- #1
Jameson
Gold Member
MHB
- 4,541
- 13
I don't know why, but I am stuck on this seemingly easy question. Here's the question and the work I've done.
---------------------
A certain model for spread of rumors states that [tex]\frac{dy}{dt} = 3y(3-2y)[/tex] , where [tex]y[/tex] is the proportion of the population that has heard the rumor at time [tex]t[/tex]. What proportion of the population has heard the rumor when it is spreading the fastest?
--------
Ok. You are given the derivative of the proportion function, so setting it equal to 0 will give you when it is changing the fastest/slowest. Solving the equation [tex]3y(3-2y) = 0[/tex] you get 0 and 1.5...
Next part is to find the original equation and evaluate it at 1.5. So I will need to separate the variables, and when I do I get:
[tex]\frac{1}{3y(3-2y)}dy = dt[/tex]
This integral (I did it on my calculator) is [tex]\frac{-\ln{\frac{\mid2x-3\mid}{\mid{x}\mid}}}{9}[/tex]
When I evaulate 1.5 I get [tex]\infty[/tex]
Help me please.
Jameson
---------------------
A certain model for spread of rumors states that [tex]\frac{dy}{dt} = 3y(3-2y)[/tex] , where [tex]y[/tex] is the proportion of the population that has heard the rumor at time [tex]t[/tex]. What proportion of the population has heard the rumor when it is spreading the fastest?
--------
Ok. You are given the derivative of the proportion function, so setting it equal to 0 will give you when it is changing the fastest/slowest. Solving the equation [tex]3y(3-2y) = 0[/tex] you get 0 and 1.5...
Next part is to find the original equation and evaluate it at 1.5. So I will need to separate the variables, and when I do I get:
[tex]\frac{1}{3y(3-2y)}dy = dt[/tex]
This integral (I did it on my calculator) is [tex]\frac{-\ln{\frac{\mid2x-3\mid}{\mid{x}\mid}}}{9}[/tex]
When I evaulate 1.5 I get [tex]\infty[/tex]
Help me please.
Jameson
Last edited by a moderator: