Download Presentation
## Polar Equations

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Polar Equations**Project by Brenna Nelson, Stewart Foster, Kathy Huynh**Converting From Polar to Rectangular Coordinates**• A point P in a polar coordinate system is represented by an ordered pair of numbers (r, θ) • (r, θ): polar coordinates • r: radius • θ: angle • A point with the polar coordinates (r, θ) can also be represented by either of the following: • (r, measure (in radians) • θ± 2kπ) or (-r, θ + π + 2kπ) where k is any integer • Polar coordinates of the pole are (0, θ) where θ can be any angle**Polar to Rectangular Coordinates**• If P is a point with polar coordinates (x,y) of P are given by • x = rcosθ cosθ = x/r • y = rsinθ sinθ = y/r tanθ = y/x • r² = x² + y² where r is the hypotenuse and x and y are the corresponding sides to the triangle • Plug the values of r and θ into the x and y equations to find the values of the rectangular coordinates**Converting from Polar to Rectangular**• Polar Coordinates (r, θ) • Given: (6, π/6) • r = 6 θ = π/6 • Use the equations x=rcosθ and y=rsinθ to find the values for x and y by plugging in the given values of r and θ • x=rcosθ y=rsinθ • x=(6)cos(π/6) y=(6)sin(π/6) x=(6) · ( ) y=(6) · (1/2) x=3 y=3 • (x, y) = (3 , 3) Insert the values for r and θ into the equations Find the numerical values from solving the found equations The found values for x and y are the rectangular coordinates**Rectangular to Polar Coordinates**• r = • tanθ = y/x so θ = • Plug the values of the x and y coordinates into the equations to find the values of the polar coordinates • Steps for conversion: • Step 1) Always plot the point (x,y) first • Step 2) If x=0 or y=0, use your illustration to find (r, θ) polar coordinates • Step 3) If x does not equal zero and y does not equal zero, then r= • Step 4) To find θ, first determine the quadrant that the point lies in**Converting from Rectangular to Polar**• Rectangular Coordinates (x, y) • Given (2, -2) • x = 2 y = -2 • By plugging the values of x and y into the polar coordinate equations r = and , you can thus find the values of r and θ. • r = • θ = • Polar coordinates (r, θ) = ( , -π/4)**Try these on your own:**• Convert r = 4sinθ from the polar equation the rectangular equation. • Convert 4xy=9 from the rectangular equation to the polar equation.**Solutions: Example 1**• Convert r = 4sinθ from the polar equation the rectangular equation. r = 4sinθ Given equation r² = 4rsinθ Multiply each side by r r² = 4y y = rsinθ x² + y² = 4y r² = x² + y² Equation of a circle x² + (y² - 4y) = 0 Subtract 4y from each side x² + (y² - 4y + 4) = 4 Complete the square in y x² + (y – 2)² = 4 Factor y This is the standard form of the equation of a circle with center (0,2) and radius 2.**Solutions: Example 2**• Convert 4xy = 9 from the rectangular equation to the polar equation. • Use x =rcosθ and y = rsinθ to substitute into the equation 4(rcosθ)(rsinθ) = 9 x = rcosθ, y =rsinθ 4r²cosθsinθ = 9 2r²(2sinθcosθ) = 9 2r²sin(2θ) = 9 • This is the standard polar equation for the rectangular equation 4xy = 9 Double Angle Formula 2sinθcosθ = sin(2θ)**Polar Equations**• Limaçons: • Gen. equation: r = a ± bcosθ (0 < a, 0 < b) r = a ± bsinθ • Rose Curves: • Gen. equation: r = a ± acos(nθ) r = a ± asin(nθ) (n petals if n is odd, 2n petals if n is even)**Polar Equations**• Circles: • Gen. equation: r = a r = cos(θ) • Lemniscates: • Gen. equation: r2 = a ± a2cos(2θ) r2 = a ± a2sin(2θ)**How to Sketch Polar Equations**• Sketch the graph of the polar equation: r = 2 + 3cosθ • The function is a graph of a limaçon because it matches the general formula: • r = a ± bcosθ**x**y 0 5 3.982 1.75 0 -.25 .518 1 0 2.299 3.031 2 .433 -.299 0 Method 1 r = 2 + 3cosθ • Convert the equation from polar to rectangular ~x = rcosθ ~ y = rsinθ x = (2 + 3cosθ)cosθ y = (2 + 3cosθ)sinθ • Substitute different values for θ to find the remaining coordinates**Method 2**r = 2 + 3cosθ • Substitute values of θ and use radial lines to plot points • Use a number of radial lines to ensure that the entire graph of the polar function is sketched • Radial line: the lines that extends from the origin, forming an angle equivalent to the radian value • Ex. Because = 90 , the radial line for is… • REMEMBER: draw arrows to show in which direction the polar function is being sketched**r**0 5 4.598 3.5 2 .5 -.598 -1 Method 2 (con’t.) r = 2 + 3cosθ • Method 2 is used to sketch the polar equation • The work is shown below: Because you know that the equation is a limaçon, you can roughly sketch the rest of the graph. NOTE: this method is only an approximation; it should not be used for calculations.**Method 3**r = 2 + 3cosθ • Using a calculator • The easiest way to graph a polar equation is to just put the equation into the calculator • The method for graphing the polar equations with the calculator are explained in a later slide.**Try these on your own:**• Graph the polar equation, r = 3cosθ, using Method 1 • Graph the polar equation, r = 2, using Method 2**0**x 3 2.25 .75 0 .75 2.25 3 0 1.299 1.299 0 -1.299 -1.299 0 y 0 r 2 2 2 2 2 2 2 Solutions: • Graph the polar equation, r = 3cosθ, using Method 1 • x = 3cosθ(cosθ) • y = 3cosθ(sinθ) • Graph the polar equation, r = 2, using Method 2**Finding Polar Intersection Points**Method 1: • Set equations equal to each other. • Solve for θ. Method 2, for θvalues not on unit circle: • Set calculator mode to polar. • Graph equations. • Find approximate intersection points using TRACE and then find exact intersection points using method 1.**=**Use Method 1 to find the intersection points for the two polar equations. • r = cos(θ) • r = sin(θ) tanθ = 1 θ= 45º , 225º θ = and**Try these on your own:**• Find the intersection points of the equations using Method 1: r = 3 + 3sin(θ) r = 2 – cos(2 θ)** Trig Property**cos2θ = 1 – sin2θ Solutions: • Find the intersection points of the equations using Method 1: r = 3 + 3sin(θ) r = 2 – cos(2θ) 3 + 3sinθ = 2 – cos(2θ) 1 + 3sinθ = −cos(2θ) 1 + 3sinθ = −2cos2θ +1 3sinθ + 2(1 – sin2θ) = 0 3sinθ + 2 – 2sin2θ = 0 Double Angle Formula cos(2θ) = 2cos2θ + 1 Factor**Solutions:**3sinθ + 2 – 2sin2θ = 0 2sin2θ – 3sinθ – 2 = 0 (2sinθ + 1)(sinθ – 2) = 0 2sinθ = −1 sinθ= 2 sinθ = −1/2 θ = and Factor Doesn’t exist • Use unit circle to solve for θ**Method 2**r = 1 + 3cosθ r = 2**Bibliography**• Sullivan, Michael. Precalculus. Upper Saddle River: Pearson Education, 2006. • Foerster, Paul. Calculus: Concepts and Applications. Emeryville: Key Curriculum Press, 2005. • http://curvebank.calstatela.edu/index/lemniscate.gif • http://curvebank.calstatela.edu/index/limacon.gif • http://curvebank.calstatela.edu/index/rose.gif • http://www.libraryofmath.com/pages/graphing-polar-equations/Images/graphing-polar-equations_gr_3.gif