2 Identify and graph θ = π/4 This is the graph of all points for any value of r along the ray at θ = π/4, with positive r values along the terminal side and negative r values point in the opposite direction. Converting to rectangular coordinates we get: y/x = tan-1 π/4 = 1 So multiplying both sides by x gives y = x. π/2 = -3π/2 90°=-270° 2π/3 =- 4π/3 120°=-240° 3π/4= -5π/4 135°=- 225° π/3 =- 5π/3 60°= -300° π/4 =- 7π/4 45°=-315° 5π/6= - 7π/6 150°=-210 ° π= - π 5 180°=- 180° π/6 = -11π/6 30°=-330° 4 3 2 1 1 2 4 3 5 0 =2π = -2π 0° = 360° 7π/6=- 5π/6 210°=- 150° 11π/6= - π/6 330°= - 30° 5π/4= -3π/4 225°=- 135° 4π/3=- 2π/3 240°= - 120° 3π/2=- π/2 270°=- 90° 7π/4= - π/4 315°= - 45° 5π/3 = -π/3 300°=- 60° Example 3/ Example 5: Identify and graph the equations rsin θ = 2 and rcos θ = -3 This one is not as easily identifiable when only looking at polar coordinates. But if we convert to rectangular coordinates it is simple. x = rcos θ and y = rsin θ, so substituting in y for rsin θ gives y = 2 and substituting x for rcos θ gives x = -3. On a polar grid, rcos θ = -3 is 3 units on the opposite side of the polar axis. π/2 = -3π/2 90°=-270° 2π/3 =- 4π/3 120°=-240° 3π/4= -5π/4 135°=- 225° π/3 =- 5π/3 60°= -300° π/4 =- 7π/4 45°=-315° rsin θ = 2 5π/6= - 7π/6 150°=-210 ° 7π/6=- 5π/6 210°=- 150° 4 3 2 1 1 rcos θ = -3 π= - π 5 180°=- 180° π/6 = -11π/6 30°=-330° 2 4 3 5 0 =2π = -2π 0° = 360° 11π/6= - π/6 330°= - 30° 5π/4= -3π/4 225°=- 135° 4π/3=- 2π/3 240°= - 120° 7π/4= - π/4 315°= - 45° 5π/3 = -π/3 300°=- 60° 3π/2=- π/2 270°=- 90° For future reference: Any polar equation in the form rsin θ = a is a horizontal line going through the rectangular coordinate point (0,a). And rcos θ = a is a vertical line going through the rectangular coordinate point (a,0). POLAR CIRCLES Equation Description r = 2asin θ Circle; radius a; center at rectangular coordinate (0,a) r = -2asin θ Circle; radius a; center at rectangular coordinate (0,-a) r = 2acos θ Circle; radius a; center at rectangular coordinate (a,0) r = -2acos θ Circle; radius a; center at rectangular coordinate (-a,0) If you have a polar equation in one of these forms, you don’t have to convert it to rectangular coordinates to graph it. SYMMETRY IN POLAR COORDINATES Tests for Symmetry Symmetry with Respect to the Polar Axis (x-axis) In a polar equation, replace θ by –θ and if an equivalent equation results, the graph is symmetric with respect to the polar axis (positive x-axis). [Recall that cos(θ) = cos(-θ)]. Symmetry with Respect to the Line θ = π/2 (y-axis) In a polar equation, replace θ by π – θ. If an equivalent equation results, the graph is symmetric with respect to the line θ = π/2 [Recall that sin(θ) = sin(π-θ)] Symmetry with Respect to the Pole (origin) In a polar equation, replace r by -r. If an equivalent equation results, the graph is symmetric with respect to the pole. Cardioid A cardioid is a curve given by one of the following polar equations: r = a(1 + cos θ) r = a(1 - cos θ) (a, π/2) (a, π/2) (0, π) (2a, 0) (0, 0) (2a, π) (a, 3π/2) (a, 3π/2) r = a(1 + sin θ) r = a(1 - sin θ) (2a, π/2) (0, π/2) (a, π) (a,0) (a, π) (a, 0) (0, 3π/2) (2a, 3π/2) Rose A curve which has the shape of a petalled flower. This curve was named rhodonea by the Italian mathematician Guido Grandi between 1723 and 1728 because it resembles a rose (MacTutor Archive). The polar equation of the rose is r = a sin (nθ) or r = a cos (nθ) If n is odd, the rose is n-petalled. If n is even, the rose is 2n-petalled. Limaçon The limaçon is a polar curve of the form (1) r = b + acosθ also called the limaçon of Pascal. It was first investigated by Dürer, who gave a method for drawing it in Underweysung der Messung (1525). It was rediscovered by Étienne Pascal, father of Blaise Pascal , and named by Gilles-Personne Roberval in 1650 (MacTutor Archive). The word "limaçon" comes from the Latin limax, meaning "snail." If b ≥ 2a, the limaçon is convex. If 2a > b > a, the limaçon is dimpled. If b=a, the limaçon degenerates to a cardioid. If b < a, the limaçon has an inner loop. If , it is a trisectrix (but not the Maclaurin trisectrix). Logarithmic Spiral The logarithmic spiral is a spiral whose polar equation is given by r = aebθ where is the distance from the origin, is the angle from the xaxis, and a and b are arbitrary constants. Copy this page to make your own polar graphs.