Question

differential equation that describes natural growth

Answer 1

write a differential equation

Answer 2

dP/dt=kP sufficient enough?

Answer 3

how did they actually derive p= Ce^kt

Answer 4

\[\frac{dP}{dt} = kP\]separate the variables and integrate
\[\frac{dP}{P} = k \,dt\]\[\int \frac{dP}{P} = \int k\, dt\]\[\ln P = k t + C\]raise \(e\) to both sides:\[P = e^{kt+C} = e^{kt}*e^C = Ce^{kt}\] (different value of \(C\) obviously)

Answer 5

ahh! totally forgot abt separable equations! thanks! :D

Answer 6

you're welcome. and that equation describes so many things...like my consumption of chocolate chip cookies, for example :-)

Answer 7

haha yes! its like how my braincells are dying every single second i study calculus haha xD

Answer 8

ah, you're thinking of it the wrong way. what you're doing is culling the herd, eliminating the weak brain cells, and raising the average level :-)

Answer 9

thats natural decay though

Answer 10

is there an equation for natural decay?

Answer 11

depends on the value of \(k\) whether it is decay or growth

Answer 12

i see.. if its negative then must be dP/dt=-kP then?

Answer 13

well, you can write the ODE either way, and find the value of \(k\) to be positive or negative, depending on what happens. certainly if the exponent of the exponential is negative, you're going to have a decreasing quantity, and an increasing quantity if it is positive...

Answer 14

if you have an exponential decay, for example, where at \(t = 0\) you have \(P(t) = P_0\) and at \(t = 1\) you have \(P(t) = P_0/2\),

\[P(0) =P_0 = C e^{kt}\]\[P_0 =C e^{k0}\]\[C = P_0\]\[P(1) = P_0 e^{k(1)}\]\[P_0/2 = P_0e^{k}\]\[\frac{1}{2} = e^k\]\[\ln \frac{1}{2} = k = -0.693\]

Plotting \[y = e^{-0.693t}\]

Answer 15

wow nice! thanks! :D

Answer 16

But if our initial conditions showed growth, we'd get a different sign on \(k\):

\[P(2) = 2*P_0\]\[2P_0 = P_0e^{2k}\]\[2=e^{2k}\]\[\ln 2 = 2k\]\[k = \frac{\ln 2}{2} = \frac{0.693}{2} = 0.3465\]

Answer 17

what does relative growth rate mean?

Answer 18

when you have a program that makes it easy to do graphs, every problem starts looking like a problem that needs one or two for illustration :-)

Answer 19

could you give me some context?

Answer 20

after solving for k i got, k= (lnP- c) /

Answer 21

*t

Answer 22

k= (lnP-c)/t

Answer 23

could you show me the whole process?

Answer 24

ln P= lne^kt+c
lnP-c=kt
k=(lnP-c)/t

Answer 25

lnP=lne^(kt+c)

Answer 26

since k would be the growth rate right?

Answer 27

I'm confused about where you are getting the equation you started with.

\[\ln P = k t + C\]is what you get after integrating both sides, right?

Answer 28

yep! :D i'm suppose to find what does the term of growth rate stand for?

Answer 29

and you don't know the value of C until you plug in the initial conditions

Answer 30

like if i were to solve for k what is is the meaning by k= (lnP-c)/t

Answer 31

ohh so i cant put it like that?

Answer 32

okay, \(k\) is the growth constant (or decay constant), and represents the number of times per unit time that the quantity grows (or decays) by a factor of \(e \approx 2.718281828...\)

You can convert \(k\) to/from a doubling or halving time by multiplying or dividing by a factor of \(\ln 2\approx 0.693\) — it's just the change of base property of logarithms in action.

So if we have something that doubles every \(t_{dbl}\) units of \(t\), we could write it as

\[P(t) = P_0(2)^{t/t_{dbl}}\]or we could equivalently write it as \[P(t) = P_0e^{kt}\]where \(\large k = \frac{\ln 2}{t_{dbl}}\)

Answer 33

the first graph is showing a quantity that doubles every 5 units of \(t\), written in the "obvious" way, and the second graph shows a quantity that doubles every 5 units of \(t\) but written with base \(e\) instead. As you can see, the two are indistinguishable...

The second form is more convenient when doing differential equations, and the first is more convenient when you already know the form of the solution and just want to write down a formula.

Answer 34

It's late, I'm calling it a night, good luck with the rest of whatever you have to do...

Answer 35

wow that is really precisely what im looking for.. thank you! :D ya i have my final exam tmr! thanks again! goodnight! :D

Answer 36

I'm sure you'll do well, but if not, I offer a full money-back guarantee on my help :-)

Answer 37

LOL! you're funny! xD i do hope i hope i pass my calculus 2 this is my very last math course that i have to take! thank god haha xD okay off u go! ill off now soon too need to take a break thank you! :D