My Math Forum Derivatives/linear approximation problem

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 April 18th, 2010, 04:15 AM #1 Member   Joined: Apr 2010 Posts: 91 Thanks: 0 Derivatives/linear approximation problem The population of a certain country grew from 2.1 million in 1970 to 3 million in 1995. a) What was the average rate of change of the population over that period? (worked out to be 36 thousand people per year). b)Suppose that the population grew exponentially. By what percentage did the population grow each year? What was the average rate of change from 1970 to 1975? What was the average rate of change from 1990 to 1995? Illustrate your answers using a sketch. I worked out no. a), but I am having problems with number b) could some please help? This question appeared under a section in my note under the heading of derivatives and linear approximations, so I am thinking you have to use that to solve the problem.
 April 26th, 2010, 03:33 PM #2 Senior Member   Joined: Jan 2009 Posts: 345 Thanks: 3 Re: Derivatives/linear approximation problem b. Population in million $3=(2.1)e^{25r}$ We need to find the rate (r), so we divide both sides by 2.1, take the natural log of both sides and divide by 25. $\frac{3}{2.1}=e^{25r} \Rightarrow \text ln\left (\frac{3}{2.1} \right) = 25r \Rightarrow \frac{\text ln\left (\frac{3}{2.1} \right)}{25} \Rightarrow \frac{0.356675}{25} = r \approx 0.0142670$ We now have a function. $p[t]= 2.1e^{(0.0142670)t}$ Using 1970 as a starting point and 1995 as an ending point over the 25 years we can calculate the average rate of change which is normally has a general form of $\frac{(f[b]-f[a])}{b-a}$ on the open interval $(a,b)$ Average rate of change $\frac {p[25]-p[0]} {25-0} \Rightarrow \frac {\left (2.1e^{(0.0142670)(25)}- 2.1e^{(0.0142670)(0)} \right)}{25} \approx 36000 people/year$

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