rickportega Posted February 7, 2008 Share Posted February 7, 2008 I have a tough problem for my calculus class I was hoping to get help for: A person meets new people at a rate of about 100 every day. After 7 days, he remembers the names of 600 of the 700 people he met. The rate of change of the # of names he remembers (dR/dt) equals 100 minus an amount that is directly proportional to R. a) write an equation expressing the assumptions above assuming at t=0 he knew no names. _________________________________________________________________________________ What I've come up with so far is dR/dt=100-kR, where k is the constant of proportionality. I'm not sure where to go from here...I think the next step would be to get the dR and the R isolated on one side of the equation, but I can't get this (seems like simple algrebra, but its tough haha). Any help would be greatly appreciated. Link to comment Share on other sites More sharing options...
Bignose Posted February 7, 2008 Share Posted February 7, 2008 Are you trying to just write an equation or are you trying to solve said equation? Because, it seems to me that you've already written an equation. The only thing you are missing at the moment is an initial or boundary condition. You need at least as many boundary/initial conditions as the order of the differential equation, so far you've written down a 1st order DE, so you need at least one boundary or initial condition. If that's all you are looking for, then you are done. If you need help solving it too, just say so in the thread here, and I can help with that too (though I won't do it for you -- doing homework for you is explicitly against the forum's rules and the spirit of the forums -- I'll give you as many hints as I can). Link to comment Share on other sites More sharing options...
thedarkshade Posted February 7, 2008 Share Posted February 7, 2008 Isn't this for HW section? Link to comment Share on other sites More sharing options...
rickportega Posted February 7, 2008 Author Share Posted February 7, 2008 Yes, I think I am supposed to elliminate k (and once I integrate, elliminate C) using the conditions at t=0, R=0, and at t=7, R=600. If I posted this in the wrong forum, I apologize. As you probably have noticed, Im new to the forum. Thanks again for all the help you can provide Link to comment Share on other sites More sharing options...
Bignose Posted February 7, 2008 Share Posted February 7, 2008 Well, I think that a small correction first. 'Eliminate' is the wrong word, but 'solve for' is right. Eliminate would be get rid of or ignore or remove, but what you want to do is find the value of k and the integration constant, not get rid of them. Ok, well, there are two completely equivalent ways to solve this. Again, I won't do it for you, but I will give you hints. Method 1) You have a differential equation with the differential alone on the left hand side and a polynomial on the right hand side. You can always divide both sides by that polynomial. Then, if you look at your integral tables, you should find some inspiration. Method 2) You can implement a change of variables. Let Z = 100-kR. Then write an expression for dZ in terms of dR. Then, rearrange so that you only have Z terms on one side -- that expression should be directly integrable. Be sure to remember to put the definition of Z back into the final expression when you're done to solve it in the original variables. You should try both methods to confirm that they are in fact equivalent. Both methods are useful, sometimes it helps to look at a table of integrals to inspire you to see what forms may be available with the problem at hand. And, change of variables can be very powerful to simplify very messy situations. Sometimes you need both methods -- a change of variables makes the messy expression into a form that is covered by an integral table. So, both methods are powerful and should be practiced. If you need further help, just ask. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now