Question

sea grass grows on a lake at a rate that is dG/dt=kG, where k is a constant.
a) find an expression for G, the amount of grassin the lake, in terms of t, the number of years, if the amount of grass is 100 tons initially, and 120 tons after one year.
b) in how many years will the amount of grass be 300 tons?

Answer 1

dG/dt = kG -- > Separate variables to common ones

(1/G) * dG = k * dt --> Integrate both sides (1/x = ln(x) ) and integral of 1 is t)

ln( G ) = kt + C -- > where C is integration contant Undo the logarithm by raising both sides to e

e^ (ln(G)) = e ^ (kt + C) -- > right side of equation is reduced to C*e^(kt). Remmerb e is just a number so one number raised to another number is still a constant so C drops out of exponent field.

G = C * e^ (kt) --> now its plug and chug for the numbers

Answer 2

a.) G is initially 100 tons and 120 tons after 1 year then...

100 = C*e^(k*0) --> e^(0) = 1 C = 100

120 = 100*e^(k*1)

ln( 120/100 ) = k

G = 100*e^(kt) --> where k is ln(120/100)

Answer 3

you should be able to figure our B from there

Answer 4

ohh ook i got confused i didnt realize that was all for a. how did you kniow you had to integrate it?

Answer 5

Well the fact that you had a derivative with G in it AND G not in a derivative means that you need to integrate. Its called a first order differential equation, from now on you should know that if what you want is in the derivative and is outside as a "stand alone" variable, then you need to integrate.

Answer 6

oh alright,thank you

Answer 7

But of course =).

Answer 8

how do i find the time for b? do i do 300=100e^kt

Answer 9

Correct, remember that we solved for k in part a.) and we are told for part b that the grass is 300 tons...

300 = 100*e^(k*t) --> solve for t

ln(300/100) = k*t --> solve for t

[ln( 300/100 ) / ln ( 120/100 ) ] = t

Answer 10

ok, just checking, i wanted to make sure i was learning it the right way, thanks again :)

Answer 11

YUPPPPPPPPP =)