Solução_Calculo_Stewart_6e

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F. TX.10 SECTION 10.3 POLAR COORDINATES ¤ 425 29. θ = −π/6 31. r =sinθ ⇔ r 2 = r sin θ ⇔ x 2 + y 2 = y ⇔ x 2 + y − 1 2 = 1 2 . The reasoning here is the same 2 2 as in Exercise 17. This is a circle of radius 1 2 centered at 0, 1 2 . 33. r =2(1− sin θ). This curve is a cardioid. 35. r = θ, θ ≥ 0 37. r =4sin3θ 39. r =2cos4θ 41. r =1− 2sinθ
F. 426 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES TX.10 43. r 2 =9sin2θ 45. r =2cos 3 2 θ 47. r =1+2cos2θ 49. For θ =0, π,and2π, r has its minimum value of about 0.5. Forθ = π and 3π , r attains its maximum value of 2. 2 2 We see that the graph has a similar shape for 0 ≤ θ ≤ π and π ≤ θ ≤ 2π. 51. x =(r)cosθ =(4+2secθ)cosθ =4cosθ +2.Now,r →∞ ⇒ (4 + 2 sec θ) →∞ ⇒ θ → π − or θ → 3π + [since we need only 2 2 consider 0 ≤ θ
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