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t6 polar coordinates
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- 2. Polar Coordinates The locationof a point P in the plane may be given by the following two numbers: P x y (r, ) O
- 3. Polar Coordinates r =the distance between P and the origin O(0, 0) The location of a point P in the plane may be given by the following two numbers: P x (r, ) r O y
- 4. Polar Coordinates r =the distance between P and the origin O(0, 0) = a signed angle between the positive x–axis and the direction to P, The location of a point P in the plane may be given by the following two numbers: P x (r, ) r O y
- 5. Polar Coordinates r =the distance between P and the origin O(0, 0) = a signed angle between the positive x–axis and the direction to P, specifically, is + for counter clockwise measurements and is – for clockwise measurements. The location of a point P in the plane may be given by the following two numbers: P x (r, ) r O y
- 6. Polar Coordinates r =the distance between P and the origin O(0, 0) = a signed angle between the positive x–axis and the direction to P, specifically, is + for counter clockwise measurements and is – for clockwise measurements. The location of a point P in the plane may be given by the following two numbers: The ordered pair (r, ) is a polar coordinate of P. P x (r, ) r O y
- 7. Polar Coordinates r =the distance between P and the origin O(0, 0) = a signed angle between the positive x–axis and the direction to P, specifically, is + for counter clockwise measurements and is – for clockwise measurements. The location of a point P in the plane may be given by the following two numbers: The ordered pair (r, ) is a polar coordinate of P. P x (r, ) r The ordered pairs (r, ±2nπ ) with n = 0,1, 2, 3… give the same geometric information hence lead to the same location P(r, ). O y
- 8. Polar Coordinates r =the distance between P and the origin O(0, 0) = a signed angle between the positive x–axis and the direction to P, specifically, is + for counter clockwise measurements and is – for clockwise measurements. The location of a point P in the plane may be given by the following two numbers: The ordered pair (r, ) is a polar coordinate of P. P x (r, ) r The ordered pairs (r, ±2nπ ) with n = 0,1, 2, 3… give the same geometric information hence lead to the same location P(r, ). We also use signed distance, i.e. with negative values of r which means we are to step backward for a distance of lrl. O y
- 9. If needed, wewrite (a, b)P for a polar coordinate ordered pair, and (a, b)R for the rectangular coordinate ordered pair. Polar Coordinates
- 10. Polar Coordinates Conversion Rules Ifneeded, we write (a, b)P for a polar coordinate ordered pair, and (a, b)R for the rectangular coordinate ordered pair.
- 11. Polar Coordinates Conversion Rules Let(x, y)R and (r, )P be the rectangular and polar coordinates of the same point P, then P x y (r, )R = (x, y)P r O The rectangular and polar coordinates relations x = y = r = If needed, we write (a, b)P for a polar coordinate ordered pair, and (a, b)R for the rectangular coordinate ordered pair.
- 12. Polar Coordinates Conversion Rules Let(x, y)R and (r, )P be the rectangular and polar coordinates of the same point P, then P x y (r, )R = (x, y)P r O x = r*cos() The rectangular and polar coordinates relations x = r*cos() y = r*sin() y = r*sin() r = If needed, we write (a, b)P for a polar coordinate ordered pair, and (a, b)R for the rectangular coordinate ordered pair.
- 13. Polar Coordinates Conversion Rules Let(x, y)R and (r, )P be the rectangular and polar coordinates of the same point P, then P x y (r, )R = (x, y)P r O x = r*cos() y = r*sin() The rectangular and polar coordinates relations x = r*cos() y = r*sin() r = √ x2 + y2 If needed, we write (a, b)P for a polar coordinate ordered pair, and (a, b)R for the rectangular coordinate ordered pair.
- 14. Polar Coordinates Conversion Rules Let(x, y)R and (r, )P be the rectangular and polar coordinates of the same point P, then P x y (r, )R = (x, y)P r O x = r*cos() y = r*sin() The rectangular and polar coordinates relations x = r*cos() y = r*sin() r = √ x2 + y2 For we have tan() = cos() = If needed, we write (a, b)P for a polar coordinate ordered pair, and (a, b)R for the rectangular coordinate ordered pair.
- 15. Polar Coordinates Conversion Rules Let(x, y)R and (r, )P be the rectangular and polar coordinates of the same point P, then P x y (r, )R = (x, y)P r x = r*cos() y = r*sin() The rectangular and polar coordinates relations x = r*cos() y = r*sin() r = √ x2 + y2 For we have tan() = y/x cos() = x/√x2 + y2 If needed, we write (a, b)P for a polar coordinate ordered pair, and (a, b)R for the rectangular coordinate ordered pair.
- 16. Polar Coordinates Conversion Rules Let(x, y)R and (r, )P be the rectangular and polar coordinates of the same point P, then P x y (r, )R = (x, y)P r O x = r*cos() y = r*sin() The rectangular and polar coordinates relations x = r*cos() y = r*sin() r = √ x2 + y2 For we have = tan–1(y/x) = cos–1 (x/√x2 + y2) tan() = y/x cos() = x/√x2 + y2 inverse trig. functions that or using If needed, we write (a, b)P for a polar coordinate ordered pair, and (a, b)R for the rectangular coordinate ordered pair.
- 17. Polar Coordinates Example A.a. Plot the following polar coordinates A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P. Find their corresponding rectangular coordinates.
- 18. Polar Coordinates Example A.a. Plot the following polar coordinates A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P. Find their corresponding rectangular coordinates. x y x = r*cos() y = r*sin() r2 = x2 + y2 tan() = y/x For A(4, 60o)P
- 19. Polar Coordinates Example A.a. Plot the following polar coordinates A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P. Find their corresponding rectangular coordinates. x y 60o 4 x = r*cos() y = r*sin() A(4, 60o)P r2 = x2 + y2 tan() = y/x For A(4, 60o)P
- 20. Polar Coordinates Example A.a. Plot the following polar coordinates A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P. Find their corresponding rectangular coordinates. For A(4, 60o)P x = r*cos() y = r*sin() (x, y)R = (4*cos(60), 4*sin(60)), r2 = x2 + y2 tan() = y/x x y 60o 4 A(4, 60o)P
- 21. Polar Coordinates Example A.a. Plot the following polar coordinates A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P. Find their corresponding rectangular coordinates. For A(4, 60o)P x = r*cos() y = r*sin() (x, y)R = (4*cos(60), 4*sin(60)), = (2, 23) r2 = x2 + y2 tan() = y/x x y 60o 4 A(4, 60o)P
- 22. Polar Coordinates Example A.a. Plot the following polar coordinates A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P. Find their corresponding rectangular coordinates. For A(4, 60o)P x = r*cos() y = r*sin() (x, y)R = (4*cos(60), 4*sin(60)), = (2, 23) for B(5, 0o)P, (x, y) = (5, 0), r2 = x2 + y2 tan() = y/x x y 60o 4 A(4, 60o)P B(5, 0)P
- 23. Polar Coordinates Example A.a. Plot the following polar coordinates A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P. Find their corresponding rectangular coordinates. x y 60o 4 For A(4, 60o)P x = r*cos() y = r*sin() (x, y)R = (4*cos(60), 4*sin(60)), = (2, 23) for B(5, 0o)P, (x, y) = (5, 0), for C and D, A(4, 60o)P B(5, 0)P 4 r2 = x2 + y2 tan() = y/x –45o C
- 24. Polar Coordinates Example A.a. Plot the following polar coordinates A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P. Find their corresponding rectangular coordinates. x y 60o 4 For A(4, 60o)P x = r*cos() y = r*sin() (x, y)R = (4*cos(60), 4*sin(60)), = (2, 23) for B(5, 0o)P, (x, y) = (5, 0), for C and D, A(4, 60o)P B(5, 0)P 4 r2 = x2 + y2 tan() = y/x –45o 3π/4 C
- 25. Polar Coordinates Example A.a. Plot the following polar coordinates A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P. Find their corresponding rectangular coordinates. x y 60o 4 For A(4, 60o)P x = r*cos() y = r*sin() (x, y)R = (4*cos(60), 4*sin(60)), = (2, 23) for B(5, 0o)P, (x, y) = (5, 0), for C and D, A(4, 60o)P B(5, 0)P 4 r2 = x2 + y2 tan() = y/x –45o 3π/4 C&D
- 26. Polar Coordinates Example A.a. Plot the following polar coordinates A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P. Find their corresponding rectangular coordinates. x y 60o 4 For A(4, 60o)P x = r*cos() y = r*sin() (x, y)R = (4*cos(60), 4*sin(60)), = (2, 23) for B(5, 0o)P, (x, y) = (5, 0), for C and D, A(4, 60o)P B(5, 0)P C(4, –45o)P = D(–4, 3π/4 rad)P 4 r2 = x2 + y2 tan() = y/x –45o 3π/4 C&D
- 27. Polar Coordinates Example A.a. Plot the following polar coordinates A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P. Find their corresponding rectangular coordinates. x y 60o 4 For A(4, 60o)P x = r*cos() y = r*sin() (x, y)R = (4*cos(60), 4*sin(60)), = (2, 23) for B(5, 0o)P, (x, y) = (5, 0), for C and D, (x, y)R = (4cos(–45), 4sin(–45)) = (–4cos(3π/4), –4sin(3π/4)) A(4, 60o)P B(5, 0)P C(4, –45o)P = D(–4, 3π/4 rad)P 4 r2 = x2 + y2 tan() = y/x –45o 3π/4 C&D
- 28. Polar Coordinates Example A.a. Plot the following polar coordinates A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P. Find their corresponding rectangular coordinates. x y 60o 4 For A(4, 60o)P x = r*cos() y = r*sin() (x, y)R = (4*cos(60), 4*sin(60)), = (2, 23) for B(5, 0o)P, (x, y) = (5, 0), for C and D, (x, y)R = (4cos(–45), 4sin(–45)) = (–4cos(3π/4), –4sin(3π/4)) = (22, –22) A(4, 60o)P B(5, 0)P C(4, –45o)P = D(–4, 3π/4 rad)P 4 r2 = x2 + y2 tan() = y/x –45o 3π/4 C&D
- 29. Polar Coordinates Example A.a. Plot the following polar coordinates A(4, 60o)P , B(5, 0o)P, C(4, –45o)P, D(–4, 3π/4 rad)P. Find their corresponding rectangular coordinates. x y 60o 4 For A(4, 60o)P x = r*cos() y = r*sin() (x, y)R = (4*cos(60), 4*sin(60)), = (2, 23) for B(5, 0o)P, (x, y) = (5, 0), for C and D, (x, y)R = (4cos(–45), 4sin(–45)) = (–4cos(3π/4), –4sin(3π/4)) = (22, –22) A(4, 60o)P B(5, 0)P C(4, –45o)P = D(–4, 3π/4 rad)P 4 Converting rectangular positions into polar coordinates requires more care. r2 = x2 + y2 tan() = y/x –45o 3π/4 C&D
- 30. Polar Coordinates b. Finda polar coordinate then list all possible polar coordinates for each of the following points (with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R.
- 31. Polar Coordinates b. Finda polar coordinate then list all possible polar coordinates for each of the following points (with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R. x y E(–4, 3)
- 32. Polar Coordinates b. Finda polar coordinate then list all possible polar coordinates for each of the following points (with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R. We have the distance formula r = x2 + y2, x y E(–4, 3)
- 33. Polar Coordinates b. Finda polar coordinate then list all possible polar coordinates for each of the following points (with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R. We have the distance formula r = x2 + y2, hence for E, r = 16 + 9 = 5. x y E(–4, 3) r=5
- 34. Polar Coordinates b. Finda polar coordinate then list all possible polar coordinates for each of the following points (with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R. There is no single formula that would give . We have the distance formula r = x2 + y2, hence for E, r = 16 + 9 = 5. x y E(–4, 3) r=5
- 35. Polar Coordinates b. Finda polar coordinate then list all possible polar coordinates for each of the following points (with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R. There is no single formula that would give . This is because has to be expressed via the inverse trig–functions hence the position of E dictates which inverse function We have the distance formula r = x2 + y2, hence for E, r = 16 + 9 = 5. would be easier to use to extract . x y E(–4, 3) r=5
- 36. Polar Coordinates b. Finda polar coordinate then list all possible polar coordinates for each of the following points (with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R. We have the distance formula r = x2 + y2, hence for E, r = 16 + 9 = 5. would be easier to use to extract . Since E is in the 2nd quadrant, the angle may be recovered by the cosine inverse function (why?). x y E(–4, 3) r=5 There is no single formula that would give . This is because has to be expressed via the inverse trig–functions hence the position of E dictates which inverse function
- 37. Polar Coordinates b. Finda polar coordinate then list all possible polar coordinates for each of the following points (with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R. We have the distance formula r = x2 + y2, hence for E, r = 16 + 9 = 5. x y E(–4, 3) r=5 would be easier to use to extract . Since E is in the 2nd quadrant, the angle may be recovered by the cosine inverse function (why?). So = cos–1(–4/5) ≈ 143o There is no single formula that would give . This is because has to be expressed via the inverse trig–functions hence the position of E dictates which inverse function
- 38. Polar Coordinates b. Finda polar coordinate then list all possible polar coordinates for each of the following points (with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R. We have the distance formula r = x2 + y2, hence for E, r = 16 + 9 = 5. x y E(–4, 3) r=5 would be easier to use to extract . Since E is in the 2nd quadrant, the angle may be recovered by the cosine inverse function (why?). So = cos–1(–4/5) ≈ 143o or that E(–4, 3)R ≈ (5,143o)P There is no single formula that would give . This is because has to be expressed via the inverse trig–functions hence the position of E dictates which inverse function
- 39. Polar Coordinates b. Finda polar coordinate then list all possible polar coordinates for each of the following points (with r > 0): E(–4, 3)R, F(3, –2)R, and G(–3, –1)R. We have the distance formula r = x2 + y2, hence for E, r = 16 + 9 = 5. x y E(–4, 3) r=5 would be easier to use to extract . Since E is in the 2nd quadrant, the angle may be recovered by the cosine inverse function (why?). So = cos–1(–4/5) ≈ 143o or that E(–4, 3)R ≈ (5,143o)P = (5,143o±n*360o)P There is no single formula that would give . This is because has to be expressed via the inverse trig–functions hence the position of E dictates which inverse function
- 40. Polar Coordinates For F(3,–2)R, x y F(3, –2,)
- 41. Polar Coordinates For F(3,–2)R, r = 9 + 4 = √13. x y F(3, –2,) r=√13
- 42. Polar Coordinates For F(3,–2)R, r = 9 + 4 = √13. x y F(3, –2,) Since F is in the 4th quadrant, the angle may be recovered by the sine inverse or the tangent inverse function. r=√13
- 43. Polar Coordinates For F(3,–2)R, r = 9 + 4 = √13. x y F(3, –2,) Since F is in the 4th quadrant, the angle may be recovered by the sine inverse or the tangent inverse function. The tangent inverse has the advantage of obtaining the answer directly from the x and y coordinates. r=√13
- 44. Polar Coordinates For F(3,–2)R, r = 9 + 4 = √13. x y F(3, –2,) Since F is in the 4th quadrant, the angle may be recovered by the sine inverse or the tangent inverse function. The tangent inverse has the advantage of obtaining the answer directly from the x and y coordinates. So = tan–1(–2/3) ≈ –0.588rad and that F(3, –2)R ≈ (√13, –0.588rad)P r=√13
- 45. Polar Coordinates For F(3,–2)R, r = 9 + 4 = √13. x y F(3, –2,) r=√13 Since F is in the 4th quadrant, the angle may be recovered by the sine inverse or the tangent inverse function. The tangent inverse has = (√13, –0.588rad ± 2nπ)P the advantage of obtaining the answer directly from the x and y coordinates. So = tan–1(–2/3) ≈ –0.588rad and that F(3, –2)R ≈ (√13, –0.588rad)P
- 46. Polar Coordinates For F(3,–2)R, r = 9 + 4 = √13. x y F(3, –2,) r=√13 Since F is in the 4th quadrant, the angle may be recovered by the sine inverse or the tangent inverse function. The tangent inverse has = (√13, –0.588rad ± 2nπ)P the advantage of obtaining the answer directly from the x and y coordinates. So = tan–1(–2/3) ≈ –0.588rad and that F(3, –2)R ≈ (√13, –0.588rad)P For G(–3, –1)R, r = 9 + 1 = √10. x y G(–3, –1) r=√10
- 47. Polar Coordinates For F(3,–2)R, r = 9 + 4 = √13. x y F(3, –2,) r=√13 Since F is in the 4th quadrant, the angle may be recovered by the sine inverse or the tangent inverse function. The tangent inverse has = (√13, –0.588rad ± 2nπ)P the advantage of obtaining the answer directly from the x and y coordinates. So = tan–1(–2/3) ≈ –0.588rad and that F(3, –2)R ≈ (√13, –0.588rad)P For G(–3, –1)R, r = 9 + 1 = √10. G is the 3rd quadrant. Hence can’t be obtained directly via the inverse– trig functions. x y G(–3, –1) r=√10
- 48. Polar Coordinates For F(3,–2)R, r = 9 + 4 = √13. x y F(3, –2,) r=√13 Since F is in the 4th quadrant, the angle may be recovered by the sine inverse or the tangent inverse function. The tangent inverse has = (√13, –0.588rad ± 2nπ)P the advantage of obtaining the answer directly from the x and y coordinates. So = tan–1(–2/3) ≈ –0.588rad and that F(3, –2)R ≈ (√13, –0.588rad)P For G(–3, –1)R, r = 9 + 1 = √10. G is the 3rd quadrant. Hence can’t be obtained directly via the inverse– trig functions. We will find the angle A as shown first, then = A + π. x y G(–3, –1) r=√10 A
- 49. Polar Coordinates Again, usingtangent inverse A = tan–1(1/3) ≈ 18.3o x y G(–3, –1) r=√10 A
- 50. Polar Coordinates Again, usingtangent inverse A = tan–1(1/3) ≈ 18.3o so = 180 + 18.3o = 198.3o x y G(–3, –1) r=√10 A
- 51. Polar Coordinates Again, usingtangent inverse A = tan–1(1/3) ≈ 18.3o so = 180 + 18.3o = 198.3o or G ≈ (√10, 198.3o ± n x 360o)P x y G(–3, –1) r=√10 A
- 52.