A model of tumour growth under chemotherapy from time t = 0 to t = t₁ > 0 is dC dt = -C log C Cmax DC 1+ D (*) where, at time t, C(t) is the size of the tumour, Cmax > 0 is the maximum size of the tumour, a constant, so that 0 < C(t) ≤ Cmax, and D(t) ≥ 0 is the rate at which the drug is administered.
A model of tumour growth under chemotherapy from time t = 0 to t = t₁ > 0 is dC dt = -C log C Cmax DC 1+ D (*) where, at time t, C(t) is the size of the tumour, Cmax > 0 is the maximum size of the tumour, a constant, so that 0 < C(t) ≤ Cmax, and D(t) ≥ 0 is the rate at which the drug is administered.
Chapter6: Exponential And Logarithmic Functions
Section6.8: Fitting Exponential Models To Data
Problem 3TI: Table 6 shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to...
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![A model of tumour growth under chemotherapy from time t = 0 to
t = t₁ > 0 is
dC
dt
= -C log
C
Cmax
dx
dt
where, at time t, C(t) is the size of the tumour, Cmax > 0 is the maximum
size of the tumour, a constant, so that 0 < C(t) ≤ Cmax, and D(t) ≥ 0 is the
rate at which the drug is administered.
Show that by making the change of variable, ax = log(C/Cmax),
equation (*) becomes
=-X
DC
1+ D
D
1+ D'
t1
S[D] = [" dt D(t).
(*)
where - < x≤ 0.
It is required to reduce the size of the tumour from Co at t = 0 to C₁ at
t = t₁. Let xo = log(Co/Cmax) and x₁ log(C₁/Cmax). For the health of the
patient, it is desired to minimise the total amount of drug administered,
which is given by the functional
=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6c8ed7d-75cc-4e27-869e-3ad6a1efc0b4%2Fe30d0ef3-9a00-4187-b475-b9fab5b6bd82%2Fbnha9th_processed.png&w=3840&q=75)
Transcribed Image Text:A model of tumour growth under chemotherapy from time t = 0 to
t = t₁ > 0 is
dC
dt
= -C log
C
Cmax
dx
dt
where, at time t, C(t) is the size of the tumour, Cmax > 0 is the maximum
size of the tumour, a constant, so that 0 < C(t) ≤ Cmax, and D(t) ≥ 0 is the
rate at which the drug is administered.
Show that by making the change of variable, ax = log(C/Cmax),
equation (*) becomes
=-X
DC
1+ D
D
1+ D'
t1
S[D] = [" dt D(t).
(*)
where - < x≤ 0.
It is required to reduce the size of the tumour from Co at t = 0 to C₁ at
t = t₁. Let xo = log(Co/Cmax) and x₁ log(C₁/Cmax). For the health of the
patient, it is desired to minimise the total amount of drug administered,
which is given by the functional
=
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