Doubling Time #2-3

Question

Suppose that the number of cells in a tumor doubles every 5 months. Considering that the tumor begins with a single cell, answer the following:

How many cells will there be after 3 years?

(Check time units and round to the nearest whole number.)

 

 How many cells will there be after 5 years?

(Check time units and round to the nearest whole number.)

Answer 1

Hi Jesse,

      I noticed the you have several Doubling half life questions. So I'll do a general treatment.

If we let y(t)=y₀e^kt where y₀ = the initial amount present, y(t) is the amount at time t and k is the growth rate. k>0 is called exponential growth and k<0 is called exponential decay.

     If you are given that the amount doubles every T months then y(T)=y₀e^kT  and y(T) =2*y₀ so we can write the equation as 2=e^kT, taking ln of both sides gives ln(2) =kT solving for k gives k=ln(2)/T. For the first problem T=5 months or 5/12 years so k=.69314*12/5=1.664 yr^-1 . Now we can find the answers y₀=1

 

Y(3)=e^1.663*3 = 147 cells rounded to the nearest whole number.

y(5)=e^1.663*5 = 4085 cells

 

Hope this helps

Jim

Answer 2

P = P₀2^m/5

 

Where:

• P = the population of the bacteria
• P₀ = the initial (starting) population = 1 cell
• m = months

 

For 3 years = 3*12 = 36 months:

P = (1)*2^36/5

P = 2^7.2

P = 147 cells

 

For 5 years = 60 months:

P = (1)*2^60/5

Can you finish it from here?