Help Calculus Problem
Calculus

Tutor: None Selected  Time limit: 1 Day 
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.
Thank you for the opportunity to help you with your question!
n'(t) = 50 e^(kt)
t =5 we have give n'(t) = 150
there fore 150 = 50 e^(5t)
solving we get e^5t = 3 ; or k =0.2197
to find number of we can write N(t) = No e^(0.2197t)
t= 12 hence N(t) = 100 * e^(0.2197*12) = 1396.66 cells (Answer)
You can use my promo code: Saroj N , at study pool and enjoy $10 absolutely free….
Please find the solution enclosed here with. In case of any doubt please feel free to ask … If you need help in any assignment of math/ science … any online exam / discussion, Please contact for quick & quality services.
Hello, thanks for your help but can you look over my answer and tell me where I went wrong:
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=100250/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.]
oh... book has gone by integration root .. i thouht you may not be aware of that
Is that my answer is correct ?
I'm asking how did we get a different answers?
see k = 0.2197 is correct as per my method .. after that i am wrong if integration is taught to you than we can proceed as follows
Integrate N'(t) to find number of cells : N(t)=c+(50/k)e^(kt). now
Initially at t=0 there are 100 cells, so N(0)=c+250/ln(3)=100; c=100250/ln(3)=  127.56.
Therefore: N(t) (50/0.2197) e^(0.2197t) 127.56
When t=12, N(12)=(50/0.2197) (e^(0.2197* 12)  127.56
solve this ; you will get N(12) = 3050.14 or 3051 (Answer)
Secure Information
Content will be erased after question is completed.