Circles #3 - Questions and Answers

Question 1 A ____ is the set of all points in a plane that are connected and equidistant from a fixed point of the plane called the ____ of the circle. Answer A circle is the set of all points in a plane that are connected and equidistant from a fixed point of the plane called the center of the circle. Question 2 Find a formula for the area of the shaded region, which represents one-fourth of an annulus. Answer Explanation: The area of a circle is given by 𝐴 = 𝜋 𝑟 2 A=πr 2 , where 𝑟 r is the radius. To find the area of the ring (annulus), subtract the area of the inner circle from the area of the outer circle. 𝐴 ring = 𝐴 outer circle − 𝐴 inner circle A ring ​ =A outer circle ​ −A inner circle ​ 𝐴 ring = 𝜋 𝑅 2 − 𝜋 𝑟 2 A ring ​ =πR 2 −πr 2 Here, 𝑅 R is the radius of the outer circle, and 𝑟 r is the radius of the inner circle. Since the shaded region represents one-fourth of the area of the ring: 𝐴 shaded = 𝐴 ring 4 A shaded ​ = 4 A ring ​ ​ 𝐴 shaded = 𝜋 ( 𝑅 2 − 𝑟 2 ) 4 A shaded ​ = 4 π(R 2 −r 2 ) ​ Final Formula: 𝐴 shaded = 𝜋 ( 𝑅 2 − 𝑟 2 ) 4 A shaded ​ = 4 π(R 2 −r 2 ) ​ Question 3 Find the area of the region that lies between both curves: 𝑟 2 = 50 sin ⁡ ( 2 𝜃 ) , 𝑟 = 5 r 2 =50sin(2θ),r=5. Answer Step 1: Find the intersection points. From the curves: 50 sin ⁡ ( 2 𝜃 ) = 5 2    ⟹    25 = 50 sin ⁡ ( 2 𝜃 )    ⟹    sin ⁡ ( 2 𝜃 ) = 1 2 50sin(2θ)=5 2 ⟹25=50sin(2θ)⟹sin(2θ)= 2 1 ​ Solving for 2 𝜃 2θ: 2 𝜃 = 𝜋 6 , 5 𝜋 6 , 7 𝜋 6 , 11 𝜋 6 . 2θ= 6 π ​ , 6 5π ​ , 6 7π ​ , 6 11π ​ . Thus, the symmetric bounds are 𝜃 = 0 θ=0 to 𝜃 = 𝜋 6 θ= 6 π ​ . Step 2: Calculate the desired area. The area 𝐴 A of a polar curve is: 𝐴 = 1 2 ∫ 𝜃 1 𝜃 2 𝑟 2 𝑑 𝜃 . A= 2 1 ​ ∫ θ 1 ​ θ 2 ​ ​ r 2 dθ. Add the area inside the cardioid 𝑟 2 = 50 sin ⁡ ( 2 𝜃 ) r 2 =50sin(2θ) and the area of the circle 𝑟 = 5 r=5: 𝐴 = 2 ( ∫ 0 𝜋 / 6 1 2 ( 50 sin ⁡ ( 2 𝜃 ) ) 𝑑 𝜃 + ∫ 0 𝜋 / 6 1 2 ( 5 2 ) 𝑑 𝜃 ) . A=2(∫ 0 π/6 ​ 2 1 ​ (50sin(2θ))dθ+∫ 0 π/6 ​ 2 1 ​ (5 2 )dθ). Step 3: Simplify the integrals: ∫ sin ⁡ ( 2 𝜃 ) 𝑑 𝜃 = − 1 2 cos ⁡ ( 2 𝜃 ) ∫sin(2θ)dθ=− 2 1 ​ cos(2θ). ∫ 𝑑 𝜃 = 𝜃 ∫dθ=θ. Substitute and solve to get: 𝐴 = 2 [ 25 2 ( − cos ⁡ ( 2 𝜃 ) ) ∣ 0 𝜋 / 6 + 1 2 ( 25 ) 𝜃 ∣ 0 𝜋 / 6 ] . A=2[ 2 25 ​ (−cos(2θ)) ​ 0 π/6 ​ + 2 1 ​ (25)θ ​ 0 π/6 ​ ]. Simplify further to obtain the final area. Question 4 Find the rate at which the area within the circle is increasing after 1 second when the ripple travels outward at 60 cm/s. Answer Step 1: Rate of change of the radius 𝑟 r: 𝑑 𝑟 𝑑 𝑡 = 60   cm/s . dt dr ​ =60cm/s. Step 2: Area of a circle 𝐴 = 𝜋 𝑟 2 A=πr 2 . Differentiate both sides with respect to 𝑡 t: 𝑑 𝐴 𝑑 𝑡 = 2 𝜋 𝑟 𝑑 𝑟 𝑑 𝑡 . dt dA ​ =2πr dt dr ​ . Step 3: Substitute 𝑟 = 60   cm r=60cm after 1 second and 𝑑 𝑟 𝑑 𝑡 = 60 dt dr ​ =60: 𝑑 𝐴 𝑑 𝑡 = 2 𝜋 ( 60 ) ( 60 ) . dt dA ​ =2π(60)(60). 𝑑 𝐴 𝑑 𝑡 = 7200 𝜋   cm 2 / s . dt dA ​ =7200πcm 2 /s. Final Answer: The area is increasing at a rate of 7200π cm²/s (approximately 22619.5   cm 2 / s 22619.5cm 2 /s). Question 5 Find the area of the blue sector. Use 𝜋 = 3.14 π=3.14 and round to the nearest hundredth. Answer Step 1: Formula for the area of a sector: 𝐴 = 𝜃 36 0 ∘ × 𝜋 𝑟 2 . A= 360 ∘ θ ​ ×πr 2 . Given: 𝜃 = 6 0 ∘ θ=60 ∘ , 𝑟 = 7 r=7, 𝜋 = 3.14 π=3.14. Step 2: Substitute the values: 𝐴 = 60 360 × 3.14 × 7 2 . A= 360 60 ​ ×3.14×7 2 . Simplify: 𝐴 = 1 6 × 3.14 × 49. A= 6 1 ​ ×3.14×49. 𝐴 = 25.64   in 2 . A=25.64in 2 . Final Answer: 25.64 in².