Sin 2 cos 4 cos 2 2 sin 4 2 cos 2 sin 4 find critical

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= (sin 2? − cos ? − 4?)′ ? = cos 2? (2?) − (− sin ?) − 4 ? = 2 cos 2? + sin ? − 4 Find critical numbers 2 cos 2? + sin? − 4 = 0 2(1 − 2 sin 2 ?) + sin? − 4 = 0 2 − 3 sin 2 ? + sin ? − 4 = 0 4 sin 2 ? − sin ? + 2 = 0 Discriminant = ? 2 − 4?? = (−1) 2 − 4(4)(2) = −31 < 0 10. A ladder 15 m long leans against the wall. The ladder slides down the wall at a rate of 1.5 m/s. What is the rate of change of the angle at the ground when the ladder is 6 m from the wall? (3 marks) There is no value for x to satisfy for ? = sin 2? − cos ? − 4? . The equation doesn’t have any critical number. We don’t have any maximum or minimum points for the given equation.
Unit # 4 Extensions cos𝜃 = 6 15 = 0.4 𝜃 = cos −1 (0.4) = 66.42 𝜃 sin𝜃 = ? 15 ? = 15 cos 𝜃 ? = (15 sin𝜃)′ ? = 15(cos 𝜃)(𝜃) ? = (15 cos 𝜃) 𝜃′ −1.5 = (15 cos 66.42 𝜃 ) 𝜃′ −1.5 = (6) 𝜃′ 𝜃 = − 1.5 6 = −0.25 The rate of change of the angle is decreasing at 0.25
Unit # 4 Extensions 11. The position of a particle as it moves horizontally is described by the given equation ? = ?𝑖?? − ???? t 0 2 ≤ ≤ π . If s is the displacement in metres and t is the time in seconds, find the maximum and minimum displacements. (3 marks) Lets find s’ ? = (sin ? − cos ?)′ ? = cos ? − (− sin ?) ? = cos ? + sin ? Find critical numbers cos ? + sin? = 0 cos ? = − sin ? sin? cos? = −1 tan ? = −1 ? = 𝜋 − 4 , ? = 2𝜋 − 4 ? = 3 4 , ? = 7 4 Since the function changes from increasing to decreasing at ? = 3 4 the maximum point at this is ? ( 3 4 ) = sin ( 3 4 ) − cos ( 3 4 ) = √2 The maximum point is ( 3 4 , √2 ) Since the function changes from decreasing to increasing at ? = 7 4 the minimum point at this is ? ( 7 4 ) = sin ( 7 4 ) − cos ( 7 4 ) = −√2 The minimum point is ( 7 4 , −√2 ) Interval ? = cos? + sin ? Increasing or Decreasing 0<t< 3 4 + Increasing 3 4 < ? < 7 4 - Decreasing 7 4 < ? < 2𝜋 + Increasing
Unit # 4 Extensions 12. Show that a rectangle with given perimeter has maximum area when it is a square. (3 marks) Let’s assume: ? = ?????ℎ, ? = ?𝑖??ℎ, ? = ???𝑖????? ?? ????????? 2(? + ?) = ? ? = ? 2 − ? − − − 1 Area of a rectangle ????′ = ( ?? 2 − ? 2 )′ ????′ = ( ? 2 − 2?) Area’ = 0, ? 2 − 2? = 0 2? = ? 2 ? = ? 4 ???? ′′ = ( ? 2 − 2?) ???? ′′ = −2 < 0 The rectangle has a maximum area of ? = ? 4 At ? = ? 4 , ? = ? 2 ? 4 = ? 4 This is a square The maximum point is ( 3 4 , √2 ) & the minimum point is ( 7 4 , −√2 ) The rectangle with a given perimeter has a maximum point area when is a square.
Unit # 4 Extensions 13. Choose any 5 of the following problems to complete. Solve the problem by following the procedure for solving an optimization problem. Show all your steps. (4 marks each x 5) a. A farmer has 1800 m of fencing to make two equal rectangular pens that share a common side. Determine the dimensions of each pen so that the area is a maximum. 𝐿?? ?????? ? = 𝐿????ℎ , ? = ?𝑖??ℎ So, ? + ? + ? + ? + ? + ? + ? = 1800 4? + 3? = 1800 3? = 1800 − 4? ? = 600 − ( 4 3 ) ? − − − 1 Area of a rectangle, 𝐴 = 2?? 𝐴 = 2? (600 − ( 4 3 ) ?) 𝐴 = 1200 ? − ( 8 3 ) ? 2 𝐴 = (1200? − ( 8 3 ) ? 2 ) 𝐴 = 1200 − 16? 3 A’ = 0 , Max and Min point
Unit # 4 Extensions 1200 − 16?

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