Please help with part (b), (c) and (d) !!! Consider the growth of a tumor in the
ID: 2902427 • Letter: P
Question
Please help with part (b), (c) and (d) !!!
Consider the growth of a tumor in the body. In the early stages the growth rate is
limited by the amount of nutrients (e.g. oxygen) that can be absorbed through its
surface. This has to be used for growth over the full volume, V (t). In this phase,
tumors are rarely dangerous. Later, using a process known as chemotaxis (the tumor
releases a chemical that lures" blood vessels toward it), blood vessels may attach to
the tumor, allowing an almost unlimited supply of nutrients.
(a) Assuming the tumor to be spherical, write a di erential equation for the growth
rate of the radius, r(t), of the tumor if it is in the rst (avascular) phase and
the rate of change of volume is proportional to the surface area divided by the
volume with constant of proportionality k1. Solve the equation if the initial radius
r(0) = 1 mm. Note the volume of a sphere of radius r is V = 4
3r3 and the surface
area is S = 4r2. Make sure you convert everything into units of radius.
(b) Suppose the tumor reaches the vascular phase (with blood vessels feeding directly
into the tumor) at time t = t0 months. The volume growth rate is then propor-
tional to the volume, with constant k2. Write a new equation assuming that the
e ect of absorption in (a) is negligible and can be neglected. Solve the modi ed
problem, starting from the beginning of the vascular phase at t = t0 and match-
ing with the solution in (a). There will now be two unknown constants in the
problem representing the two rates, k1 and k2.
(c) Once the tumor has been discovered at t > t0, a weekly dose of radiotherapy kills
a volume of cells on the surface of the tumor proportional to the surface area
(constant k3). Include a term for this in the model and determine how big k3
needs to be to shrink the tumor. Comment on this result in terms of when it may
be possible.
(d) Comment using an equation on what would have happened if the tumor was
discovered (and radiotherapy administered) well before the vascular phase, i.e.
t < t0?
Explanation / Answer
a)
dV/dt = k1* A
V= 4 /3 pi*r^3
dV/dt = 4 pi* r^2 dr/dt
so ,
4pi* r^2 dr/dt = k1 *4pi* r^2
dr/dt = k1
solving it with r0 = 1mm
r = k1t + r0
r = 1+ k1* t ..............( r is in mm)
b)
now,
dV/dt = k2 V
4pi* r^2 dr/dt = 4 /3 pi*r^3
dr/dt = r/3,
solving it,
ln r = 3*k2 t + c
r = A exp( 3* k2* t)
according to , answer in part 1,
at t= t0,
r0 = 1+k1*t0
so ,
1+ kt0 = A exp ( 3*k2*t0)
A= (1+k1*t0) exp(- 3k2* t0)
so ,
r = (1+k1*t0) exp(- 3k2* t0) *exp( 3* k2* t)
r= (1+k1*t0) exp( 3 k2(t- t0))
c)
for third case ,for radiotherapy term
dV/dt = - k3*A,
so ,
dr/dt = - k3,
r= -k3*t
so ,
after t>t0,
net , r = (1+k1*t0) exp( 3 k2(t- t0)) -k3*t
for r= 0 ,
k3 = (1+k1*t0) exp( 3 k2(t- t0)) /t
d)
if tmour has been discovered before t<t0,
net ,
dr/dt would have been equal to k1-k3
i.e, r= (k1-k3) t +1,
so ,it could have been killed by choosing a proper value or k3