# Honors 2201-4 Fall 2003 Homework 8 Solutions Section 3.2

```Honors 2201-4
Homework 8 Solutions
Fall 2003
Section 3.2
(21)
dy
= −8(y − 10)
dt
y(0) = 20
dz
dy
=
dt
dt
dz
= −8z
dt
z = ce−8t = y − 10
y = 10 + ce−8t
20 = 10 + ce0 = 10 + c
c = 10
y = 10 + 10e−8t
z = y − 10
(34)
u = 10 − 8e−kt
u(1) = 4
4 = 10 − 8e−k
−6
= e−k
−8
3
ln
= −k
4
4
k = ln
3
(41)
dT
= −k(T − 21)
dt
T (0) = 26
y = T − 21
1
T (1) = 25
dy
dT
=
dt
dt
dy
= −ky
dt
y = ce−kt = T − 21
T = 21 + ce−kt
Evaluate at t = 0 to find c.
26 = 21 + c ⇒ c = 5
T = 21 + 5e−kt
Evaluate at t = 1 to find k.
25 = 21 + 5e−k
4
= e−k
5 4
−k = ln
5
5
k = ln
4
−t ln (5/4)
T = 21 + 5e
Find td , such that T (td ) = 37.
37 = 21 + 5e−td ln (5/4)
16
= e−td ln (5/4)
5
5
16
= −td ln
ln
5
4
ln (16/5)
td = −
≈ −5.213
ln (5/4)
The time of death was approximately 5 hours and 13 minutes before 8:50am, or 3:37am.
This suggests that the janitor committed the crime.
(44)
The initial value problem is
dy
= −ky
dt
y(0) = y0
where k &gt; 0. The negative sign indicates that the substance is decaying, so y is decreasing
over time. The solution of the initial value problem is
y = y0 e−kt
2
The half-life, t1/2 , is the time it takes for y to decay from the level y0 to the level y0 /2.
y0
= y0 e−kt1/2
2
1
= e−kt1/2
2
1
−kt1/2 = ln
2
ln 2
t1/2 =
k
y=
(45) We can use the half life formula found in problem (44) to solve for the decay constant
k.
15 =
ln 2
ln 2
⇒k=
k
15
Then, we can use the solution to the initial value problem in (44) to determine the time, td ,
that superman is in danger, i.e. we need to determine td , such that y(td ) = .1y0 = y0 /10.
y = y0 e−kt
1
y0 = y0 e−td ln 2/15
10
1
= e−td ln 2/15
10
ln 2
1
= −td
ln
10
15
15 ln 10
td =
≈ 49.83
ln 2
Superman is in danger for approximately 49 hours and 50 minutes.
Project 1: Glucose in the Bloodstream
(a) The overall change in the glucose concentration, dG/dt, is equal to the difference between
the total rate of input of glucose into the bloodstream, c, and the total rate of output of
glucose from the bloodstream. We are told in the introduction that the total rate of output of
glucose from the bloodstream is proportional to the amount present in the bloodstream with
proportionality constant r. Thus, the rate of output is rG. Putting these pieces together,
we see that the overall rate of change is c − rG (rate of input - rate of output). Therefore,
the amount of glucose present at time t, G(t), can be modeled by a differential equation of
the form dG/dt = c − rG with G(0) = G0 .
3
(b) In this part you are asked to solve the initial value problem
dG/dt = c − rG with G(0) = G0 .
dG
dy
= −r
dt
dt
y = c − rG
dy
= −r(c − rG) = −ry
dt
y = ae−rt = c − rG
c − ae−rt
G=
r
Use the initial condition G(0) = G0 to solve for the constant a.
c − ae0
c−a
=
r
r
a = c − rG0
c + (rG0 − c)e−rt
=
r
G0 =
G=
c − (c − rG0 )e−rt
r
(c) The level of glucose in the bloodstream is constant when the rate of change is zero, or
dG/dt = 0. This happens when c−rG is always equal to zero. Using the fact that G(0) = G0
is the initial level of glucose in the blood, we find that if c = rG0 , the total rate of input will
equal the total rate of output, and the glucose level in the bloodstream will remain constant
at its initial level of G0 .
(d) We are given c = 10, G0 = 2, r = 3. Therefore, the general equation for the amount of
glucose in the bloodstream t hours after the infusion starts is
G(t) =
10 + (3(2) − 10)e−3t
10 − 4e−3t
=
3
3
After 2 hours the level of glucose is
G(2) =
10 − 4e−3(2)
10 − 4e−6
=
≈ 3.33
3
3
4
Let t3 be the amount of time it takes for the glucose level to reach 3 grams. t3 must satisfy
the following equation.
10 − 4e−3t3
3
9 = 10 − 4e−3t3
1
= e−3t3
4
1
= 3t3
− ln
4
ln 4
t3 =
≈ 0.462
3
3=
The time is approximately 27 minutes and 44 seconds.
(e) We are given G0 = 2, r = 4 and G(3) = 3.5. We want to solve for c.
3.5 =
c + (4(2) − c)e−4(3)
c + (8 − c)e−12
=
4
4
−12
14 = 8e
+ 1 − e−12 c
c=
14 − 8e−12
≈ 14
1 − e−12
The doctor must infuse the glucose into the patients body at a rate of 14 grams per hour.
5
```