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Current time:0:00Total duration:4:53

Exponential models & differential equations (Part 2)

AP.CALC:
FUN‑7 (EU)
,
FUN‑7.F (LO)
,
FUN‑7.F.1 (EK)
,
FUN‑7.F.2 (EK)
,
FUN‑7.G (LO)
,
FUN‑7.G.1 (EK)

Video transcript

in the last video we established that if we say a pop the rate of change of a population with respect to time is going to be proportional to the population we were able to solve that differential equation find a general solution which involves an exponential that the population is going to be equal to some some constant times e to some other constant times time in the last video we assumed time was days so let's just apply this just to just feel good that we can truly model population in this way so let's use it with some concrete numbers and once again you've probably done that before you probably started with the assumption that you can model with an exponential function and then you use some information some some conditions to figure out what the constants are we probably did this earlier in precalculus or algebra class but let's just do it again just so we can feel feel that this thing right over here is useful so let's give you some information let's say that at time equal zero the population is equal to 100 insects or whatever we're measuring the population of and let's say that at time equals let's say at time equals 50 the population so after 50 days the population is 200 so notice it doubled after 50 days so given this information can we solve for C and K and encourage you to pause the video and try to work through it on your own so this first initial condition is pretty straightforward to use because when T is equal to 0 P is a hundred so we could say we could say let's just use so based on this first piece of information we could say that 100 must be equal to C C times e times e to the K times 0 well that's that's just going to be e to the 0 well e to the 0 is just 1 so this is just the same thing as C times 1 and just like that we have figured out what C is so now we can write we can now write that the population is going to be equal to to 100 100 e to the KT and so you can see Express this way RC is always going to be our initial population so e to the KT e to the K T and now we can use the second piece of information so our population is 200 let's write that down so our population is 200 when time is equal to 50 after 50 days so 200 is equal to 100 e 100 e to the K times 50 right T is now 50 so we can let me just write that times K times 50 now we could divide both sides by 100 and we will get 2 is equal to e to the 50k e to the 50k then we could take the natural log of both sides natural log on the left hand side we get the natural log of 2 and on the right hand side the natural log of e to the 50k well that's just going to be that's just that's the power that you need to raise e to to get e to the 50k well that's just going to be 50k this is going to be 50 K all I did is took the natural log of both sides notice that this equation that I've just written expresses the same thing natural log of 2 is equal to 50 K that means e to the 50k is equal to 2 which is exactly what we had written there and now we can solve for K divide both sides by 50 and we are left with K is equal to the natural log of 2 natural log of 2 over 50 over 50 and we're done we can now write the particular solution that meets that meets these conditions so we can now write that our population and I can even write our population as a function of time is going to be equal to is going to be equal to 100 100 times e to the now K is natural log of 2 over 50 so I'll write that so natural log of 2 over 50 and then that times T and so if you assume population is going or rate of change of population is going to be proportional to population and you assume this information right over here this function is what is going to describe your growth of population