The growth rate of an organism is often measured using carbon biomass as the "currency." The von Bertalanffy growth model is
m′ = ax^2 − bx^3
where m is its biomass, x is some characteristic length of the organism, a is its biomass assimilation rate, and b is its biomass use rate. Thus, it assimilates nutrients proportional to its area, and uses nutrients proportional to its volume.
Assume m = rhox^3 [and rho constant] and rewrite the model in terms of x.
Determine the dimensions of the constants a, b, and rho.
Select time and length scales rho/a and a/b, respectively, and reduce the problem to dimensionless form.
If x(0) = 0, find the length x and time t.
Does this seem like a reasonable model?

a) dx/dt = (a/3ρ) - (bx/3ρ)

b) a has units of g/cm².s

b has units of g/cm³.s

ρ has units of g/cm³

c) (ρb/a²)(dx/dt) = (b/3a) - (b²x/3a²)

d) x(t) = (a/b) (1 - e⁻ᵇᵗ/³ᵖ)

e) The model makes sense because the function obtained is a growing function, that is, this function increases with time. And the organism being studied too should increase in length and biomass with time. So, yes the model makes a lot of sense.

Step-by-step explanation:

m' = dm/dt

m' = ax² - bx³

m = biomass

a = biomass assimilation rate

b = biomass use rate

x = characteristic length of the organism

a) m = ρx³ (ρ is constant)

dm/dt = (dm/dx) × (dx/dt) (chain rule)

dm/dx = 3ρx²

dm/dt = 3ρx² (dx/dt)

m' = dm/dt = 3ρx² (dx/dt)

m' = ax² - bx³

3ρx² (dx/dt) = ax² - bx³

Divide through by 3ρx²

dx/dt = (a/3ρ) - (bx/3ρ)

b) m = ρx³

m is in the units of mass, say grams; x is in the units of length, say centimetres (cm)

ρ = m/x³ that is, units of (g/cm³)

From the differential equation,

dx/dt = (a/3ρ) - (bx/3ρ)

dx/dt has units of length/time = cm/s,

So, a/ρ = cm/s

a = ρ × cm/s = (g/cm³) × (cm/s) = g/cm².s

bx/ρ = cm/s;

b = (ρ/x) × (cm/s) = (g/cm⁴) × (cm/s) = g/cm³.s

c) Time scale = ρ/a

Length scale = a/b

To make the equation dimensionless, we need to multiply through by time/length scale, since the differential equation is in a length/time scale

Time/length = (ρ/a) ÷ (a/b) = (ρ/a) × (b/a) = (ρb/a²)

Multiply through the differential equation

dx/dt = (a/3ρ) - (bx/3ρ)

(ρb/a²)(dx/dt) = (ρb/a²)(a/3ρ) - (ρb/a²)(bx/3ρ)

(ρb/a²)(dx/dt) = (b/3a) - (b²x/3a²)

d) dx/dt = (a/3ρ) - (bx/3ρ)

dx/[(bx/3ρ) - (a/3ρ)] = -dt

∫ dx/[(bx/3ρ) - (a/3ρ)] = -∫ dt

(3ρ/b) In [(bx/3ρ) - (a/3ρ)] = - t + C (C is the constant of integration)

At t = 0, x = 0

(3ρ/b) In [-(a/3ρ)] = C

C = (3ρ/b) In (-a/3ρ)

(3ρ/b) In [(bx/3ρ) - (a/3ρ)] = -t + (3ρ/b) In (-a/3ρ)

In [(bx/3ρ) - (a/3ρ)] = -tb/3ρ + In (-a/3ρ)

(bx/3ρ) - (a/3ρ) = (-a/3ρ) e⁻ᵇᵗ/³ᵖ

bx/3ρ = (a/3ρ) - (a/3ρ) e⁻ᵇᵗ/³ᵖ

bx/3ρ = a/3ρ (1 - e⁻ᵇᵗ/³ᵖ)

x = (a/b) (1 - e⁻ᵇᵗ/³ᵖ)

e) The model makes sense because the function obtained is a growing function, that is, this function increases with time. And the organism being studied too should increase in length and biomass with time. So, yes the model makes a lot of sense.