##### Help Calculus Problem

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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.

Jul 22nd, 2015

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)

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Jul 22nd, 2015

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=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.]

Jul 22nd, 2015

oh... book has gone by integration root .. i thouht you may not be aware of that

Jul 22nd, 2015

Is that my answer is correct ?

Jul 22nd, 2015

Jul 22nd, 2015

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=100-250/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)

Jul 22nd, 2015

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Jul 22nd, 2015
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Jul 22nd, 2015
Oct 19th, 2017
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