In a follow-up study to the tumor growth study described in Example 4, mice were
ID: 2855036 • Letter: I
Question
In a follow-up study to the tumor growth study described in Example 4, mice were infected with a relatively aggressive line of proliferating clonogenic cells that grew exponentially with a doubling time of approximately 1.8 days. Each mouse was given a dose of 20mg/kg of cisplatin per treatment with the following results: At the time of the therapy, the average tumor size was approximately 0.6cm . After treatment, 99.10% of the proliferating cells became quiescent cells and decayed with a half-life of approximately 4.4 days. a. Write a function V(t) that represents the size of the tumor (proliferating plus quiescent cells) t days after therapy. b. Determine at what point in time the tumor starts to regrow and therapy should be readministered.
Explanation / Answer
V(t)=p(t)+Q(t)
Proliferating cells are increasing at an exponential rate and have initial volume of p(0)=0.01×0.60= 0.006cm^3
Hence p(t)=0.006e^at
Since doubling time is 1.8 days
P(1.8)=2(0.006) t=1.8
0.006e^1.8a=0.012
Some for a
e^1.8a=2
a=ln2/1.8=0.38
Hence p(t)= 0.006e^0.38t
Similarly we have Q(0)=0.91(0.6)=0.546
Q(t)=0.546e^by
To solve b
Given half life is 4.4days ,
Q(4.4)=0.6(0.546)
0.546e^4.4b=0.6(0.546)
e^4.4b=0.60
b=ln(0.6)/4.4=-0.11
Q(t)=0.546e^-0.11t
This v(t)=0.006e^0.38t+0.546e^-0.11t
b)v(t) is increasing or decreasing.compute it's derivative
v'(t)=0.0022e^0.38t-0.60e^-0.11t
v'(0)= - 0.5978
0.0022e^0.38t-0.060e^-0.11t=0
e^0.49t=27.3
0.49t=ln27.3
t=ln27.3/0.49
t=6.74 days
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