The point of intersection gives θ π 6 therefore the

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Nature of Mathematics
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Chapter 7 / Exercise 19
Nature of Mathematics
Smith
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The point of intersection gives θ = π 6 . Therefore, the area A 1 + A 3 = 1 2 " Z π/ 6 0 (2 sin θ ) 2 + Z π/ 2 π/ 6 (1) 2 # = Z π/ 6 0 (1 - cos 2 θ ) + 1 2 θ π/ 2 π/ 6 = ( θ ) | π/ 6 0 - sin 2 θ 2 π/ 6 0 + 1 2 ( θ ) π/ 2 π/ 6 = π 6 - 3 4 + 1 2 · 2 π 6 = π 3 - 3 4
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Nature of Mathematics
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Chapter 7 / Exercise 19
Nature of Mathematics
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560 CHAPTER 9. PARAMETRIC EQUATIONS AND POLAR COORDINATES A 1 + A 3 = π 3 - 3 4 The shaded area is A 1 + A 4 . r = 1 ............... (3) r = 2 cos θ ...... (4) From eq.3 and 4 1 = 2 cos θ, cos θ = 1 2 θ = cos - 1 1 2 = π 3 . The points of intersection gives θ = π 3 . A 1 + A 4 = 1 2 " Z π/ 3 0 (1) 2 + Z π/ 2 π/ 3 (2 cos θ ) 2 # = 1 2 ( θ ) | π/ 3 0 + Z π/ 2 π/ 3 (1 + cos 2 θ ) = π 6 + ( θ ) | π/ 2 π/ 3 + sin 2 θ 2 π/ 2 π/ 3 = π 6 + π 2 - π 3 + - sin 2 π/ 3 2 = π 6 + π 6 + - 3 4 ! = π 3 - 3 4 A 1 + A 4 = π 3 - 3 4 The area of the quadrant with unit radius is π r 2 4 = π 4 . Therefore A 1 + A 3 + A 4 = π 4 ..... eq. 5 A 1 + A 3 = π 3 - 3 4 ....... eq. 6 A 1 + A 4 = π 3 - 3 4 ....... eq. 7 Substitute the value of A 1 +A 3 from eq.6 into eq.5, π 3 - 3 4 + A 4 = π 4 . Therefore , A 4 = π 4 - π 3 - 3 4 ! A 4 = 3 4 - π 12 ......... eq. 8 Substitute the value A 4 in eq.7, A 1 + 3 4 - π 12 ! = π 3 - 3 4 . A 1 = π 3 - 3 4 ! - 3 4 - π 12 ! . A 1 = 5 π 12 - 3 2 ......... eq. 11 Therefore we found the area of the shaded region, A 3 and A 4 . To find A 1 +A 2 : The above shaded region is A 1 + A 2 . r = 2 sin θ ............ eq. 9 . r = 2 cos θ ........... eq. 10 . From eq.9 and eq.10, 2 sin θ = 2 cos θ θ = π 4 . The point of intersection gives θ = π 4 . Therefore A 1 + A 2 = 1 2 " Z π/ 4 0 (2 sin θ ) 2 + Z π/ 2 π 4 (2 cos θ ) 2 # = Z π 4 0 2sin 2 θdθ + Z π/ 2 π/ 4 2cos 2 θdθ = Z π/ 4 0 (1 - cos 2 θ ) + Z π/ 2 π/ 4 (1 + cos 2 θ ) = ( θ ) | π/ 4 0 - sin 2 θ 2 π/ 4 0 + ( θ ) | π/ 2 π/ 4 + sin 2 θ 2 π/ 2 π/ 4 = π 4 - 1 2 + π 4 - 1 2 = π 2 - 1 . A 1 + A 2 = π 2 - 1 ....... eq. 12 The above region is A 1 . But from eq . 12 , A 1 + A 2 = π 2 - 1 . A 1 = 5 π 12 - 3 2 from eq . 11 . Therefore, A 2 = π 2 - 1 - 5 π 12 - 3 2 ! A 2 = π 12 - 1 + 3 2 A 1 = 5 π 12 - 3 2 ; A 2 = π 12 - 1 + 3 2 A 3 = 3 4 - π 12 ; A 4 = 3 4 - π 12 . 31. To find the points of intersection, we graph the function y = 1 - 2 sin x - 2 cos x and look for the roots. We find x 1 . 9948 and x 5 . 8592. Note that the point (0 , 0) is not an intersec- tion point since the two graphs pass through the origin at different values of θ . 32. To find the points of intersection, we graph the function y = 1 + 3 cos x - ( - 2 + 5 sin x ) and look for the roots. We find x 1 . 0808 and x = π . Note that the other places where the graphs overlap are not intersection points since the two graphs pass through these points at different values of θ .
9.5. CALCULUS AND POLAR COORDINATES 561 33. To find the points of intersection in this case, we just set the two expressions for r equal to each other and solve for θ : 1 + sin θ = 1 + cos θ sin θ = cos θ This occurs when θ = π/ 4 or θ = 5 π/ 4. 34. To find the points of intersection in this case, we just set the two expressions for r equal to each other and solve for θ : 1 + 3 sin θ = 1 + cos θ 3 sin θ = cos θ tan θ = 1 3 This occurs when θ = π/ 6 or θ = 7 π/ 6.

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