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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 May 1st, 2014, 08:33 AM #1 Newbie   Joined: May 2014 From: Europe Posts: 2 Thanks: 0 High school level Hey guys, this is the english translation of my math problem: In the middle of 2013 the world population was 7.1 billion. The growth rate was then 1% per year. Assume that the growth rate in a decade goes down from 1% to 0.8% per year. Describe this situation with a differential equation.
 May 2nd, 2014, 02:02 AM #2 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,131 Thanks: 716 Math Focus: Physics, mathematical modelling, numerical and computational solutions Kudos to you for doing differential equations in high school! This is actually a really easy question once you get to know about differential equations, so I'll try and explain every stage of getting the answer Differential equations are all about changes in a variable, usually over time or space (although it can be anything really). So to answer this question we need to think about changes in population over time. Every year the population increases by 1%. Therefore, the number of extra people being born into the world is going to depend on how many people are already alive. Basically if more people are around, more people will be born because you are finding 1% of a bigger number. For example, if only 100 people are around, 1 person will be born each year. If 1000 people are around, 10 people will be born. Let's call the number of new people born into the world each year $\displaystyle \frac{dP}{dt}$, where $\displaystyle P$ represents the population at any given time, $\displaystyle t$. Just in case you don't know about derivatives, the $\displaystyle d$s are just a mathematical notation that refers to a "change" in something and are kinda "glued" to the letter next to it. Therefore, $\displaystyle dP$ means the change in population and $\displaystyle dt$ means the change in time. In one big chunk, the $\displaystyle \frac{dP}{dt}$ reads as "the change in population with respect to the change in time" or the slightly shorter form "the change in population with respect to time". We know that the change in population is 1% of the current population from the question. We can turn the percentage into a normal number dividing by 100 to give 0.01. The change in population is then $\displaystyle 0.01\times P = 0.01P$, which is equal to $\displaystyle \frac{dP}{dt}$. Therefore $\displaystyle \frac{dP}{dt} = 0.01P$. This is called a differential equation because it contains a derivative (the $\displaystyle \frac{dP}{dt}$ thing). Solving differential equations is generally much harder than solving regular equations with normal variables in them. The question also mentions a reduction in the percentage growth from 1% to 0.8%. If this is the case, you just swap the 0.01 for a 0.008. $\displaystyle \frac{dP}{dt} = 0.008P$. I hope this helps. Let me know if you need to the find the population as a function of time! Last edited by Benit13; May 2nd, 2014 at 02:05 AM.

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