Application of Geometric Sequences and Series: Populations

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inexplicable.ashes
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Application of Geometric Sequences and Series: Populations

The population of India in 1992 was 886,362,000 with a growth rate of 1.6%.

a.) Calculate India's population in 2001 using a constant annual growth rate of 1.6%.
b.) Calculate India's population in 2001 if the population is growing continuously at 1.6%.

I am unsure as to what the difference is between an annual growth rate and a continuous growth rate and how to reflect the difference in a formula.
Help is appreciated

stapel_eliz
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inexplicable.ashes wrote:I am unsure as to what the difference is between an annual growth rate and a continuous growth rate and how to reflect the difference in a formula.

Ouch! They were supposed to have given you those formulas (and explained them)!

The "annual" (or any other fixed time period) growth is supposed to use the compound-interest sort of formula:

. . . . .$A\, =\, P\left(1\, +\, \frac{r}{n}\right)^{nt}$

...where "A" is the ending amount, "P" is the beginning amount, "r" is the rate of growth (or decay) for a unit of time (usually "annual", so "yearly"), "n" is the number of compoundings per unit time (usually years), and "t" is the total amount of time (expressed in the same time units).

In your case, P = 886,362,000, r = 0.016, n = 1, and t = 2001 - 1992 = 9.

For "continuous" growth, you use the "continously compounded" form, given in terms of the natural exponential "e":

. . . . .$A\, =\, Pe^{rt}$

...where "A", "P", "r", and "t" are defined as previously, and "e" is just the natural exponential. You'd find your answer using the same values, but plugging them straight into your calculatur. (You'll probably be using a "second" function for the key labelled "LN".)