98AB1a R = 16/3
98AB1b h = 16^(1/3)
98AB1c Vx = 8pi
98AB1d k = 2(2)^(1/2)
98AB2a/98BC2a lim x->inf f(x) = inf, lim x->-inf f(x) = 0
98AB2b/98BC2b f(-1/2) = -1/e
98AB2c/98BC2c Rf:y in [-1/e, inf)
98AB2d/98BC2d y(-1/b) = -1/e
98AB3a t in (0, 30) U (45, 50)
98AB3b 36/25 ft/sec^2
98AB3c many possibilites: try (v(40+5)-v(40))/5 = -3ft/sec^2
98AB3d MSUM = 2530 ft (total distance traveled)
98AB4a f'(1) = 1/2
98AB4b y-4=(1/2)(x-1) so f(1.2) is about 4+(1.2-1)/2 = 4.1
98AB4c f(x) = (x^3 + x + 14)^(1/2)
98AB4d (1.2) is about 4.114
98AB5a/98BC5a sketch F(t)
98AB5b/98BC5b Favg is about 87 deg F
98AB5c/98BC5c t is about 5.231 or 18.769 hr
98AB5d/98BC5d 510 cents
98AB6a do imp diff
98AB6b y = 0.165 or y = 2
98AB6c x is about -0.500
98BC1a R = 96/5
98BC1b Vx = 576pi/5
98BC1c k is about 0.995
98BC3a P3(x) = 5-3x+x^2/2!+4x^3/3! and f(0.2) is about 4.425
98BC3b Q4(x) = 5-3x^2+x^4/2
98BC3c R3(x) = 5x-3x^2/2+x^3/3!
98BC3d no can do
98BC4a sketch slope field
98BC4b f(0.2) is about 3.015
98BC4c f(x) = 3e^(x^2/4) so f(0.2) is about 3.030
98BC6a x(t) = (2t+1)^(1/2) - 5
98BC6b dy/dt=(3((2t+1)^(1/2)-5))^2 - 3)/(2t+1)^(1/2)
98BC6c position: (-2, -2),
magnitude of the velocity vector is about 3.018.
How did I do?
TIA,
On Sun, 17 May 1998, AHSAP wrote:
> From a relatively new AP AB teacher - could someone please send me the answers
> they got for the 1998 exam so I can compare my answers with the rest of the
> world?
A. Jorge Garcia Teacher/Professor Mathematics/CompSci BaldwinSHS/NassauCC
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