2010-05-20, 03:36 PM
2a e^(.03t) = a e^(.04t)
Find t that makes this true. You should have too much trouble solving this.
To see why I got this let G be the population of grey moths with G(0)=2a:
dG/dt =.03 G
^Change in grey moths per time
dG / G =.03 dt <~ I multiplied both sides by dt then divided both sides by G.
Integrate both sides:
ln(G) = .03 t +C
put a base of e on both of these.
G = e^(.03t +C)
G = e^(.03t) * e^© <~ this last part is just a constant, so we'll replace it with C
G = C * e^(.03t)
Use the fact that G(0) = 2a to get that C = 2a
G = 2a e^(.03t)
you do the same to get the population for B (the black moths)
Then just set them equal to each other and solve for t.
Find t that makes this true. You should have too much trouble solving this.
To see why I got this let G be the population of grey moths with G(0)=2a:
dG/dt =.03 G
^Change in grey moths per time
dG / G =.03 dt <~ I multiplied both sides by dt then divided both sides by G.
Integrate both sides:
ln(G) = .03 t +C
put a base of e on both of these.
G = e^(.03t +C)
G = e^(.03t) * e^© <~ this last part is just a constant, so we'll replace it with C
G = C * e^(.03t)
Use the fact that G(0) = 2a to get that C = 2a
G = 2a e^(.03t)
you do the same to get the population for B (the black moths)
Then just set them equal to each other and solve for t.