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Unformatted text preview: lue of =
.0015 + C
b) Solve DE:be the an initial the dt ) rthe original tvalue of the (years). Since
x
y
the + IC: (0) x + C.
Apply ln r= ln= is continuous 05.
V (y)compounding 0.05 ) C = 0.we have the DE:
t.
x
Interest rate at time dVr(t) = 0.05 0.0015t.
t:
Solution:
= r(t)Vdr = .05 0.0015t)V
= (0 0.
(0) = 0. Since
6. (b) Let V (tinterest rate r(t): the investment0015
(a) IVP for ) be the value of
(dollars) at time tr(years).05 .
dt
dt
the compounding = 2000.
initial rate is 5%
with the IC: V (0) is continuous we have the DE:year
rate decreases 0.15% per
Z
Z
dV
Solve DE: dr =
0.0015 dt ) r = 0.0015t + C .
= r(t)V = (0.05 0.0015t)V
Z1
dt
Apply IC: r(0) = 0.05 )ZC = 0.05.
dV = (0.05 0.0015t) dt
Solve DE:
Interest IC: V time : r(t)
V
with therate at (0) =t2000. = 0.05 0.0015t.
2
ln value the investment (dollars)
(b) Let V (t) be the V  =of 0.05t 0.00075t + C1 at time t (years). Since
2
Z1
the compounding V continuoust we00075t the DE:
is = Z 0.05 0. have
Ce
dV = (0.05 0.0015t) dt
Solve DE:
dV
V = 2000 ) C = 2000.
Apply IC: V (0)
= r(t)V = (0.05 0.0015t)V
ln V  = 0.05t 0.00075t2 + C1
dt
2
Value of investment at time.05:t V.00075t2 2000e0.05t 0.00075t .
t 0 (t) =
V 2000. 0
= Ce
with the IC: V (0) =
(c) As shown in class, the value of the ﬁxed rate investment will be Vf (t) =
Apply 04t. V
Z
2000e0.IC: Z (0) = 2000 ) C = 2000. variable rate is better than the ﬁxed
From the graph below, the
1
dV = time year .(t) =t) dt e0.05t 0.00075t2.
0 0015 2000
Solve DE:
rate for bothV5 year and(0.05t: V investments.
Value of investment at 10
2
(c) With the interest  = ﬁxedt at of00075ﬁxedCrate investment will be Vf (t) =
As shown in ln V rate 0.05 0. the2 t appropriate DE is
class, the value 4%, the + 1
c)
0.05t 0.00075t
2000e0.04t. From V =graph below, the variable rate is better than the ﬁxed
the Ce
rate for both 5 year and 10 =dV investments.
Apply IC: V (0) = 2000 ) C year = 0.04V,
2000.
dt
2
Value of investment at time t: V (t) = 2000e0.05t 0.00075t .
(c) which we can...
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 Winter '14

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