A quantity N is increasing with respect to time, t. It is increasing in such a way that N = ae^(bt) where a and b are constants. Given when t = 0, N = 20, and t = 8, N = 60, find the value: of a and b, and of dN/dt when t = 12

N = ae^bt When N = 20, t = 0. We can substitute this in.20 = ae^{b x 0} We know that b x 0 = 0, and that e⁰ = 1, so we can replace these.20 = a x 1 This means that a = 20
N = 20e^bt When N = 60, t = 8. We can substitute this in.60 = 20e^8b Now, we can rearrange, to get the e^8b on one side, and everything else on the other.(60/20 = 3)3 = e^8b Now, we can use laws of logs to find 8b, and therefore b. We know e^{ln 3} = 3, so:8b = ln 3 b = (ln 3)/8
Now, we know that N = 20e^{(1/8)(ln3)(t)}We can differentiate: For y = Ae^kt, dy/dx = kAe^kt. Therefore, dN/dt = (1/8)(ln3)(20)e^{(1/8)(ln3)(t)}If we substitute in t = 12, we can find a value for dN/dt using the calculator. This comes out to 14.3