Question

Calculus
Population y grows according to the equation
dy/dt=ky, where k is a constant and t is measured in years. If the population doubles every 10 years, then the value of k is ?

Answer 1

You'll have to integrate \[\frac{ dy }{ dt } = k y \implies \frac{ dy }{ y } = k dt \implies \int\limits \frac{ dy }{ y } = \int\limits k dt\]

Answer 2

(Yes, it's a differential equation)

Answer 3

ln absolute value (y)=kt+c

Answer 4

then I use e as a base for both side

Answer 5

Yup :)

Answer 6

y=2e^(kt)

Answer 7

I meant y=Ce^(kt)

Answer 8

not sure what to do afte that

Answer 9

I meant after that

Answer 10

Well if you remember from earlier population and growth problems, the initial population is c0 at t = 0 (initial conditions). so we have y = c0 which implies ln c0 = 0+C hence our constant C = lnc0 \[lny = kt + \ln c_0\] now we look at when the population doubles every 10 years. Can you figure that out/ finish it off?

Answer 11

I am not sure I can finish this one on my own......

Answer 12

No worries, all we have to note is that t will be 10, y = 2 c0 as it's doubling, it can be tricky at first, these problems do require lots of practice :P

Answer 13

You can plug that all into our equation above, and then it's just logarithms and algebra. \[\ln(2 c_0) = 10k+\ln( c_0)\]

Answer 14

Solve for k and that's your answer, I hope that made some sense :)

Answer 15

thanks

Answer 16

.069 is the solution and I have it on my choices to pick from Thanks alot

Answer 17

Your welcome :)