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PARAMETRIC EQUATIONS AND POLAR COORDINATES

10. PARAMETRIC EQUATIONS AND POLAR COORDINATES. PARAMETRIC EQUATIONS & POLAR COORDINATES. 10.4 Areas and Lengths in Polar Coordinates. In this section, we will: Develop the formula for the area of a region whose boundary is given by a polar equation. AREAS IN POLAR COORDINATES.

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PARAMETRIC EQUATIONS AND POLAR COORDINATES

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  1. 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

  2. PARAMETRIC EQUATIONS & POLAR COORDINATES 10.4 Areas and Lengths in Polar Coordinates • In this section, we will: • Develop the formula for the area of a region • whose boundary is given by a polar equation.

  3. AREAS IN POLAR COORDINATES Formula 1 • We need to use the formula for the area of a sector of a circle A = ½r2θwhere: • r is theradius. • θ istheradianmeasure of thecentral angle.

  4. AREAS IN POLAR COORDINATES • Formula 1 follows from the fact that the areaof a sector is proportional to its central angle:A =(θ/2π)πr2= ½r2θ

  5. AREAS IN POLAR COORDINATES • LetRbe the regionbounded by the polar curver =f(θ) and by the raysθ = a and θ = b, where: • f is a positive continuous function. • 0 <b –a ≤ 2π

  6. AREAS IN POLAR COORDINATES • We divide the interval [a, b] into subintervalswith endpoints θ0, θ1,θ2, …, θn, and equalwidth ∆θ. • Then, the rays θ= θidivide R into smallerregions withcentral angle ∆θ= θi – θi–1.

  7. AREAS IN POLAR COORDINATES • If we choose θi* in the i th subinterval [θi–1,θi] then the area∆Aiof the i th region isthe area of thesector ofacirclewithcentral angle∆θand radiusf(θ*).

  8. AREAS IN POLAR COORDINATES Formula 2 • Thus, from Formula 1, we have: ∆Ai≈ ½[f(θi*)]2∆θ • So, an approximation to the total area A of R is:

  9. AREAS IN POLAR COORDINATES • It appears that theapproximation in Formula 2 improves as n →∞.

  10. AREAS IN POLAR COORDINATES • However, thesumsin Formula 2 are Riemann sums for the functiong(θ) = ½[f(θ)]2. • So,

  11. AREAS IN POLAR COORDINATES Formula 3 • Therefore,it appears plausible—and can, in fact, be proved—that the formula for the areaA ofthe polar regionR is:

  12. AREAS IN POLAR COORDINATES Formula 4 • Formula 3 is often written aswith the understanding that r = f(θ). • Note the similarity between Formulas 1 and 4.

  13. AREAS IN POLAR COORDINATES Formula 4 • When we apply Formula 3 or 4, it is helpful tothink of the area as being swept out byarotating ray through O that starts with angle a and ends with angle b.

  14. AREAS IN POLAR COORDINATES Example 1 • Find the area enclosed by one loop of the four-leaved rose r = cos 2θ. • The curve r = cos 2θwas sketched in Example 8 in Section 10.3

  15. AREAS IN POLAR COORDINATES Example 1 • Notice thatthe regionenclosed bytherightloop is sweptout bya ray that rotatesfromθ = –π/4 to θ = π/4.

  16. AREAS IN POLAR COORDINATES Example 1 • Hence, Formula 4 gives:

  17. AREAS IN POLAR COORDINATES Example 2 • Find the area of the region that lies inside thecircle r = 3 sin θ andoutside the cardioidr = 1 + sin θ.

  18. AREAS IN POLAR COORDINATES Example 2 • The values of a and b in Formula 4 aredeterminedby finding the points of intersectionof the two curves.

  19. AREAS IN POLAR COORDINATES Example 2 • They intersect when 3 sin θ= 1 + sin θ, whichgives sin θ = ½. • So, θ = π/6 and 5π/6.

  20. AREAS IN POLAR COORDINATES Example 2 • The desired area can befound by subtractingthe area inside the cardioid between θ=π/6 and θ=5π/6 fromthe area inside the circlefrom π/6to5π/6.

  21. AREAS IN POLAR COORDINATES Example 2 • Thus,

  22. AREAS IN POLAR COORDINATES Example 2 • As the region is symmetric about thevertical axis θ = π/2, we can write:

  23. AREAS IN POLAR COORDINATES • Example 2 illustrates the procedure for findingthe area of the region bounded by twopolarcurves.

  24. AREAS IN POLAR COORDINATES • In general, letR be a region that is bounded bycurves with polarequations r = f(θ), r = g(θ), θ= a, θ= b,where: • f(θ) ≥g(θ)≥ 0 • 0 < b – a < 2π

  25. AREAS IN POLAR COORDINATES • The area A of R is found by subtracting the area inside r = g(θ) from the area inside r = f(θ).

  26. AREAS IN POLAR COORDINATES • So, using Formula 3, we have:

  27. CAUTION • The fact that a single point has manyrepresentations in polar coordinatessometimes makes it difficult to find all thepoints of intersection of two polar curves.

  28. CAUTION • For instance, it is obvious from this figure thatthe circle and the cardioid have threepoints ofintersection.

  29. CAUTION • However, in Example 2, we solved theequations r = 3 sin θ andr = 1 + sin θandfound only two such points: (3/2, π/6)and (3/2, 5π/6)

  30. CAUTION • The origin is alsoa point of intersection. • However, wecan’t find it by solving the equations of thecurves. • The origin has no singlerepresentation in polar coordinates that satisfiesboth equations.

  31. CAUTION • Notice that, when represented as (0, 0) or (0, π), the origin satisfies r = 3 sin θ. • So, itlies on the circle.

  32. CAUTION • When represented as (0, 3 π/2), it satisfies r = 1 + sin θ. • So, it liesonthe cardioid.

  33. CAUTION • Think of two points moving along the curvesas the parameter valueθincreases from 0 to 2π. • On one curve, the origin is reached at θ = 0 and θ = π. • On the other, it is reached at θ = 3π/2.

  34. CAUTION • The points don’t collide at the origin since they reach the origin at different times. • However, the curves intersect there nonetheless.

  35. CAUTION • Thus, to find allpoints of intersection of twopolar curves, it is recommended that youdrawthe graphs of both curves. • It is especiallyconvenient to use a graphing calculator orcomputer to help with this task.

  36. POINTS OF INTERSECTION Example 3 • Find all points of intersection of the curvesr = cos 2θ and r = ½. • If we solve the equations r = cos 2θ and r = ½, we get cos 2θ= ½. • Therefore, 2θ= π/3, 5π/3, 7π/3, 11π/3.

  37. POINTS OF INTERSECTION Example 3 • Thus, the values of θ between 0 and 2πthatsatisfyboth equations are: θ= π/6, 5π/6, 7π/6, 11π/6

  38. POINTS OF INTERSECTION Example 3 • We have found four points ofintersection: (½, π/6), (½, 5π/6), (½, 7π/6), (½, 11π/6)

  39. POINTS OF INTERSECTION Example 3 • However, you can see that the curves have four other points of intersection: (½, π/3), (½, 2π/3), (½, 4π/3), (½, 5π/3)

  40. POINTS OF INTERSECTION Example 3 • These can be found usingsymmetry or bynoticing that another equation of the circle isr = -½. • Then, we solve r = cos 2θand r = -½.

  41. ARC LENGTH • To find the length of a polar curver = f(θ), a ≤ θ≤ b, we regard θ as a parameter andwrite the parametric equations of the curve as: • x =r cos θ=f(θ)cosθ • y =r sin θ=f (θ)sin θ

  42. ARC LENGTH • Using the Product Rule and differentiating with respect to θ, we obtain:

  43. ARC LENGTH • So, using cos2 θ+ sin2 θ= 1,we have:

  44. ARC LENGTH Formula 5 • Assuming that f’ is continuous, we can use Theorem 6 in Section 10.2to write the arc length as:

  45. ARC LENGTH Formula 5 • Therefore, the length of a curve with polar equation r = f(θ), a ≤ θ≤ b, is:

  46. ARC LENGTH Example 4 • Find the length of the cardioidr = 1 + sinθ • We sketched itin Example 7 inSection 10.3

  47. ARC LENGTH • Its full length isgivenby the parameterinterval0≤θ≤ 2π. • So, Formula 5 gives:

  48. ARC LENGTH • We could evaluate this integral by multiplying and dividing the integrand byor we could use a computer algebra system. • In any event, we find that thelength of thecardioid isL = 8.

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