DIFFERENTIAL EQUATION

[sophie.tran999]

Can anyone help me with this question, i get stuck with a)
Q: in a certain bacterial culture, the rate of increse is proportional to the number of bacteria present.
a) if the number doubles in 3 hours, find the hourly growth rate.
i can deal with b) and c).

[b^3]

Let N represent the number of bacteria present
let t represent time
then
so at t=0 lets start of with n Bacteria
then at t=3, N=2n
solve the differential equation
flip it


solve for c, plug in t=0, and N=n
so
so the hourly growth rate will be k
so plug in t=3 and N=2n



So the hourly growth rate is
Wait I'm still wrong

[sophie.tran999]

thx, thats wat i did afetr readin the question 1st time, but look, the question askes to find the GROWTH RATE, not find t in term of N
the anwser is : dN/dt = 0.23N

[b^3]

1/3loge(2)=0.23 so that means that dN/dt=0.23N, I think that answers it right?

[b^3]

since growth rate =dN/dt

[sophie.tran999]

i think so. damn me lol
thx

[b^3]

Yeh it's right and works, I had myself stuck in a loop for a min there.

[sophie.tran999]

solve for c, plug in t=0, and N=n, this is important part. i missed this one
instead od plugging t=0 and N=o first, i just pluged t=3 and N=2N.