Suppose that at 5 days, the number of cells is growing at a rate of 150 cells per day. Determine a number of cells after 12 days if there were 100 cells initially.

Question

Under certain conditions, the number of cancer cells N(t)  at time( t) increases at a rate of N'(t) =50e^kt . Suppose that at 5 days, the number of cells is growing at a rate of 150 cells per day. Determine a number of cells after 12 days if there were 100 cells initially.

Answer 1

Integrate N'(t): N(t)=c+(50/k)e^(kt). N'(5)=50e^(5k)=150, so e^5k=3, 5k=ln(3), k=ln(3)/5=0.2197.

Initially at t=0 there are 100 cells, so N(0)=c+250/ln(3)=100; c=100-250/ln(3)=-127.56. Therefore:

N(t)=250(e^(ln(3)t/5)-1)/ln(3)+100.

When t=12, N(12)=250(e^(2.4ln(3))-1)+100

ln(3)=1.0986 approx, so N(12)=250(e^(2.4*1.0986)-1)/1.0986+100=3051 cells.

[When t=5, N(5)=555 cells.]