APcal 2004-Solutions

# T 7 c 1 15 t 10 82 4sin 2 dt 5 d 1 1 15 10

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Unformatted text preview: 15 t ∫10 82 + 4sin 2 dt 5 (d) 1 1 15 10 [ F (15) − F (10)] = 4sin − 4 sin 5 5 2 2 ≈ 81.899 cars/min. ≈ 1.518 cars/min2. Question 2 ∫0 [ f ( x) − g ( x)] dx = ∫0 2 x(1 − x) − 3( x − 1) 1 1 (a) Area = (b) x dx ≈ 1.133 . 2 2 Volume= π ∫ ( 2 − g ( x) ) − ( 2 − f ( x) ) dx = 0 1 ( 1 π ∫ 2 − 3( x − 1) x 0 (c) 1 ∫ 0 ( h( x ) − g ( x) ) 2 ) 2 2 − ( 2 − 2 x(1 − x) ) dx 1 ( dx = ∫ kx(1 − x) − 3( x − 1) x 0 ≈ 16.179 . ) 2 dx = 15 . 1 Notes: 1. Do not solve. 3 4 FREE-RESPONSE SOLUTIONS ~ 2004 AB Question 3 dv et et e2 =− =− ⇒ a(2) = − . 2 dt 1 + e 2t 1 + e4 1 + ( et ) 1 (a) a(t ) = (b) v(2) = 1 − tan −1 ( e 2 ) < 0 and, from Part (a), a (2) < 0 . Since v(2) and a(2) have the same sign, the speed is increasing. (c) 2 v(t ) = 0 ⇒ tan −1 ( et ) = 1 ⇒ et = tan(1) ⇒ t = ln ( tan(1) ) ≈ 0.443 . There is a local max at t = .443 since v is positive to the left and negative to the right. t = .443 is the only critical number in the domain, therefore y reaches an absolute maximum at this time. 3 (d) ( ) y (2) = −1 + ∫ 1 − tan −1 ( et ) dt 2 0 ≈ −1.360 . The particle is moving away from the origi...
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