Solving the Tumor Growth Problem: Find B_0, a & More

  • Thread starter Thread starter missteresadee
  • Start date Start date
  • Tags Tags
    Growth
missteresadee
Messages
2
Reaction score
0
Problem: A tumor may be regarded as a population of multiplying cells. It is found empirically that the "birth rate" of cells in a tumor decreases exponentially with time, so that birth rate is given by B(t)=(B_0)*e^(-a*t) where B_0 and a are constants. Consequently, the governing equation for the tumor population P(t) at any time t is
dP/dt=(B_o)e^(-at)P

Solve this differential equation exactly. Suppose the initially at t=0, there are 10^6 tumor cells, and that P(t) is then increase at a rate of 3*10^5 cells per month. After 6 months, the tumor has double in population. Find a, and find the behavior of the tumor population as t goes to infinity

I know from the given information, I can find B_0 by using this equation
3*10^5=(B_0)*e^(-a*0)*10^6
given rate of change is 3*10^5; t=0, and p=10^6
B_0=3/10

I am having trouble finding my constant C, a...
 
Physics news on Phys.org
missteresadee said:
Problem: A tumor may be regarded as a population of multiplying cells. It is found empirically that the "birth rate" of cells in a tumor decreases exponentially with time, so that birth rate is given by B(t)=(B_0)*e^(-a*t) where B_0 and a are constants. Consequently, the governing equation for the tumor population P(t) at any time t is
dP/dt=(B_o)e^(-at)P

Solve this differential equation exactly. Suppose the initially at t=0, there are 10^6 tumor cells, and that P(t) is then increase at a rate of 3*10^5 cells per month. After 6 months, the tumor has double in population. Find a, and find the behavior of the tumor population as t goes to infinity

I know from the given information, I can find B_0 by using this equation
3*10^5=(B_0)*e^(-a*0)*10^6
given rate of change is 3*10^5; t=0, and p=10^6
B_0=3/10

I am having trouble finding my constant C, a...

What do you get if you solve the differential equation to work out P(6 months) in terms of a, and set that equal to 2P_0?
 
Yup! I figured it out, I had two system of equations with two unknowns which I can solve =) I was trying to use maple program to solve this equation, and finally figured out how to display the answers in approximation instead of weird exact numbers. THANKS!
 
I'm assuming I'm in your class (lol).

Can you explain how you were able to solve this? I tried solving for B0, got 3/10. Tried solving for P, got e^(-Be^(-at))/a

Plugging B and 2P into the equation for P gives a lambertW function which I'm assuming is far beyond the scope of DE.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top