Solutions For Calculus Midterm Exam #2
1. (a)
0 1 2 3 4 5 6
−2
−1.5
−1
−0.5 0 0.5 1 1.5
2x 10−4
(b) CI = 3 is the equilibrium and it is locally stable. (Actually, CI is globally stable.) (c)
h′(c) = −q
V = −0.2 400 < 0 is a negative constant function. Therefore,
h′(CI) < 0.
By teh stability criterion, we can determine that the equilibrium CI = 3 is locally stable.
(d)
dC
dt = q
V (CI− C)
Z 1
CI− C = Z q
V dt
− ln |CI− C| = q V t + k ln |CI− C| = −q
V t − k
|CI− C| = ke−Vqt CI− C = ke−Vqt
C = CI− ke−Vqt C0 = CI− k
k = CI− C0
C(t) = CI− (CI− C0)e−Vqt
1
(e)
C(TR) = CI− (CI− C0)e−VqTR C(TR) − CI = −(CI − C0)e−VqTR
−(CI− C0)e−VqTR = e−1(C0− CI)
−q
V TR = −1 TR = V
q = 400
0.2 = 2000.
(f)
C(TR) − CI = e−1(C0− CI)
TR is not the exact time that the system needs to return to the equisetum, but the time that return to the point that is near the equisetum. And in this case, the ration of C(TR) − CI and C0− CI is a constant e−1. Therefore we have the same TRfor different incoming concentration.
2. (a) (i)
bS(0) − a = 0.2 × 1000 − 200 = 100 > 0 The disease can spread.
(ii)
S(0) = N − I(0) − R(0)
= 10000 − 1 − 0
= 9999
bS(0) − a = 0.01 × 9999 − 100
= 99.99 − 100
< 0 The disease can not spread.
(b)
dI
dS = a b Z
dI = Z
(a b
1
S − 1) dS
I = a
b ln |S| − S + C I0 = a
b
C = I0 + S0− a
b ln(S0)
= N −a
b ln(S0) I = a
b ln S − S + N −a
b ln(S0)
= N − S + a b ln(S
S0
)
2
(c)
I′ = −1 + a b
S0
Sˆ 1 S0
= 0 when
S =ˆ a b. (d)
Imax = I( ˆS)
= I(a b)
= N −a b +a
b ln(a/b S0
).
(e) Set x = a/b.
Imax(x) = N − x + x ln( x S0
) Imax(x) = −1 + ln(′ x
S0
) + x(S0
x )( 1 S0
)
= ln(x S0
)
< 0 for
x = a b < S0.
(f) The maximum rates of infection will occur for combinations of higher values of b and lower values of a, which does make sense.
3. (a)
dN
dt = 200 − N − bNX + 2X = 200 + 2X − (1 + bX)N
dX
dt = bNX − 3X = X(bN − 3)
dN
dt = 0 when 200+2X1+bX = N
dX
dt = 0 when X = 0 or N = 3b
If X = 0, then N = 200. If N = 3b, then 200 + 2X = 3b + 3X, X = 200 − 3b. Therefore, equilibria are (N, X) = (200, 0) and (N, X) = (3b, 200 − 3b).
(b)
3
b < 200 b > 3
200.
3
4. (a)
0 1−1/c1 1
1−1/c2
(b)
dp1
dt = c1p1(1 − p1) − p1 = p1(c1− 1 − c1p1)
dp2
dt = c2p2(1 − p1− p2) − p2− c1p1p2 = p2[c2(1 − p1) − c2p2− 1 − c1p1]
dp1
dt = 0 when p1 = 0 or p1 = 1 − c1
1
dp2
dt = 0 when p2 = 0 or p2 = (1 − p1) − c1
2 − cc1
2p1
Therefore, equilibria are (0, 0), (0, 1 − c1
2), (1 − c1
1, 0), and (1 − c1
1,c1
1 −cc1
2).
(c) The nontrivial equilibrium ( ˆp1, ˆp2) is (1−c1
1,c1
1−cc1
2). For the coexistence of two species, we need
ˆ
p1 = 1 − c1
1 > 0 ˆ
p2 = c1
1 − cc1
2 > 0 ˆ
p1+ ˆp2 = 1 − c1
1 +c1
1 −cc1
2 = 1 − cc1
2 < 1 That is, c2 > c21.
5. (a)
100 3
400 7
500 10
= 0.202 (b)
100 3
700 7
800 10
= 0.091 (c) Please read page 803 in your textbook.
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