# Universal law of growth

• January 13th 2011, 09:18 AM
mathsohard
Universal law of growth
I need a serious help please

Let Y be the metabolic rate of an organism, Yc the metabolic rate of a single cell, Nc(t) the total number of cells at time t, mc the mass of a cell, and Ec the energy required to create a new cell. The cell properties, Ec, mc, and Yc, are assumed to be constant and invariant with respect to the size of the organism. Thus

Y = YcNc + Ec(dNc/dt).

Let m be the total body mass of the organism at time t, and m = mcNc.(Note that Nc is the total number of cells in a body and is proportional to mass m, while the total number of capillaries Nn is proportional to 3/4 power of m.) Y=Y0(m)^(3/4)

a.show that the above equation can be written as
dm/dt = am^(3/4) - bm
with a = Y0mc / Ec and b = Yc/Ec

b.Let m = M be the mass of a matured organism, when it stops growing (dm/dt = 0). Find M, and show that the above equation can be rewritten as
dm/dt = am^(3/4)[1-(m/M)^(1/4)].

c.Let r = (m/M)^(1/4), and R = 1-r. Then the above equation becomes
dR/dt = -(a/4M^(1/4))R.

Solve this simple ordinary differential equation and show that a plot of ln(R(t)/R(0)) vs. the no-dimensinal time at/(4M^(1/4)) should yield a straight line with a slope -1 for any organism regardless of its size.

d. Based on this scaling for time t, argue that, for a mammal, the interval between heartbeats should scale with its size as M^(1/4).

• January 13th 2011, 11:34 AM
dwsmith
Quote:

Originally Posted by mathsohard
I need a serious help please

Let Y be the metabolic rate of an organism, Yc the metabolic rate of a single cell, Nc(t) the total number of cells at time t, mc the mass of a cell, and Ec the energy required to create a new cell. The cell properties, Ec, mc, and Yc, are assumed to be constant and invariant with respect to the size of the organism. Thus

Y = YcNc + Ec(dNc/dt).

Let m be the total body mass of the organism at time t, and m = mcNc.(Note that Nc is the total number of cells in a body and is proportional to mass m, while the total number of capillaries Nn is proportional to 3/4 power of m.) Y=Y0(m)^(3/4)

a.show that the above equation can be written as
dm/dt = am^(3/4) - bm
with a = Y0mc / Ec and b = Yc/Ec

b.Let m = M be the mass of a matured organism, when it stops growing (dm/dt = 0). Find M, and show that the above equation can be rewritten as
dm/dt = am^(3/4)[1-(m/M)^(1/4)].

c.Let r = (m/M)^(1/4), and R = 1-r. Then the above equation becomes
dR/dt = -(a/4M^(1/4))R.

Solve this simple ordinary differential equation and show that a plot of ln(R(t)/R(0)) vs. the no-dimensinal time at/(4M^(1/4)) should yield a straight line with a slope -1 for any organism regardless of its size.

d. Based on this scaling for time t, argue that, for a mammal, the interval between heartbeats should scale with its size as M^(1/4).

$\displaystyle Y0m^{3/4}=YcNc+Ec\frac{dNc}{dt}\Rightarrow \frac{Y0m^{3/4}}{Ec}=\frac{YcNc+Ec\frac{dNc}{dt}}{Ec}\Rightarro w \frac{Y0m^{3/4}}{Ec}=bNc+\frac{dNc}{dt}\Rightarrow\cdots$

Since every question is related to the above equation, finishing this should allow you to do the rest.
• January 13th 2011, 11:54 AM
mathsohard
Is it possible to go little further for me ??? because I still have some difficulties dealing with derivatives :(
• January 13th 2011, 12:05 PM
dwsmith
Quote:

Originally Posted by mathsohard
Is it possible to go little further for me ??? because I still have some difficulties dealing with derivatives :(

Multiply by mc

$\displaystyle\frac{Y0m^{3/4}mc}{Ec}=bNcmc+mc\frac{dNc}{dt}$

$\displaystyle\Rightarrow am^{3/4}=b(Ncmc=m)+\frac{d(Ncmc=m)}{dt}\Rightarrow am^{3/4}-bm=\frac{dm}{dt}$
• January 13th 2011, 12:28 PM
mathsohard
for d, how can I interpret my equations??? I got up to C.
• January 13th 2011, 12:31 PM
dwsmith
Quote:

Originally Posted by mathsohard
for d, how can I interpret my equations??? I got up to C.

• January 13th 2011, 12:36 PM
mathsohard
dR/dt = -(a/4M^(1/4))R.

Solve this simple ordinary differential equation and show that a plot of ln(R(t)/R(0)) vs. the no-dimensinal time at/(4M^(1/4)) should yield a straight line with a slope -1 for any organism regardless of its size.

I solved this................... ln(R(t)/R(0)) = -(a/4M^(1/4))t thus yield a straight line with a slope -1 vs the no-dimensinal time at/(4M^(1/4))
• January 13th 2011, 12:38 PM
dwsmith
Quote:

Originally Posted by mathsohard
dR/dt = -(a/4M^(1/4))R.

Solve this simple ordinary differential equation and show that a plot of ln(R(t)/R(0)) vs. the no-dimensinal time at/(4M^(1/4)) should yield a straight line with a slope -1 for any organism regardless of its size.

I solved this................... ln(R(t)/R(0)) = -(a/4M^(1/4))t thus yield a straight line with a slope -1 vs the no-dimensinal time at/(4M^(1/4))

What is capital M?
• January 13th 2011, 12:47 PM
mathsohard
mass of a matured organism, oh so since it is all matured it should scale with its size as M^1/4 huh ??
• January 13th 2011, 12:49 PM
dwsmith
Quote:

Originally Posted by mathsohard
mass of a matured organism, oh so since it is all matured it should scale with its size as M^1/4 huh ??

I am not a biology person so it beats me.