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3 polar equations 3 polar equations Presentation Transcript

  • Polar Equations
  • Polar Equations x = r*cos(θ)We start with an example about lines y = r*sin(θ)and we list the conversion rules for r = √ x2 + y2easy references. tan(θ) = y/x
  • Polar Equations x = r*cos(θ)We start with an example about lines y = r*sin(θ)and we list the conversion rules for r = √ x2 + y2easy references. tan(θ) = y/xExample A.a. Convert the linear equation y = 3x + 4 to the polarform r = f(θ).
  • Polar Equations x = r*cos(θ)We start with an example about lines y = r*sin(θ)and we list the conversion rules for r = √ x2 + y2easy references. tan(θ) = y/xExample A.a. Convert the linear equation y = 3x + 4 to the polarform r = f(θ).Replacing y with r*sin(θ) and x with r*cos(θ), we getr*sin(θ) = 3r*cos(θ) + 4
  • Polar Equations x = r*cos(θ)We start with an example about lines y = r*sin(θ)and we list the conversion rules for r = √ x2 + y2easy references. tan(θ) = y/xExample A.a. Convert the linear equation y = 3x + 4 to the polarform r = f(θ).Replacing y with r*sin(θ) and x with r*cos(θ), we getr*sin(θ) = 3r*cos(θ) + 4Solving the r in terms of θ we have that r = 4 1*sin(θ) – 3*cos(θ)
  • Polar Equations x = r*cos(θ)We start with an example about lines y = r*sin(θ)and we list the conversion rules for r = √ x2 + y2easy references. tan(θ) = y/xExample A.a. Convert the linear equation y = 3x + 4 to the polarform r = f(θ).Replacing y with r*sin(θ) and x with r*cos(θ), we getr*sin(θ) = 3r*cos(θ) + 4Solving the r in terms of θ we have that r = 4 1*sin(θ) – 3*cos(θ)Similarly, given a line Ax + By = C its polar form is r = C A*cos(θ) + B*sin(θ)
  • Polar Equations x = r*cos(θ)b. Graph the polar equation y = r*sin(θ) 4 r = √ x2 + y2r = tan(θ) = y/x –3sin(θ) – 2cos(θ)
  • Polar Equations x = r*cos(θ)b. Graph the polar equation y = r*sin(θ) 4 r = √ x2 + y2r = tan(θ) = y/x –3sin(θ) – 2cos(θ)By its form, we know the graph is a straight line.
  • Polar Equations x = r*cos(θ)b. Graph the polar equation y = r*sin(θ) 4 r = √ x2 + y2r = tan(θ) = y/x –3sin(θ) – 2cos(θ)By its form, we know the graph is a straight line.The x intercept is at θ = 0 with r(0) = –2
  • Polar Equations x = r*cos(θ)b. Graph the polar equation y = r*sin(θ) 4 r = √ x2 + y2r = tan(θ) = y/x –3sin(θ) – 2cos(θ)By its form, we know the graph is a straight line.The x intercept is at θ = 0 with r(0) = –2 and at θ = π/2the y intercept is r(π/2) = –4/3.
  • Polar Equations x = r*cos(θ)b. Graph the polar equation y = r*sin(θ) 4 r = √ x2 + y2r = tan(θ) = y/x –3sin(θ) – 2cos(θ)By its form, we know the graph is a straight line.The x intercept is at θ = 0 with r(0) = –2 and at θ = π/2the y intercept is r(π/2) = –4/3. Its graph is as shown. (–2, 0)R (0,–4/3,)R
  • Polar Equations x = r*cos(θ)b. Graph the polar equation y = r*sin(θ) 4 r = √ x2 + y2r = tan(θ) = y/x –3sin(θ) – 2cos(θ)What is the slope of this line?By its form, we know the graph is a straight line.The x intercept is at θ = 0 with r(0) = –2 and at θ = π/2the y intercept is r(π/2) = –4/3. Its graph is as shown. (–2, 0)R (0,–4/3,)R
  • Polar Equations x = r*cos(θ)b. Graph the polar equation y = r*sin(θ) 4 r = √ x2 + y2r = tan(θ) = y/x –3sin(θ) – 2cos(θ)What is the slope of this line?By its form, we know the graph is a straight line.The x intercept is at θ = 0 with r(0) = –2 and at θ = π/2the y intercept is r(π/2) = –4/3. Its graph is as shown.The slope in question is (–2, 0) –4/3 –2 Rm= 2 = 3 (0,–4/3,)R
  • Polar Equations x = r*cos(θ)b. Graph the polar equation y = r*sin(θ) 4 r = √ x2 + y2r = tan(θ) = y/x –3sin(θ) – 2cos(θ)What is the slope of this line?Convert it to an x&y equation.By its form, we know the graph is a straight line.The x intercept is at θ = 0 with r(0) = –2 and at θ = π/2the y intercept is r(π/2) = –4/3. Its graph is as shown.The slope in question is (–2, 0) –4/3 –2 Rm= 2 = 3 (0,–4/3,)R
  • Polar Equations x = r*cos(θ)b. Graph the polar equation y = r*sin(θ) 4 r = √ x2 + y2r = tan(θ) = y/x –3sin(θ) – 2cos(θ)What is the slope of this line?Convert it to an x&y equation.By its form, we know the graph is a straight line.The x intercept is at θ = 0 with r(0) = –2 and at θ = π/2the y intercept is r(π/2) = –4/3. Its graph is as shown.The slope in question is (–2, 0) –4/3 –2 Rm= 2 = 3Clear the denominator of the (0,–4/3,) Rpolar equation, we have that–3r*sin(θ) – 2r*cos(θ) = 4
  • Polar Equations x = r*cos(θ)b. Graph the polar equation y = r*sin(θ) 4 r = √ x2 + y2r = tan(θ) = y/x –3sin(θ) – 2cos(θ)What is the slope of this line?Convert it to an x&y equation.By its form, we know the graph is a straight line.The x intercept is at θ = 0 with r(0) = –2 and at θ = π/2the y intercept is r(π/2) = –4/3. Its graph is as shown.The slope in question is (–2, 0) –4/3 –2 Rm= 2 = 3Clear the denominator of the (0,–4/3,) Rpolar equation, we have that–3r*sin(θ) – 2r*cos(θ) = 4 or –3y – 2x = 4.
  • Polar EquationsPolar Equations of Conic Sections
  • Polar EquationsPolar Equations of Conic SectionsThe cross–sectional boundaries of acone are known as the conic sections. Conic Sections (Wikipedia}
  • Polar EquationsPolar Equations of Conic SectionsThe cross–sectional boundaries of acone are known as the conic sections.They are circles, ellipses, parabolasand hyperbolas. Conic Sections (Wikipedia}
  • Polar EquationsPolar Equations of Conic SectionsThe cross–sectional boundaries of acone are known as the conic sections.They are circles, ellipses, parabolasand hyperbolas. In the standardpositions, i.e. not tilted, they are thegraphs of 2nd degree equations Conic SectionsAx2 + By2 + Cx + Dy = E. (Wikipedia}
  • Polar EquationsPolar Equations of Conic SectionsThe cross–sectional boundaries of acone are known as the conic sections.They are circles, ellipses, parabolasand hyperbolas. In the standardpositions, i.e. not tilted, they are thegraphs of 2nd degree equations Conic SectionsAx2 + By2 + Cx + Dy = E. (Wikipedia}Each non–circular conic section maybe defined geometrically by a constant e > 0, its eccentricity.
  • Polar EquationsPolar Equations of Conic SectionsThe cross–sectional boundaries of acone are known as the conic sections.They are circles, ellipses, parabolasand hyperbolas. In the standardpositions, i.e. not tilted, they are thegraphs of 2nd degree equations Conic SectionsAx2 + By2 + Cx + Dy = E. (Wikipedia}Each non–circular conic section may Dbe defined geometrically by a constant e > 0, its eccentricity. Specifically,a point F and a line D are given, F
  • Polar EquationsPolar Equations of Conic SectionsThe cross–sectional boundaries of acone are known as the conic sections.They are circles, ellipses, parabolasand hyperbolas. In the standardpositions, i.e. not tilted, they are thegraphs of 2nd degree equations Conic SectionsAx2 + By2 + Cx + Dy = E. (Wikipedia}Each non–circular conic section may Dbe defined geometrically by a constant P e > 0, its eccentricity. Specifically,a point F and a line D are given, let P Fbe a point
  • Polar EquationsPolar Equations of Conic SectionsThe cross–sectional boundaries of acone are known as the conic sections.They are circles, ellipses, parabolasand hyperbolas. In the standardpositions, i.e. not tilted, they are thegraphs of 2nd degree equations Conic Sections Ax2 + By2 + Cx + Dy = E. (Wikipedia}Each non–circular conic section may Dbe defined geometrically by a constant P e > 0, its eccentricity. Specifically, a point F and a line D are given, let P Fbe a point and that PF, PD be thedistances from to F and P to Drespectively.
  • Polar EquationsPolar Equations of Conic SectionsThe cross–sectional boundaries of acone are known as the conic sections.They are circles, ellipses, parabolasand hyperbolas. In the standardpositions, i.e. not tilted, they are thegraphs of 2nd degree equations Conic Sections Ax2 + By2 + Cx + Dy = E. (Wikipedia}Each non–circular conic section may Dbe defined geometrically by a constant P e > 0, its eccentricity. Specifically, PF a point F and a line D are given, let P Fbe a point and that PF, PD be thedistances from to F and P to Drespectively.
  • Polar EquationsPolar Equations of Conic SectionsThe cross–sectional boundaries of acone are known as the conic sections.They are circles, ellipses, parabolasand hyperbolas. In the standardpositions, i.e. not tilted, they are thegraphs of 2nd degree equations Conic Sections Ax2 + By2 + Cx + Dy = E. (Wikipedia}Each non–circular conic section may Dbe defined geometrically by a constant P PD e > 0, its eccentricity. Specifically, PF a point F and a line D are given, let P Fbe a point and that PF, PD be thedistances from to F and P to Drespectively. A non–circular conic A conic section with PF = e*PDsection consists of all points P such that where e ≈ ½
  • Polar Equations Directrix PThe point F is called the focus and the PDline D is called the directrix. PF Focus A conic section where PF = e*PD
  • Polar Equations Directrix PThe point F is called the focus and the PDline D is called the directrix. PF FocusFor 0 < e < 1, we have ellipses,for e = 1, we have parabolas, A conic section and for 1 < e, we have hyperbolas. where PF = e*PD
  • Polar Equations Directrix P PDThe point F is called the focus and theline D is called the directrix. PF FocusFor 0 < e < 1, we have ellipses,for e = 1, we have parabolas, A conic section and for 1 < e, we have hyperbolas. where PF = e*PDThe entire graph for e = ½ is shown below.
  • Polar Equations Directrix P PDThe point F is called the focus and theline D is called the directrix. PF FocusFor 0 < e < 1, we have ellipses,for e = 1, we have parabolas, A conic section and for 1 < e, we have hyperbolas. where PF = e*PDThe entire graph for e = ½ is shown below.Note that it has two foci and two directrices. D2 D1 P F1 An ellipse F2 with e=½ i.e. PF = ½*PD
  • Polar Equations Directrix P PDThe point F is called the focus and theline D is called the directrix. PF FocusFor 0 < e < 1, we have ellipses,for e = 1, we have parabolas, A conic section and for 1 < e, we have hyperbolas. where PF = e*PDThe entire graph for e = ½ is shown below.Note that it has two foci and two directrices.Ellipses and hyperbolas have two focus–directrix pairs,i.e. each focus is paired with a directrix. D2 D1 P F1 An ellipse F2 with e=½ i.e. PF = ½*PD
  • Polar Equations Directrix P PDThe point F is called the focus and theline D is called the directrix. PF FocusFor 0 < e < 1, we have ellipses,for e = 1, we have parabolas, A conic section and for 1 < e, we have hyperbolas. where PF = e*PDThe entire graph for e = ½ is shown below.Note that it has two foci and two directrices.Ellipses and hyperbolas have two focus–directrix pairs,i.e. each focus is paired with a directrix. D2 D1 P PD1 PF1 F1 An ellipse F2 with e=½ i.e. PF = ½*PD
  • Polar Equations Directrix P PDThe point F is called the focus and theline D is called the directrix. PF FocusFor 0 < e < 1, we have ellipses,for e = 1, we have parabolas, A conic section and for 1 < e, we have hyperbolas. where PF = e*PDThe entire graph for e = ½ is shown below.Note that it has two foci and two directrices.Ellipses and hyperbolas have two focus–directrix pairs,i.e. each focus is paired with a directrix. D2 D1 PD2 P PF2 PD1 PF1 F1 An ellipse F2 with e=½ i.e. PF = ½*PD
  • Polar Equations Directrix P PDThe point F is called the focus and theline D is called the directrix. PF FocusFor 0 < e < 1, we have ellipses,for e = 1, we have parabolas, A conic section and for 1 < e, we have hyperbolas. where PF = e*PDThe entire graph for e = ½ is shown below.Note that it has two foci and two directrices.Ellipses and hyperbolas have two focus–directrix pairs,i.e. each focus is paired with a directrix.Hence, if P is any point on an ellipsethen PF1 = e*PD1, D 2 P D 1 PD 2 PF2 = e*PD2, PF PF 2 PD 1 1but PF1 ≠ e*PD2 F F 1 An ellipse 2 with e=½and PF2 ≠ e*PD1. i.e. PF = ½*PD
  • Polar Equations (0,0) (0,0) (0,0) (0,0) e = 0.5 e = 0.5, 0.8 e = 0.5, 0.8, 1 e = 0.5, 0.8, 1, 2We give thepolar forms ofthe conicsections belowwithout proofs.
  • Polar EquationsThe Intersection of Polar Equations
  • Polar EquationsThe Intersection of Polar Equations As noted before, unlike therectangular coordinate system where each point isaddressed by a unique ordered pair (x, y), the polarcoordinates (r, θ+2nπ)P give the same location as (r,θ)P where n = ±1, ±2..
  • Polar EquationsThe Intersection of Polar Equations As noted before, unlike therectangular coordinate system where each point isaddressed by a unique ordered pair (x, y), the polarcoordinates (r, θ+2nπ)P give the same location as (r,θ)P where n = ±1, ±2.. One consequence of this is thata polar graph may be represented by infinitely many“different” equations.
  • Polar EquationsThe Intersection of Polar Equations As noted before, unlike therectangular coordinate system where each point isaddressed by a unique ordered pair (x, y), the polarcoordinates (r, θ+2nπ)P give the same location as (r,θ)P where n = ±1, ±2.. One consequence of this is thata polar graph may be represented by infinitely many“different” equations. For example, the polarequations θ = π/4 + nπ give the same graph as θ = π/4,the diagonal line y = x.
  • Polar Equations The Intersection of Polar Equations As noted before, unlike the rectangular coordinate system where each point is addressed by a unique ordered pair (x, y), the polar coordinates (r, θ+2nπ)P give the same location as (r, θ)P where n = ±1, ±2.. One consequence of this is that a polar graph may be represented by infinitely many “different” equations. For example, the polar equations θ = π/4 + nπ give the same graph as θ = π/4,Fact: The graph of = x.θ + 2nπ) = 0 for n = ±1, ±2.. the diagonal line y f(r,is the same as the graph of polar equation f(r, θ) = 0
  • Polar Equations The Intersection of Polar Equations As noted before, unlike the rectangular coordinate system where each point is addressed by a unique ordered pair (x, y), the polar coordinates (r, θ+2nπ)P give the same location as (r, θ)P where n = ±1, ±2.. One consequence of this is that a polar graph may be represented by infinitely many “different” equations. For example, the polar equations θ = π/4 + nπ give the same graph as θ = π/4,Fact: The graph of = x.θ + 2nπ) = 0 for n = ±1, ±2.. the diagonal line y f(r,is the same as the graph of polar equation f(r, θ) = 0Proof: Let (a, b)P be a point on the graph of f(r, θ) = 0so that f(a, b) = 0.
  • Polar Equations The Intersection of Polar Equations As noted before, unlike the rectangular coordinate system where each point is addressed by a unique ordered pair (x, y), the polar coordinates (r, θ+2nπ)P give the same location as (r, θ)P where n = ±1, ±2.. One consequence of this is that a polar graph may be represented by infinitely many “different” equations. For example, the polar equations θ = π/4 + nπ give the same graph as θ = π/4,Fact: The graph of = x.θ + 2nπ) = 0 for n = ±1, ±2.. the diagonal line y f(r,is the same as the graph of polar equation f(r, θ) = 0Proof: Let (a, b)P be a point on the graph of f(r, θ) = 0so that f(a, b) = 0. Let g(r, θ) = f(r, θ + 2nπ) for some n
  • Polar Equations The Intersection of Polar Equations As noted before, unlike the rectangular coordinate system where each point is addressed by a unique ordered pair (x, y), the polar coordinates (r, θ+2nπ)P give the same location as (r, θ)P where n = ±1, ±2.. One consequence of this is that a polar graph may be represented by infinitely many “different” equations. For example, the polar equations θ = π/4 + nπ give the same graph as θ = π/4,Fact: The graph of = x.θ + 2nπ) = 0 for n = ±1, ±2.. the diagonal line y f(r,is the same as the graph of polar equation f(r, θ) = 0Proof: Let (a, b)P be a point on the graph of f(r, θ) = 0so that f(a, b) = 0. Let g(r, θ) = f(r, θ + 2nπ) for some nthen g(a, b – 2nπ) = f(a, b – 2nπ + 2nπ) = f(a, b) = 0.
  • Polar Equations The Intersection of Polar Equations As noted before, unlike the rectangular coordinate system where each point is addressed by a unique ordered pair (x, y), the polar coordinates (r, θ+2nπ)P give the same location as (r, θ)P where n = ±1, ±2.. One consequence of this is that a polar graph may be represented by infinitely many “different” equations. For example, the polar equations θ = π/4 + nπ give the same graph as θ = π/4,Fact: The graph of = x.θ + 2nπ) = 0 for n = ±1, ±2.. the diagonal line y f(r,is the same as the graph of polar equation f(r, θ) = 0Proof: Let (a, b)P be a point on the graph of f(r, θ) = 0so that f(a, b) = 0. Let g(r, θ) = f(r, θ + 2nπ) for some nthen g(a, b – 2nπ) = f(a, b – 2nπ + 2nπ) = f(a, b) = 0.Hence (a, b – 2nπ)P is on the graph of g(r, θ) = 0which is the same as (a, b) .
  • Polar EquationsWhen we solve algebraically for the intersection pointsof graphs of rectangular equations, we solve thesystem of the given equations.
  • Polar EquationsWhen we solve algebraically for the intersection pointsof graphs of rectangular equations, we solve thesystem of the given equations. But to solve for theintersection points of polar equations, we mustconsider other possible polar equations that give risethe same graphs.
  • Polar EquationsWhen we solve algebraically for the intersection pointsof graphs of rectangular equations, we solve thesystem of the given equations. But to solve for theintersection points of polar equations, we mustconsider other possible polar equations that give risethe same graphs.Example B.a. Find the intersection of the polar graphs r = sin(2θ)and θ = π/4. Graph.
  • Polar EquationsWhen we solve algebraically for the intersection pointsof graphs of rectangular equations, we solve thesystem of the given equations. But to solve for theintersection points of polar equations, we mustconsider other possible polar equations that give risethe same graphs.Example B.a. Find the intersection of the polar graphs r = sin(2θ)and θ = π/4. Graph.
  • Polar EquationsWhen we solve algebraically for the intersection pointsof graphs of rectangular equations, we solve thesystem of the given equations. But to solve for theintersection points of polar equations, we mustconsider other possible polar equations that give risethe same graphs.Example B.a. Find the intersection of the polar graphs r = sin(2θ)and θ = π/4. Graph.If we just set θ = π/4, then r = sin(π/2) = 1 we wouldobtain only one solution (1, π/4).
  • Polar EquationsWhen we solve algebraically for the intersection pointsof graphs of rectangular equations, we solve thesystem of the given equations. But to solve for theintersection points of polar equations, we mustconsider other possible polar equations that give risethe same graphs.Example B.a. Find the intersection of the polar graphs r = sin(2θ)and θ = π/4. Graph.If we just set θ = π/4, then r = sin(π/2) = 1 we wouldobtain only one solution (1, π/4). (0, π/4)
  • Polar EquationsWhen we solve algebraically for the intersection pointsof graphs of rectangular equations, we solve thesystem of the given equations. But to solve for theintersection points of polar equations, we mustconsider other possible polar equations that give risethe same graphs.Example B.a. Find the intersection of the polar graphs r = sin(2θ)and θ = π/4. Graph.If we just set θ = π/4, then r = sin(π/2) = 1 we wouldobtain only one solution (1, π/4). However, there aretwo other points. (0, π/4)
  • Polar EquationsWhen we solve algebraically for the intersection pointsof graphs of rectangular equations, we solve thesystem of the given equations. But to solve for theintersection points of polar equations, we mustconsider other possible polar equations that give risethe same graphs.Example B.a. Find the intersection of the polar graphs r = sin(2θ)and θ = π/4. Graph.If we just set θ = π/4, then r = sin(π/2) = 1 we wouldobtain only one solution (1, π/4). However, there aretwo other points. At θ = –3π/4, we obtain (0, π/4)the point (1, –3π/4) and the pole (0, #).which always have to be checkedseparately.
  • Polar Equationsb. Find the intersection points of the polar graphs r = θand θ = π/4 where 0 ≤ r. Graph.
  • Polar Equationsb. Find the intersection points of the polar graphs r = θand θ = π/4 where 0 ≤ r. Graph.For n = 0, 1, 2,.. the equations θ = π/4 + nπ give thesame graph as θ = π/4.
  • Polar Equationsb. Find the intersection points of the polar graphs r = θand θ = π/4 where 0 ≤ r. Graph.For n = 0, 1, 2,.. the equations θ = π/4 + nπ give thesame graph as θ = π/4. Hence the intersection pointswhere 0 ≤ r are (π/4 + nπ, π/4 + nπ).
  • Polar Equationsb. Find the intersection points of the polar graphs r = θand θ = π/4 where 0 ≤ r. Graph.For n = 0, 1, 2,.. the equations θ = π/4 + nπ give thesame graph as θ = π/4. Hence the intersection pointswhere 0 ≤ r are (π/4 + nπ, π/4 + nπ). θ = π/4 (9π/4,9π/4) (π/4, π/4) o (5π/4, 5π/4) (13π/4, 13π/4) r=θ
  • Polar Equationsb. Find the intersection points of the polar graphs r = θand θ = π/4 where 0 ≤ r. Graph.For n = 0, 1, 2,.. the equations θ = π/4 + nπ give thesame graph as θ = π/4. Hence the intersection pointswhere 0 ≤ r are (π/4 + nπ, π/4 + nπ). We check thatthe point (0,0) is also θ = π/4a solution.The graphs areshown here. (9π/4,9π/4) (π/4, π/4) o (5π/4, 5π/4) (13π/4, 13π/4) r=θ
  • Polar Equationsb. Find the intersection points of the polar graphs r = θand θ = π/4 where 0 ≤ r. Graph.For n = 0, 1, 2,.. the equations θ = π/4 + nπ give thesame graph as θ = π/4. Hence the intersection pointswhere 0 ≤ r are (π/4 + nπ, π/4 + nπ). We cheek thatthe point (0,0) is also θ = π/4a solution.The graphs areshown here. (9π/4,9π/4) (π/4, π/4)In general, we need to o (5π/4, 5π/4)graph when solving for (13π/4, 13π/4)intersection points ofpolar equations to makesure that we have all the r=θsolutions.