Honors AP Calculus / Mr. Hansen
Posted 12/12/2000

Name: _________________________

1.

chaotic
no
The system was said to be deterministic. Although the system’s behavior is essentially unpredictable, the system does follow rules.

2.(a)

dM/dt = kM

(b)

dM/dt = kM
dM/M = kdt
ln |M| = kt + C₁
|M| = exp(kt + C₁) = Ce^{k t}
General solution: M = Ce^{k t} since M ³ 0 (M = mass)
[Note: On AP, you could write Ce^{k t} as general solution without showing work (unless specifically told to) since you are supposed to know that Ce^{k t} is always the general solution in a problem involving exponential growth or decay.]
Initial condition: 8 = Ce^k · 0 Þ C = 8.
Half-life condition: 4 = Ce^k · 4.2
4 = 8e^k · 4.2
0.5 = e^4.2k
ln 0.5 = 4.2k
k = (ln 0.5)/4.2 » –0.165
Particular solution: M = 8 exp((ln 0.5)t/4.2)

(c)

Set graphing window to [1, 10] ´ [1, 10], store (ln 0.5)/4.2 into variable K, and define Y₁ = KY. Run the SLPFLD program. After the display is complete, press 2nd QUIT and define Y₂ = 8e^(KX). Note that this is calculator notation; it is not acceptable to use the notation 8e^(kx) on the AP exam. Press GRAPH and transcribe the resulting graph onto your paper. If you wish to see the slope field with the particular solution overlaid, please visit me in Math Lab or in my office.

(d)

See table below. Note: Values were truncated for showing work but were kept at full precision internally. It is important to maintain full precision to avoid cumulative errors in your final answer.

t_i

M_i = M(t_i)

dM/dt(t_i, M_i) [see note]

M_i+1 » M_i + dM/dt(t_i, M_i) · Dt

0

0

8

kM = –0.165···(8)
= –1.320···

8 + (–1.320···)(0.5) = 7.339···

1

0.5

7.339···

kM = –0.165···(7.339···)
= –1.211···

7.339··· + (–1.211···)(0.5) = 6.734···

2

1.0

6.734···

kM = –0.165···(6.734···)
= –1.111···

6.734··· + (–1.111···)(0.5) = 6.1785···

3

1.5

6.179