Chapter 8: First Order Differential Equations
406 22.
Using the logistic growth model
c) = f′ ∫+bf bdt
7000 = P = 10 +−690e−kt 1 0 + 6 9 0 e = 6 9 8 0 k e − =8 k 7000 =8 k e − 300 10 + 690e−8k 70 3 70 3 −10
k
23.
a) y(2) = 28.9587 b) y(2) = 34.9444
23.
a) y(2) = 1.8421
ln (
40
=ebtf = bt = ef
aN b f(t) = −aN + aN e−bt+Fe−bt 0 b b = − )aN 1−e−bt+Fe−bt o ( ) b −bT + Fe−bT d) F1=− aN o b 1− e ( e) C = F0+
b) y(2) = 1.8422
F1
Technology Connection
• 586
Left to the student
3.
Page
Extended Life Science Connection 589
1.
a) Left to the student b) There is no growth rates to consider only decay due to loss of moisture c) ∫
−aN
= = =
= =
c) M = aNT = −(F1−F0p)bNT (1 −p)N = = (F1 −F0p)lnp 1− p
0 bT d) c( T) = c− e
e) c = c e−bT 0 c c0 = e−bT=p
2.
e−bT + Fe−bT o )
−aN 1 e−bT + Fe−bT o ) b ( − e−bT ) −aN 1 F1 − F0e−bT (− baN (1−p) F1 −F0p = − b = −(F1−F0p)b a (1 −p)N
c0 = K c(t) = c e−bt 0
−bT ln p − T
(1 −
baN 1−e−T(−lnp/t) +Fe−T(lnp/T) 0 − ) − baN (1−p)+F0p b a) The mass of the leafs consumed has to equal the the number of larva times the rate at which the consume the leafs times the times they spend consuming the leaves. b) F1
dc = ∫ −bdt c lnc = −bt + K c(t) = Ke−bt
lnp = b =
−aN ebt+C b −aN +Ce−bt b +C aN − b
F0
)
− 1 ln 40 ( 2070) 8 = 0.4933
Left to the student
−aNebtdt
= f
=
• Page
∫ebt
G(t) =
1 40 690 ( 3 ) − 8k =
−aN
−(pF0−F1)lnp 1− p 4.
a) M
=
a) Left to the student
=
b) There are no growth rates only decay rates due to loss of moisture and mass eaten by the larva
=
−(pF0−F1)lnp 1− p 0.324 . 0346 0307 ln 0324) − 0352 . . 0.352 [( . ) (− ) ] ( 1−0.324 0.352 0.01196grams
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