APcal 2004-Solutions

# 24814 free response solutions 2004 bc question 5 a

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Unformatted text preview: ( 2t + 1) 3 + cos ( t 2 ) ⇒ y′′(4) = [ y′(t ) ] x′(t ) dt t =4 a (4) = ( x′′(4), y′′(4) ) = ( 2.303, 24.814 ) . Notes: 1. To avoid mistakes and save time, do not simplify. Question 4 See AB Question 4. ≈ 24.814 . FREE-RESPONSE SOLUTIONS ~ 2004 BC Question 5 (a) This is a logistic model with carrying capacity 12, so for both initial conditions, P (0) = 3 and P (0) = 20 , lim P (t ) = 12 . t →∞ (b) (c) P grows from 3 approaching 12. P is growing the fastest when P is half the carrying capacity, that is, P = 6. dY Y t = 1 − ⇒ dt 5 12 dY 1 t t2 1 = ∫ 1 − dt ⇒ ln Y = t − + C . ∫ Y 5 12 5 24 1 t2 t− 1 t2 5 24 Y (0) = 3 ⇒ C = ln 3 . ln Y = t − + ln 3 ⇒ Y (t ) = 3e . 5 24 1 (d) t2 t− t2 5 24 = 0. t − → −∞ ⇒ lim Y (t ) = lim 3e t →∞ t →∞ 24 9 10 FREE-RESPONSE SOLUTIONS ~ 2004 BC Question 6 (a) 2 π π 5 2 f (0) = sin = ; f ′(0) = 5cos = ; 2 4 2 4 25 2 125 2 π π f ′′(0) = −25sin = − ; f ′′′(0) = −125sin = − . 2 2 4 4 P( x) = 2 52 25 2 2 125 2 3 + x− x− x. 2 2 2 ⋅ 2! 2 ⋅ 3! f ( ) ( x) 522 2 . =− 22! 2 ⋅ 22! 22 (b) (c) (d) 1 π 54 sin 5c + 4 54 1 1 4 1 1...
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