Getting a New Perspective
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1 Section 6.3 Polar Coordinates Getting a New Perspective We have worked etensively in the Cartesian coordinate system, plotting points, graphing equations, and using the properties of the Cartesian plane to investigate functions and solve problems. In this section, we introduce a different coordinate system that will give you a new perspective of the plane. In this system, points do not have unique ordered pairs associated with them and some complicated-looking graphs have very simple equations. Objective #1: Plot points in the polar coordinate system. Solved Problem #1 1a. Plot the point with polar coordinates (3, 315 ). Draw an angle measuring 315 in standard position. Then plot a point 3 units from the origin (pole) along the terminal side of the angle. Pencil Problem #1 1a. Plot the point with polar coordinates (, 45 ). 1b. Plot the point with polar coordinates (, ). 1b. Plot the point with polar coordinates ( 1, ). Draw an angle measuring radians in standard position. Then plot a point units from the origin (pole) along a line in the opposite direction from the terminal side of the angle. Since the terminal side of the angle points to the left, this means that we move units to the right. Copyright 014 Pearson Education Inc. 57
2 Precalculus 5e 1c. Plot the point with polar coordinates ç -1, -. ø Draw an angle measuring - radians in standard position. Then plot a point 1 unit from the origin (pole) along a line in the opposite direction from the terminal side of the angle. Since the terminal side of the angle points down, this means that we move units up. 1c. Plot the point with polar coordinates ç -, -. ø Objective #: Find multiple sets of polar coordinates for a given point. Solved Problem # Pencil Problem # a. Find a representation of ç 5, in which r is 4 ø positive and < < 4. æ 3 a. Find a representation of ç 10, in which r is 4 ø positive and < < 4. Add to the angle and do not change r. æ 8 æ 9 ç 5, = 5, + = 5, + = 5, 4 ø ç ç ç 4 ø 4 4 ø 4 ø 58 Copyright 014 Pearson Education Inc.
3 Section 6.3 b. Find a representation of ç 5, in which r is 4 ø negative and 0 < <. æ 3 b. Find a representation of ç 10, in which r is 4 ø negative and 0 < <. Add to the angle and replace r with r. æ 4 æ 5 ç 5, = - 5, + = - 5, + = - 5, 4 ø ç ç ç 4 ø 4 4 ø 4 ø c. Find a representation of ç 5, in which r is 4 ø positive and < < 0. æ 3 c. Find a representation of ç 10, in which r is 4 ø positive and < < 0. Subtract from the angle and do not change r. æ 8 æ 7 ç 5, = 5, - = 5, - = 5, - 4 ø ç ç ç 4 ø 4 4 ø 4 ø Objective #3: Convert a point from polar to rectangular coordinates. Solved Problem #3 3a. Find the rectangular coordinates of the point with polar coordinates (3, ). = rcos = 3cos = 3( - 1) = - 3 y = rsin = 3sin = 3(0) = 0 The rectangular coordinates are ( 3, 0). Pencil Problem #3 3a. Find the rectangular coordinates of the point with polar coordinates ç,. 3 ø Copyright 014 Pearson Education Inc. 59
4 Precalculus 5e 3b. Find the rectangular coordinates of the point with polar coordinates ç - 10,. 6 ø æ 3 = rcos = - 10cos = ç = - ø æ1 y = rsin = - 10sin = - 10 ç = ø 3b. Find the rectangular coordinates of the point with polar coordinates ç - 4,. ø The rectangular coordinates are (-5 3, - 5). Objective #4: Convert a point from rectangular to polar coordinates. Solved Problem #4 4a. Find polar coordinates of the point with rectangular coordinates (1, - 3). The point (1, - 3) is in quadrant IV. Find r. Pencil Problem #4 4a. Find polar coordinates of the point with rectangular coordinates (, ). r = + y = 1 + (- 3) = 1+ 3 = 4 = Find. Since quadrant IV, - 1 tan 3 = and lies in = - = - = The polar coordinates are æ 5 ç,. 3 ø 60 Copyright 014 Pearson Education Inc.
5 Section 6.3 4b. Find polar coordinates of the point with rectangular coordinates (0, 4). 4b. Find polar coordinates of the point with rectangular coordinates (5, 0). The point (0, 4) lies on the negative y-ais 4 units 3 below the origin. Thus, r = 4 and =. The polar coordinates are æ 3 ç 4,. ø Objective #5: Convert an equation from rectangular to polar coordinates. Solved Problem #5 Pencil Problem #5 5a. Covert 3- y = 6 to a polar equation. 5a. Covert 3+ y = 7 to a polar equation. Replace with rcos and y with rsin and solve for r. 3- y = 6 3rcos - rsin = 6 r(3cos - sin ) = 6 6 r = 3cos - sin Copyright 014 Pearson Education Inc. 61
6 Precalculus 5e 5b. Covert + ( y+ 1) = 1 to a polar equation. 5b. Covert ( - ) + y = 4 to a polar equation. Replace with rcos and y with rsin and solve for r. + ( y+ 1) = 1 ( rcos ) + ( rsin + 1) = 1 r cos + r sin + r sin + 1 = 1 r (cos + sin ) + rsin = 0 r + rsin = 0 rr ( + sin ) = 0 r = 0 or r + sin = 0 r =-sin The graph of r = 0 is the pole. The graph of r =- sin also includes the pole, so it is not necessary to include the equation r = 0. The polar equation is r =- sin. Objective #6: Convert an equation from polar to rectangular coordinates. Solved Problem #6 6a. Convert r = 4 to a rectangular equation. Square each side in anticipation of using = +. r y r = 4 r = 4 + y = 16 The rectangular equation is + y = 16, which we recognize as the equation of a circle centered at the origin with radius 4. Pencil Problem #6 6a. Convert r = 8 to a rectangular equation. 6 Copyright 014 Pearson Education Inc.
7 Section b. Convert = to a rectangular equation. 4 y Take the tangent of each side and use tan =. Then solve for y. 3 tan = tan 4 6b. Convert = to a rectangular equation. y =-1 y =- The rectangular equation is y =-, which we recognize as the equation of a line passing through the origin with slope 1. 6c. Convert r =- sec to a rectangular equation. 6c. Convert r = 4csc to a rectangular equation. Use a reciprocal identity to rewrite the secant in terms of cosine. Multiply each side by cos and replace rcos with. r =-sec - r = cos r cos =- =- The rectangular equation is =-, which we recognize as the equation of a vertical line with -intercept. 6d. Convert r = 10sin to a rectangular equation. 6d. Convert r = 1cos to a rectangular equation. Multiply both sides by r and then replace + y and rsin with y. r = 10sin = 10rsin + y = 10y + y - 10y = 0 + ( y - 10y + 5) = 5 r + ( y- 5) = 5 r with The rectangular equation is + ( y- 5) = 5, which we recognize as the equation of a circle centered at (0, 5) with radius 5. Copyright 014 Pearson Education Inc. 63
8 Precalculus 5e 64 Copyright 014 Pearson Education Inc.
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