Homework 3 Solutions
Math 314, Fall 2016
Problem 1.
The Excel charts for this problem are shown in Homework3Problem1.xlsx. The allometric
exponents are as follows:
Brain Weight: 0.7553 (very close to 3/4)
Life Span: 0.2156 (very close to 1/5, though perhaps consistent with 1/4)
Gestation Period: 0.2433 (very close to 1/4)
Problem 2.
In a growing organism, metabolic energy is used both to maintain existing cells and to create
new cells. We can model this relationship using the differential equation
Y = Yc N + Ec
dN
dt
(1)
where
• Y is the metabolic rate of the organism,
• N is the number of cells of the organism,
• Yc is the metabolic rate of each cell, and
• Ec is the amount of energy required to make a new cell.
Here Y and N vary over an organism’s life, but Yc and Ec are assumed to be constant.
Part (a)
In this equation, the Yc N term represents the rate at which energy is being expended to
maintain the existing cells. In particular, Yc is the metabolic rate of each cell, and N is the
number of cells, so multiplying these gives the total metabolic rate of all the existing cells.
The second term Ec (dN/dt) represents the rate at which energy is being used to make new
cells. This the product of the energy Ec required to make a new cell and the rate dN/dt at
which new cells are being made.
Part (b)
Let M denote the mass of the growing organism, and suppose that Y = aM 3/4 for some
constant a (Kleiber’s allometric law). Then equation (1) above can be written
aM 3/4 = Yc N + Ec
1
dN
dt
(2)
But M = N Mc , where Mc is the (constant) mass of each cell, and hence N = M/Mc . Since
Mc is constant, it follows that
1 dM
dN
=
.
dt
Mc dt
Substituting this into (2) gives
M
1 dM
3/4
aM
= Yc
+ Ec
Mc
Mc dt
and rearranging yields
Ec
dM
= aMc M 3/4 − Yc M.
dt
(4)
Part (c)
The adult mass Mmax of the organism should be the equilibrium solution to equation (4).
Setting dM/dt equal to 0 gives
0 = aMc M 3/4 − Yc M
and solving for M yields
M =
aMc
Yc
4
Thus
Mmax =
aMc
Yc
.
4
.
(5)
Part (d)
From equation (5), we see that
Yc =
aMc
1/4
Mmax
Substituting this into equation (4) gives
Ec
dM
aMc M
= aMc M 3/4 −
1/4
dt
Mmax
Dividing through by Ec and factoring the right side gives
dM
aMc
1/4
=
M 3/4 Mmax
− M 1/4 .
1/4
dt
Mmax Ec
This has the form
dM
1/4
− M 1/4 .
= 4k M 3/4 Mmax
dt
where
k =
aMc
1/4
4 Mmax Ec
2
(6)
(7)
Part (d)
We can use separation of variables to solve equation (6). Rearranging gives
dM
1/4
4M 3/4 Mmax − M 1/4
so
Z
= k dt
Z
dM
1/4
4M 3/4 Mmax − M 1/4
=
k dt
the integral on the right is kt + C. For the integral on the left, substituting u = M 1/4 yields
Z
1/4
dM
Mmax − M 1/4 + C
=
−
ln
1/4
4M 3/4 Mmax − M 1/4
and thus
1/4
− ln Mmax
− M 1/4 = kt + C
Then
1/4
Mmax
− M 1/4 = Ae−kt
for some constant A, so
1/4
M = Mmax
− Ae−kt
3
4