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Pre-AP PrecalculusModule 9Polar EquationsWhen converting from a Cartesian coordinate system to a Polar one, the origin is renamed the pole, and the positive x-axis is called the polar axis. In addition the Cartesian coordinates x,y become Polar coordinates r,θ, where r is the distance between the point and the pole, and θ is the angle formed by the polar axis and the ray from the pole through the point. If r<0, then the point is plotted 180° from the stated angle.Example 1 – Plot the points with the following polar coordinates: (a)3,5π3, (b) 2, -π4, (c) 3,0, and (d) -2,π4.Example 2 – Consider a point with polar coordinates 2,π4. Find two polar coordinates r,θ that map the same point, one with a positive r and one with a negative r.1647825544195Example 3 – Plot the point P with polar coordinates 3,π6, and find other polar coordinates r, θ of this same point for which: (a) r>0, 2π≤θ<4π; (b) r<0, 0≤θ<2π; and (c) r>0, -2π≤θ<0. A point with polar coordinates r,θ, θ in radians, can also be represented by either of the following: r,θ+2πk or -r,θ+π+2πk, k any integer. The polar coordinates of the pole are 0,θ, where θ can be any angle.If P is a point with polar coordinates r, θ, the rectangular coordinates x,y of P are given by rcosθ, rsinθ.Example 4 – Find the rectangular coordinates of the points with the following polar coordinates: (a) 6,π6 and (b) -4, -π4.Example 5 – Find the polar coordinates of the points whose rectangular coordinates are (a) 0,3 and (b) -3,0.Example 6 – Find the polar coordinates of a point whose rectangular coordinates are 2,-2.Example 7 – Find the polar coordinates of a point whose rectangular coordinates are -1,-3.r=x2+y2 and θ=tan-1yx if x,y is in Quadrant I or IVtan-1yx+π if x,y is in Quadrant II or III.This is due to the range of inverse tangent.Example 8 – Transform the equation r=6cosθ from polar coordinates to rectangular coordinates.Example 9 – Transform the equation 4xy=9 from rectangular coordinates to polar coordinates. Complete the following exercises on a separate sheet of paper.In exercises 1 – 8, plot each point given in polar coordinates, and find other polar coordinates r,θ of the point for which a r>0, -2π≤θ<0, (b) r<0, 0≤θ<2π, and (c) r>0, 2π≤θ<4π.5,2π34,3π4-2, 3π-3,4π1,π22, π-3, -π4-2, -2π3In exercises 9 – 18, the polar coordinates of a point are given. Find the rectangular coordinates of each point.3,-π24,3π2-2,0-3,π6,150°5,300°-2,3π4-2,-2π36.3, 3.88.1, 5.2In exercises 19 – 26, the rectangular coordinates of a point are given. Find the polar coordinates for each point with r>0 and 0≤θ<2π.0,2-1,01,-1-3,33,1-2, -231.3, -2.1-0.8,-2.1In exercises 27 – 34, convert each Cartesian equation into its corresponding polar equation.2x2+2y2=3x2+y2=xx2=4yy2=2x2xy=14x2y=1x=4y=-3In exercises 35 – 42, convert each polar equation into its corresponding Cartesian equation. r=cosθr=sinθ+1r2=cosθr=sinθ-cosθr=2r=4r=41-cosθr=33-cosθA unique aspect to polar equations is that the input is θ and the output is r. Thus, polar coordinates r, θ are reversed from rectangular coordinates x,y.Example 1 – Identify and graph the equation r=3. Example 2 – Identify and graph the equation θ=π4. Example 3 – Identify and graph the equation rsinθ=2.Example 4 – Use a graphing calculator to graph the polar equation rsinθ=2.Example 5 – Identify and graph the equation rcosθ=-3.Let a be a nonzero real number. Then the graph of rsinθ=a is a horizontal line a units above the pole if a>0 and a below the pole if a<0. The graph of rcosθ=a is a vertical line aunits to the right of the pole if a>0 and a to the left of the pole if a<0.Example 6 – Identify and graph the equation r=4sinθ.Example 7 – Identify and graph the equation r=-2cosθ.Let a be a positive real number.r=2asinθ graphs a circle with radius a and center 0,a in rectangular coordinates.r=2acosθ graphs a circle with radius a and center a,0 in rectangular coordinates.r=-2asinθ graphs a circle with radius a and center 0,-a in rectangular coordinates.r=-2acosθ graphs a circle with radius a and center -a, 0 in rectangular coordinates.Tests for SymmetryIn a polar equation, if replacing θ with –θ results in an equivalent equation, the graph is symmetric with respect to the polar axis.In a polar equation, if replacing θ with π-θ results in an equivalent equation, the graph is symmetric with respect to the line θ=π2.In a polar equation, if replacing r with -r results in an equivalent equation, the graph is symmetric with respect to the pole.θr2π33π45π6π7π65π4θr4π33π25π37π411π62πθr0π12π6π4π3π2Example 8 – Graph r=1-sinθ.left201295θr0π12π6π4π3π2θr2π33π45π6π7π65π4θr4π33π25π37π411π62πExample 9 – Graph r=3+2cosθ.381005715θr0π12π6π4π3π2θr2π33π45π6π7π65π4θr4π33π25π37π411π62πExample 10 – Graph r=1+2cosθ.2857512065Lima?ons are characterized by equations of the form r=a+bcosθ, r=a-bcosθ, r=a+bsinθ, and r=a-bsinθ, where a>0 and b>0.If a=b, the Lima?on is a cardioid.If a<b, the Lima?on has an inner loop.If a>b, the Lima?on has no inner loop.θr0π12π6π4π3π2θr2π33π45π6π7π65π4θr4π33π25π37π411π62πExample 11 – Graph r=2cos2θ.0-1905θr0π12π6π4π3π2θr2π33π45π6π7π65π4θr4π33π25π37π411π62πExample 12 – Graph r=2sin2θ.left59055θr0π12π6π4π3π2θr2π33π45π6π7π65π4θr4π33π25π37π411π62πExample 13 – Graph r=2cos3θ.left57785θr0π12π6π4π3π2θr2π33π45π6π7π65π4θr4π33π25π37π411π62πExample 14 – Graph r=2sin3θ.0-1905Rose curves are characterized by equations of the form r=acosnθ and r=asinnθ, where a≠0 and n≠0. If n<0 use the odd and even identities to eliminate the negative coefficient from the argument.If n is even, the rose has 2n petals of length a.If n is odd , the rose has n petals of length a. For r=acosnθThe first petal lies along the polar axis if a>0. The first petal lies along θ=π if a<0.The rest of the petals are equally spaced around the pole. The rose is symmetric with respect to the polar axis.For r=asinnθThe first petal lies along θ=π2n if a>0.The first petal lies along θ=-π2n if a<0.The rest of the petals are equally spaced around the pole.The rose is symmetric with respect to the line θ=π2.θr4π33π25π37π411π62πθr2π33π45π6π7π65π4θr0π12π6π4π3π2Example 15 – Graph r2=4sin2θ.01905Lemniscates are characterized by equations of the form r2=a2sin2θ and r2=a2cos2θ, where a≠0.r2=a2cos2θ graphs a two-petal rose such that each petal is of length a, with one petal along θ=0 and the other along θ=π.θr4π33π25π37π411π62πθr2π33π45π6π7π65π4θr0π12π6π4π3π2r2=a2sin2θ graphs a two-petal rose such that each petal is of length a, with one petal along θ=π4 and the other along θ=5π4.Example 16 – Graph r=θ.0-635Example 17 – Graph r=2+2sinθ.An important distinction must be made. x2+y2=1 is an implicit equation that graphs the unit circle. To graph x2+y2=1 in the calculator, two separate equations must be graphed: y=±1-x2, which means a circle cannot be written as a function in the rectangular coordinates since every x-value in the domain could possibly yield two different y-values. However, in polar coordinates, the unit circle can be graphed with one equation r=1, and since every angle θ could only have one output, namely a radius of 1, the unit circle can be represented as a function in polar coordinates. We will find it useful in Calculus to represent nonfunctions in rectangular coordinates as functions in polar plete the following exercises on a separate sheet of paper.In exercises 1 – 8, transform each polar equation to an equation in rectangular coordinates. Identify and graph the equation by hand.r=2θ=-π4rcosθ=4rsinθ=-2r=2sinθr=-4cosθrcscθ=8rsecθ=-4In exercises 9 – 20, identify and graph each polar equation by hand.r=1+sinθr=2-2cosθr=2-cosθr=4+2sinθr=1-2sinθr=2+4cosθr=2sin3θr=3cos4θr2=sin2θr2=9cos2θr=3+cosθr=4cos3θIn exercises 21 – 26, graph each pair of polar equations on the same polar grid. Find the polar coordinates of the point(s) of intersection and label the point(s) on the graph.r=8cosθ; r=2secθr=8sinθ; r=4cscθr=sinθ; r=1+cosθr=3; r=2+2cosθr=1+sinθ; r=1+cosθr=1+cosθ;r=3cosθAn imaginary number is the square root of a negative number. Using i for the unit imaginary number -1, we get -a2=ai.A complex number is the sum of a real number and an imaginary number, a+bi. Real numbers and imaginary numbers fall within the set of complex numbers. You can represent complex numbers as points on a Cartesian coordinate system called the complex plane with the horizontal real axis and vertical imaginary axis. We can also represent complex numbers in polar form as well.i=-1i2=-1z=a+bi is a complex number with a being the real part of z and b being the imaginary part of z.z=a+bi is plotted as a,b on the complex plane.a-bi is the complex conjugate of a+bi.a+bia-bi=a2+b2 is always real.Since a+bi can be graphed as a point with rectangular coordinates a,b, it can also be graphed as r, θ in polar coordinates, such that a=rcosθ and b=rsinθ. Thus z=a+bi=rcosθ+risinθ=rcosθ+isinθ=rcisθ. We refer to rcisθ as the polar form of a complex number or cis notation. r is called the modulus or magnitude, and θ is called the argument. The absolute value of a complex number is the modulus of that number: z=a+bi=r.z=a+bi=rcisθa=rcosθb=rsinθr2=a2+b2θ=tan-1ba or θ=tan-1ba+π depending on which quadrant z falls into.Example 1 – Transform the complex number z=-5+7i into polar form.Example 2 – Transform z=5cis144° into Cartesian form.Example 3 – If z1=3cis83° and z2=2cis41°, find the product z1z2.If z1=r1cisθ1 and z2=r2cisθ2, then z1z2=r1r2cisθ1+θ2.Example 4 – Find the reciprocal of z=2cis29°.If z=rcisθ, then 1z=1rcis-θ.Example 5 – If z1=5cis71° and z2=2cis29°, find z1z2.If z1=r1cisθ1 and z2=r2cisθ2, then z1z2=r1r2cisθ1-θ2.Example 6 – If z=2cis29°, find z5.DeMoivre’s TheoremIf z=rcisθ, then zn=rncisnθ.If z=rcisθ, then nz=nrcisθn+360°nk, for all integers 0≤k<n. There will be n distinct roots.Example 7 – If z=8cis60°, find the cube roots of z, plete the following exercises on a separate sheet of paper.In exercises 1 – 12, write the complex number in polar form, r cis θ.-1+i1-i3-i1+i3-4-3i-3+4i5+7i-11-2i1i-i-8In exercises 13 – 22, write the complex number in Cartesian form, a+bi.8cis34°11cis247°6cis120°8cis150°2cis225°32cis45°5cis180°9cis90°3cis270°2cis0°In exercises 23 – 26, find (a) z1z2, (b) z1z2, (c) z12, and (d) z23.z1=3cis47°, z2=5cis36°z1=2cis154°, z2=3cis27°z1=4cis238°, z2=2cis51°z1=6cis19°, z2=4cis96°In exercises 27 – 36, find all of the indicated roots.Cube roots of 27cis120°Cube roots of 8cis15°Fourth roots of 16cis80°Fourth roots of 81cis64°Square roots of iSquare roots of -iCube roots of 8Cube roots of -27Sixth roots of -1Tenth roots of 1 ................
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