Simple Modelling Question (ODE, 1st-order, population growth)

• Apr 26th 2011, 06:17 PM
ragnar
Simple Modelling Question (ODE, 1st-order, population growth)
I'm looking at MIT's open course-ware class on differential equations and reading through the first problem set, found here: http://ocw.mit.edu/courses/mathemati..._03S10_ps1.pdf. First, I've only done math. I've never had any science classes (except in high school, and at that the classes basically just told you that an atom has a positron or whatever).

Question 1: So while I got 0.a right, the answer sheet says that the units of k(t) are time^-1. WTF does that mean? Just that k(t) is a function of time? I would have thought that the units of k(t) would be, say, (kilo-oryx)/time (rate of change of thousands of oryx over whatever unit of time).

Question 2: k(t) is the growth rate of the population as time varies, right? So I would have thought that the answer to 0.b would be just x' = (k_0)/(a + t)^2, since x'(t) is also the rate of change of the population as time. But the answer sheet says it's x' = (x*k_0)/(a+t)^2. Another WTF?

[Edit: Note the question I'm referring to starts at the top of the second page of the (.pdf) .]
• Apr 26th 2011, 06:28 PM
pickslides
Quote:

Originally Posted by ragnar
the answer sheet says that the units of k(t) are time^-1. WTF does that mean?

you'll find this used when talking about a rate i.e metres/second = metres $\times$ seconds $^{-1}$
• Apr 26th 2011, 06:43 PM
ragnar
Okay. I think understanding this point will make me understand the answer to both of my questions. Why do we then call it time^-1 rather than (kilo-oryx) x time^-1? I thought k(t) was supposed to represent the growth rate. I.e., plug in a time t, get out a rate of growth k(t). Rate of growth is (quantity change)/(time change), right?
• Apr 26th 2011, 07:19 PM
pickslides
Quote:

Originally Posted by ragnar
Rate of growth is (quantity change)/(time change), right?

Sounds good to me!