Discover the Constant Relative Rate of Population Growth: C=50.25 and k=0.0359

[r_swayze]

A population grows at a constant relative rate. After 10 months the population grows to 72, and after 18 months the population grows to 96. Find C and k.

the answer I got was c=50.25 and k=.0359

can somebody just double check for me? I don't know if the method I was used was correct.

 

[CompuChip]

If you have doubts about your method, you better post the method :)

I get (up to two decimals) the same answers, but that might be a co-incidence.

[Also, please be clear about your notation. I just assumed that C would be the initial population and k the rate of growth.]

 

[r_swayze]

ok here's the method I used, its very long and we haven't had to solve these type of problems this way in class so I was apprehensive about it

C = Initial Population K = Relative Growth Rate

P(10) = Ce^(10k) = 72 P(18) = Ce^(18k) = 96
e^(10k) = 72/C e^(18k) = 96/C
10k = ln(72/C) 18k = ln(96/C)
k = (ln(72/C))/10 k = (ln(96/C))/18

k = k
(ln(72/C))/10 = (ln(96/C))/18

18(ln(72/C)) = 10(ln(96/C))
18ln72 - 18lnC = 10ln96 - 10lnC
76.98 - 18lnC = 45.64 - 10lnC
-10lnC + 18lnC = 76.98 - 45.64
8lnC = 31.34
lnC = 3.92
C = 50.4 (I got 50.25 because I didnt round so I could get a more accurate answer)

then I plugged in C to find k

(ln(72/50.4))/10 = .0356 = k

was my method correct? I felt like I was making a simple problem, harder than what it was, but idk

shouldnt C be a whole number since it is a population?

 

[CompuChip]

By the looks of it, your method is correct, although it is not the most clear way to write down what you are doing. In particular, I was wondering where the exponential is coming from on the first line, although I think in the end it doesn't matter because taking the wrong base number (e instead of something else) produces a common factor which drops out when you equate the two.

Here is what I would expect.
The formula for growth at a constant relative rate is
P(n) = C (1 + k)^n
because for n = 0 the population is C, and for each time step you have to multiply the previous population P(n - 1) by (1 + k).

Then it is given that
P(10) = 72 = C (1 + k)^(10)
P(18) = 96 = C (1 + k)^(18)
which gives you two equations in two unknowns.

What you could do is divide them, and get
\frac{P(18)}{P(10)} = \frac{96}{72} = \frac{C (1 + k)^{18}}{C (1 + k)^{10}} = (1 + k)^8
and you see that C drops out. So now it's easy to solve for k:
(1 + k)^8 = 4/3
so
1 + k = \sqrt[8]{4/3} = 1,0366...
\qquad\implies k = 0,0366...

To find C, you only need to plug it back into one of the equations, for example
P(10) = 72 = C * (1,0366...)^(10)
(try not to round, e.g. if you use a calculator try to use it's ANS function to plug in 1 + k) and so
C = 72 / (1,0366...)^(10) = 50.25...

Note how the formula one starts with is more intuitive (I see an exponential in your formula, while it says the growth is at constant relative rate; my formula explicitly has this behaviour by multiplying by the growth percentage at every step) and easier to work with (you don't need logarithms, just some root).

That C and P(n) are in general not integer numbers is common in such exercises. You need to remember that this is merely a model for some "real" process, which makes assumptions (constant growth), tries to describe something discrete (counting the population every month) by a continuous function (you can calculate C (1 + k)^n for any n), and all we want of it is to give us more or less the right numbers at n = 0, 1, 2, ... although we will never be able to use it to predict any exact historic or future data.