Results 1 to 9 of 9

Math Help - Find the number of years it takes for half of the population to have been exposed to

  1. #1
    Junior Member
    Joined
    Apr 2010
    Posts
    57

    Find the number of years it takes for half of the population to have been exposed to

    An infectious disease spreads through a large population according to the model
    dy/dt = (1-y)/4

    a) Solve this differential equation, assuming y(0)=0
    b) Find the number of years it takes for half of the population to have been exposed to the disease.

    a) I got
    y = (-1/4)Ce^t + 1

    How do I find b)? Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    I get y = 1-Ce^{\frac{t}{4}}

    now use y(0)=0 to find C

    Then we can discuss b)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by ewkimchi View Post
    An infectious disease spreads through a large population according to the model
    dy/dt = (1-y)/4

    a) Solve this differential equation, assuming y(0)=0
    b) Find the number of years it takes for half of the population to have been exposed to the disease.

    a) I got
    y = (-1/4)Ce^t + 1

    How do I find b)? Thanks
    What have you tried? Where are you stuck?

    Step 1: \frac{dt}{dy} = \frac{4}{1 - y}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Apr 2010
    Posts
    57

    I found C, now what do I do?

    C = 1

    So I plugged it back into the equation and got

    y = 1 - Ce^(t/4)

    Do I make y = 1/2 and solve for t?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by ewkimchi View Post
    C = 1

    So I plugged it back into the equation and got

    y = 1 - Ce^(t/4)

    Do I make y = 1/2 and solve for t?
    Yes.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2012
    From
    United States
    Posts
    3

    Angry Re: I found C, now what do I do?

    I have the same question for my homework but mine says that y is the percent, so y would be 50. Everything else about your question is the same so I'd guess the y value unit is too.
    Also, this is the same as I got when I tried to solve it, y = 1 - Ce^(t/4), where C=1, but I'm running into a big problem. If you try to insert a y value you start by subtracting 1 from both sides.
    So let's say y=50 (50 percent), 50=1-1e^(t/4)--> 49=-e^(t/4).
    Divide each side by -1, -49=e^(t/4).
    To solve for t, you must get rid of the e which is where I run into problems. To get rid of the e, take the natural log of each side to cancel the e. Ln(-49)=t/4 --> Ln(-49)/4=t... BUT you can't take the natural log of a negative number.

    I noticed that no one has addressed this problem yet as I too have to find when half of the population is infected.
    Last edited by NickA28; April 10th 2012 at 09:43 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Mar 2012
    From
    Sheffield England
    Posts
    440
    Thanks
    76

    Re: Find the number of years it takes for half of the population to have been exposed

    I got 1-e^-t/4
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Apr 2012
    From
    United States
    Posts
    3

    Re: I found C, now what do I do?

    I'm sorry. I was just told that on our homework sheet the question wording was wrong. It should be y equals the portion of the population, not the percent. So y= .50, not 50

    Quote Originally Posted by NickA28 View Post
    I have the same question for my homework but mine says that y is the percent, so y would be 50. Everything else about your question is the same so I'd guess the y value unit is too.
    Also, this is the same as I got when I tried to solve it, y = 1 - Ce^(t/4), where C=1, but I'm running into a big problem. If you try to insert a y value you start by subtracting 1 from both sides.
    So let's say y=50 (50 percent), 50=1-1e^(t/4)--> 49=-e^(t/4).
    Divide each side by -1, -49=e^(t/4).
    To solve for t, you must get rid of the e which is where I run into problems. To get rid of the e, take the natural log of each side to cancel the e. Ln(-49)=t/4 --> Ln(-49)/4=t... BUT you can't take the natural log of a negative number.

    I noticed that no one has addressed this problem yet as I too have to find when half of the population is infected.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Mar 2012
    From
    Sheffield England
    Posts
    440
    Thanks
    76

    Re: I found C, now what do I do?

    y is the proportion so for 50% you should put y=0.5
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: August 10th 2010, 03:38 AM
  2. half of the number
    Posted in the Math Puzzles Forum
    Replies: 8
    Last Post: August 20th 2009, 09:20 AM
  3. Replies: 1
    Last Post: May 24th 2009, 05:16 AM
  4. Replies: 2
    Last Post: February 26th 2008, 11:43 PM
  5. [SOLVED] T-Test: How to take into account data number for each population member?
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: June 7th 2005, 07:08 AM

Search Tags


/mathhelpforum @mathhelpforum