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

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The discussion centers on solving a differential equation that models tumor growth, where the birth rate of tumor cells decreases exponentially over time. The equation is given as dP/dt=(B_0)e^(-at)P, with initial conditions of 10^6 cells and a growth rate of 3*10^5 cells per month. Participants share their progress in determining the constants B_0 and a, with one confirming B_0 as 3/10. They discuss using a system of equations to find the population after six months, which doubles, and the challenges of solving the equation, including the use of the Lambert W function. The conversation highlights the complexities of differential equations in biological modeling.
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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...
 
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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