Question

The population of a town grows at a rate proportional to the population present at time t. The initial population of 50 increases by 15% in 10 years. What will be the population in 30 years? how fast is the population growing at t=30?

Answer 1

\[\frac{dP}{dt}=kA\]

Answer 2

kP i mean

Answer 3

\[P'-kP=0\]

Answer 4

\[e^{\int{}{}p(x)dx}=e^{kt}\]

Answer 5

+c of course

Answer 6

actually not yet

Answer 7

\[\frac{d}{dx}(e^{-kt}y)=0\]

Answer 8

\[e^{-kt}y=c\]

Answer 9

\[e^{-kt}P=c\]

Answer 10

P(0)=500

Answer 11

\[e^{-k(0)}*500=c\]=
\[500=c\]

Answer 12

P(10)=500+(.15*500)=575

\[e^{-10k}*500=575\]

Answer 13

\[e^{-10k}=\frac{575}{500}\]

Answer 14

\[-10k=ln(\frac{575}{500})\]

Answer 15

\[-10k=ln(575)-ln(500)\]

\[k=\frac{ln(575)-ln(500)}{-10}\]

Answer 16

I would have just done this:\[\begin{align}
\frac{dP}{dt}&=kP\\
\therefore\int\frac{dP}{P}&=\int k dt\\
\therefore\log(P)&=kt+c\\
\therefore P&=e^{kt+c}=e^c\times e^{kt}=Ae^{kt}\\
\end{align}\]when t=0, P=50, therefore:\[50=Ae^0=A\]\[P=50e^{kt}\]

Answer 17

then, in 10 years the population increases by 15%, which means it becomes 50*1.15 = 57.5

Answer 18

therefore:\[57.5=50e^{10k}\]\[e^{10k}=\frac{57.5}{50}\]\[10k=\log(\frac{57.5}{50})\]\[k=\frac{1}{10}\log(\frac{57.5}{50})\]

Answer 19

yeah it's 575 i mean

Answer 20

so you can now work out the P and dP/dt when t=30

Answer 21

yeah how'd you get 10k though i'm wondering

Answer 22

because i did it through linear and it came out negative

Answer 23

ok, you agree we got to this point earlier where k was some unknown constant?\[P=50e^{kt}\]

Answer 24

yes

Answer 25

then you are told that when t=10, P=50 increased by 15%=57.5
agreed?

Answer 26

yes your way but if you do it with the linear equation

y'-x(t)y=g(t)

Answer 27

y'+p(x)y=g(x)

Answer 28

ok, using your way you ended up with:\[e^{-kt}P=c\]if you multiply both sides by \(e^{kt}\) you get:\[P=ce^{kt}\]which is the same form as mine.

Answer 29

if you leave it as so and solve

Answer 30

\[e^{-10k}(575)=500\]

Answer 31

you mean leave it in your form?

Answer 32

yes idk if it's just the answers are the sme

Answer 33

\[e^{-10k}=\frac{500}{575}\]

\[-10k=ln(\frac{500}{575}\]

Answer 34

cause when you get farther you're taking the diference of two logs correct? does the -10 correct that

Answer 35

it should lead to the same answers. from your equation we get:\[k=-\frac{\log(\frac{500}{575})}{10}=-\frac{\log(500)-\log(575)}{10}=\frac{\log(575)-\log(500)}{10}\]

Answer 36

and this is identical to my result above of:\[k=\frac{1}{10}\log(\frac{57.5}{50})\]