c. Use MINIMUM to find the least value: (0.2) − (0.1) = 7.2654 − 3.2492 ≈ d 25 40.162 0.2 − 0.1 0.2 − 0.1 − = − ≈ (0.3) (0.2) 13.764 7.2654 d d − = − ≈ d. (0.1) (0) 3.2492 0 32.492 d d 64.986 0.1 − 0 0.1 − 0 _π 0 0 2 0.3 − 0.2 0.3 − 0.2 (0.4) − (0.3) 30.777 d d −13.764 θ is least when ≈ 0.83 . L 0.4 − 0.3 = 0.4 − 0.3 ≈ 170.13 d. ≈ 3 + 4 ≈ 9.86 L The first differences represent the average feet . e. (0.83) cos (0.83) rate of change of the beam of light against sin up will result in a ladder that measured in feet per second. For won’t fit around the corner. Answers will vary. example, between = 0 seconds and = 0.1 t t seconds, the average rate of change of the 51. (t ) = 10 tan(π ) beam of light against the wall is 32.492 feet d t 50. a. per second. y Note that 5 rounding x π – π 3π 2 2 –5 = tan y x (t ) = 10 tan(π b. y d π is undefined at = t ) 2 t 5 and = 3π , or in general at t 2 ⎧ is an odd integer ⎫ . At these k k x π – π 3π ⎨ ⎬ 2 2 2 ⎩ ⎭ instances, the length of the beam of light approaches infinity. It is at these instances in the rotation of the beacon when the beam –5 of light being cast on the wall changes from = y x + π –cot one side of the beacon to the other. 2 Yes, the two functions are equivalent. c.
c. Use MINIMUM to find the least value: 25 d. d (0.1) − d (0) = 3.2492 − 0 ≈ 32.492 0.1 − 0 0.1 − 0 (0.2) − d (0.1) = 7.2654 − 3.2492 ≈ 40.162 0.2 − 0.1 0.2 − 0.1 d (0.3) − d (0.2) = 13.764 − 7.2654 ≈ 64.986 0 _π 0 2 0.3 − 0.2 0.3 − 0.2 d (0.4) − d (0.3) 30.777 −13.764 L is least when θ ≈ 0.83 . = 0.4 − 0.3 0.4 − 0.3 ≈ 170.13 d. L ≈ 3 + 4 ≈ 9.86 feet . e. The first differences represent the average cos (0.83) sin (0.83) rate of change of the beam of light against 50. a. Note that rounding up will result in a ladder that won’t fit around the corner. Answers will vary. d (t ) = 10 tan(π t ) 51. measured in feet per second. For example, between t = 0 seconds and t = 0.1 seconds, the average rate of change of the beam of light against the wall is 32.492 feet per second. y 5 x – π π 3π 2 2 b. d (t ) = 10 tan(π t ) is undefined at t = π 2 –5 y = tan x y 5 and t = 3π , or in general at 2 ⎧ k k is an odd integer ⎫ . At these x – π π 3π ⎨ 2 ⎬ 2 2 ⎩ ⎭ instances, the length of the beam of light approaches infinity. It is at these instances in the rotation of the beacon when the beam of light being cast on the wall changes from one side of the beacon to the other. c. –5 y = –cot x + π 2 Yes, the two functions are equivalent.
62514-0132318806_ISM_02-1.docx