008 polar coordinates
- 1. LECTURE UNIT 007 Polar Coordinates The coordinates of a point in a plane are its distance from a fixed point and its direction from a fixed line. P (r, θ) Where: r = radius vector or modulus = unique distance from pole to point P (+) if laid at the terminal side of θ. r (-) if laid off the prolongation through 0 from the ide terminal side θ. ls ina θ = amplitude or argument or vectorial angle T erm = angle from the polar axis to the line joining the pole/origin 0 to the θ point P 0 0o (+) if measured counterclockwise. Polar axis (-) if measured clockwise. The coordinate of P are written as P (r, θ) or just (r, θ) Polar Graph Paper o (90 ) π/2 (120o) 2 π/3 π/3 (60o) (150o) 5 π/6 π/6 (30o) 0 (0o) (180o) π 1 2 3 4 (210o) 7π/6 11π/6 (330o) (240o) 4 π/3 5 π/3 (300o) (270o) 3 π/2 Plot: 1. (4, 60o) 2. (-4, 240o) P(4, 60o) 4 240o 60o 0 0 -4 P(-4, 240o) Relations Between Rectangular and Polar Coordinates y P (r, θ) (x, y) Relationship between rectangular coordinates and polar coordinates x = r cosθ r = x2 + y2 r y y y = r sinθ θ = tan-1 x θ 0 x x “Success is a day to day progressive journey towards a pre-determined worthwhile goal.”
- 2. Change the following to rectangular coordinates. Change the following to polar coordinates. 1. (2, 120o) 9. (-3, 2) o 2. (-5, 0 ) 10. (2 3, 2) o 3. (3, -135 ) 11. x2 + y2 - 8x = 0 4. r = 9 sin 2θ 12. 9(x2 - 1)2 + y2 = 9 5. r = cos 2θ 6. r = 4 13. x2 + y2 + 2x - 4y = 0 2 - 3 sin2θ 14. xy = 8 7. r = 4 3 sinθ - 5 cosθ 15. 4x2 - y2 - 3y = 0 8. r = 2 cscθ 16. x2 + 3y2 - 5x = 0 17. 3x + y 3 = 6 Curve Tracing on Polar Coordinates I. The graph of r = k, where k is a constant and greater than 0 (k > 0) is a with center at the pole. 90o (a.) r = 2 0o II. The graph of r = 2a sinθ or r = 2a cosθ is a that passes through the pole. c (a, 90o) c (a, 0o) r=a r=a 90o 90o (a.) r = 4 sinθ (b.) r = 4 cosθ 2a = 4 2a = 4 a=2 a=2 0o c (2, 90o) c (2, 0o) 0o r=a=2 r=a=2 III. The graph of r = a (1 + sinθ) or r = a (1 + cosθ) is a - - . (a.) r = 3 (1 - cosθ) 90o θ 0o 90o 180o 270o r 0 3 6 3 3 180o 0o 6 3 270o (a.) r = 1 + sinθ θ 0o 90o 180o 270o 90o r 1 2 1 0 2 180o 0o 1 1 270o “Success is a journey not a destination - half the fun is getting there.” - Gita Bellin
- 3. IV. The graph of r2 = a2 sin2θ or r2 = a2 cos2θ is a . o o 90 90 45o 180o 0o 180o 0o 270o 270o (a.) r2 = 4 sin2θ (b.) r2 = 3 cos2θ a2 = 4 a2 = 3 a=2 a = 1.7 90o 90o 2 180o 0o 180o 1.7 1.7 0o 2 270o 270o V. The graph of r = a sinnθ or r = a cosnθ is an n . for n is an odd integer, it is a n-leaved rose for n is an even integer, it is a 2n-leaved rose for n = 1, there is one petal and it is circular 90o (a.) r = 3 sin4θ 67.5o 45o The graph is a 8 leaved rose 4θ = 90o 22.5o θ = 22.5o 180o 3 0o 270o (b.) r = 2 cos2θ 90o The graph is a 4 leaved rose 2θ = 90o 45o θ = 45o 2 180o 2 2 0o 2 270o VI. The graph of r = a + b sinθ or r = a + b cosθ is a - - . If |a| < |b| the graph is a limacon with inner loop |a| = |b| the graph is a cardiod |a| > |b| the graph is a limacon without a loop (a.) r = 1 - 2 cosθ (b.) r = 4 + sinθ with inner loop without inner loop 90o 90o 5 1 180o 0o 180o -4 4 0o -3 -1 -1 -3 270o 270o “We need to be polished, going through hard times, suffering times, before the greatness in us is revealed.”
- 4. Intersection of Polar Coordinate Graphs Find the points of intersection and sketch the graph on the same set of axis 18. r = 1 and r = 2 sinθ 19. r = tanθ and r = cotθ 20. r = 2 (1 + cosθ) and r = 3 21. r = 2 sinθ and r = 2 cosθ 22. r = 1 + 2 sinθ and r = cscθ Polar Equations of Conics Focus at the pole and Directrix perpendicular to the polar axis. ep r= + 1 - ecosθ Focus at the pole and Directrix parallel to the polar axis. ep r= + 1 - esinθ Where: is the eccentricity Note: = 1 parabola < 1 ellipse > 1 hyperbola Sketch the graph: 4 26. r = 8 23. r = 1 - cosθ 1 - 2 sinθ 24. r = 4 4 27. r = 1 + sinθ 2 + cosθ 30 28. r = 6 25. r = 5 - 10 cosθ 3 + cosθ PARAMETRIC EQUATIONS If y = g(t) and x = h(t) then this equation are parametric equations where t is the parameter. Change the following to rectangular coordinates and then graph. 29. x = 3t + 1 , y = 2t - 5 30. x = t2 + 2t , y = t - 3 31. x = cost , y = sint 32. Show that x = a cost , y = b sint represents an ellipse “Life is not about running away from mistakes or giving up when we fail. It is about learning how to handle them to finally reach our goals.”
- 5. POLAR COORDINATES Example 1: Change (2, 120o) to rectangular coordinates. x = 2 cos120o y = 2 sin120o x = -1 y= 3 (2, 120o) = (-1, 3) Example 7: Change r = 4 to rectangular coordinates. 3 sinθ - 5 cosθ 4 r= ( 3y ) - ( 5x ) r r Simplifying: 3y - 5x = 4 Example 9: Change (-3, 2) to polar coordinates. 2 r = (-3)2 + (2)2 tanθ = - 3 r = 13 θ = 146.33o (-3, 2) = ( 13, 146.33o) Example 10: Change x2 + y2 - 8x = 0 to polar coordinates. (r cosθ)2 + (r sinθ)2 - 8(r cosθ) = 0 r2 cos2θ + r2 sin2θ - 8r cosθ = 0 r2 (cos2θ + sin2θ) - 8r cosθ = 0 r2 - 8r cosθ = 0 r (r - 8 cosθ) = 0 r = 8 cosθ Find the points of intersection and sketch the graph on the same set of axis Example 18: r = 1 and r = 2 sinθ Substituting; 90o r = 2 sinθ 1 = 2 sinθ 150o 30o 1 sinθ = 2 0o θ = 30o , 150o 210o 330o o If θ = 30 , r = 1 r=2 If θ = 150o, r = 1 Hence; the point of intersection are (1, 30o) and (1, 150o) Example 19: r = tanθ and r = cotθ 90o Substituting; 135o 45o tanθ = cotθ 1 0o 90o 180o 270o tanθ = tan 0 0 tanθ 8 8 180o 0o cot 0 0 8 8 2 tan θ = 1 r = cotθ tanθ = ±1 225o 135o 270o tanθ = +1 tanθ = -1 r = tanθ o o θ = 45 and 225 θ = 135o and 315o Hence; the point of intersection are (1, 45o), (1, 225o), (-1, 135o) and (-1, 315o) “Failure is only the opportunity to begin again more intelligently.” - Henry Ford
- 6. 4 Example 24: r= 1 + sinθ = 1 Parabola θ 0o 90o 180o 270o r 4 2 4 8 4 Example 27: r= 2 + cosθ Dividing the right member by 2, we obtain r= 2 1 + ( 1 ) cosθ 2 =½ < 1 ellipse θ 0o 90o 180o 270o r 1.33 2 4 2 Example 29: Change x = 3t + 1 , y = 2t - 5 to rectangular coordinates and then graph. x = 3t + 1 (eq.1) y = 2t - 5 (eq.2) From eq.1 x-1 t= Substitute to eq.2 (17/2, 0) 3 We have x-1 y=2 3( -5 ) (0, -17/3) Simplifying; 3y - 2x + 17 = 0 (Line) y To graph, simplify to intercept form x + =1 a b x + y (17/2) (-17/3) =1 Example 31: Change x = cost , y = sint to rectangular coordinates and then graph. Squaring both sides of the equation x2 = cos2t , y2 = sin2t Substituting; x2 + y2 = cos2t + sin2t But; cos2t + sin2t = 1 Hence; x 2 + y2 = 1 (Circle with center at the origin) “The only man who makes no mistakes is the man who never does anything. Do not be afraid of mistakes providing you do not make the same one twice.” - Roosevelt