關於數學CALCULUS功課

Let y be the number of the population
Set up the differential equation
dy/dt = (2^3)y
∫(1/y)dy = 8 ∫dt
lny = 8t+c where c is a constant
y=Ae^(8t) where A is a constant -----------(1)

Using initial condition and (1), when t=0, y=20
20=A(1)
y=20e^(8t) ---------(2)

so expression for the number of cells after t hours is y=20e^(8t)

Using (2), for the number of cells after 3 hours, put t=3, the number required is 20e^(8*3)

Using (2), differentiate both sides w.r.t. t,
dy/dt = 180e^(8t)
for the rate of growth after 9 hours, put t=9, required rate of growth is 180e^(8*9)

Using (2), put y=20000
20000=20e^(8t)
ln1000=8t
t=(ln1000)/8
so the population reach 20,000 cells when t=(ln1000)/8

relative growth rate就..應該係8倍, 但係唔知寫8y定係就咁寫8

You can find a similar example here:

http://math.ucsd.edu/~wgarner/math4c/textbook/chapter4/expgrowthdecay.htm