Unformatted text preview: 2 MA 36600 LECTURE NOTES: FRIDAY, JANUARY 23 The rate of change of the quantity of salt in the solution is the diﬀerence of the rate which ﬂows in and the
rate which ﬂows out. Hence our initial value problem is
dQ
1
Q(t)
= ·r−
· r,
Q(0) = Q0 .
dt
4
100
One method of ﬁnding QL is to solve the initial value problem for Q(t), then consider the behavior as t
increases without bound. We can solve the diﬀerential equation using an integrating factor. Note that this
equation is a ﬁrst order linear equation:
r
r
r
r
dQ
+
Q=
=⇒
p(t) =
,
g (t) = .
dt
100
4
100
4
The integrating factor is
t
r
µ(t) = exp
p(τ ) dτ = exp
t = ert/100 .
100
Multiplying the diﬀerential equation above by this function yields the equation
r
dQ
d rt/100
µ(t)
+ µ(t) p(t) Q = µ(t) g (t)
=⇒
e
Q(t) = ert/100 .
dt
dt
4
Integrating both sides gives
ert/100 Q(t) = 25 ert/100 + C =⇒ Q(t) = 25 + C e−rt/100 . =⇒ QL = lim Q(t) = 25 lbs. For the initial condition, set t = 0 to ﬁnd C = Q0 − 25. Hence the solution to the initial value problem is
Q(t) = 25 + (Q0 − 25) e−rt/100 t→∞ We give a second method of ﬁnding QL . Consider the diﬀerential equation Q + p(t) Q = g (t), and take
the limit as t increases without bound: QL + p(t) QL = g (t). Since QL is a constant, its derivative is zero.
Hence
g (t)
r/4
0 + p(t) QL = g (t)
=⇒
QL =
=
= 25 lbs.
p(t)
r/100
This trick works in many cases: Consider the ﬁrst order diﬀerential equation
dy
= G(t, y ).
dt
Say that we wish to ﬁnd an equilibrium solution
yL = lim y (t).
t→∞ If we take the limit as t increases without bound, we ﬁnd that yL = G(t, yL ). Since yL is a constant, we
actually seek solutions yL to the equation G(t, yL ) = 0 which hold for all time t.
Interest Rate Problems. Say that you wish to invest $1000 for the next year. You have two banks to
choose from: Bank A will give interest compounded quarterly, whereas Bank B will give interest compounded
daily. The rate of interest is determined by the Federal Reserve, say at r = 8% APR. Which bank gives you
the better return on your investment?
We make sense of these terms. Let S (t) denote the sum of money in the bank at time t. Note that
S (0) = $1000. If this interest rate r is compounded m times per year then the rate is r/m for that one
round, and so we have
r mτ
S (t + τ ) = 1 +
S (t)
m
after a time period τ years elapses. We wish to consider τ = 1 year. Note that m = 4 for interest compounded
quarterly, and m = 365 for interest compounded daily. In fact,
r m
lim 1 +
= er
m→∞
m
so that we may approximate daily compounding with continuous compounding i.e., m = ∞. (In particular,
r mt
dS
S (t) = 1 +
S (0) ≈ ert S (0)
=⇒
≈ r S.
m
dt
This means we can approximate interest problems with ﬁrst order linear diﬀerential equations.) Consider
the ratio: ...
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue.
 Spring '09
 EdrayGoins
 Rate Of Change

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