My Math Forum  

Go Back   My Math Forum > College Math Forum > Differential Equations

Differential Equations Ordinary and Partial Differential Equations Math Forum

LinkBack Thread Tools Display Modes
May 1st, 2014, 08:33 AM   #1
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.
neko is offline  
May 2nd, 2014, 02:02 AM   #2
Senior Member
Joined: Apr 2014
From: Glasgow

Posts: 2,142
Thanks: 726

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.
Benit13 is offline  

  My Math Forum > College Math Forum > Differential Equations

high, level, school

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
How Do I Take My High School Maths To The Next Level. And... pilgrimnewton New Users 4 June 26th, 2014 12:01 AM
High Level math in the real world rage New Users 16 January 18th, 2013 08:49 AM
simple high level uses of quaternions and matrices josh3light New Users 0 May 30th, 2009 09:49 AM
high school algebra squerlyq Algebra 8 January 6th, 2009 09:10 AM
XYZ HIGH SCHOOL symmetry Algebra 4 May 14th, 2007 07:28 AM

Copyright © 2019 My Math Forum. All rights reserved.