Thread: 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).

I am completely lost,,,, please help ,,,, Thank you

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).

I am completely lost,,,, please help ,,,, Thank you

Since every question is related to the above equation, finishing this should allow you to do the rest.

Is it possible to go little further for me ??? because I still have some difficulties dealing with derivatives

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

for d, how can I interpret my equations??? I got up to C.

Originally Posted by mathsohard

for d, how can I interpret my equations??? I got up to C.

What is your C?

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))

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?

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

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.