Whats the difference

[naveed_786110]

Q. The population of a country is growing exponentially at a constant rate of 2% per year. How much time this population will take to double itself?

Solution:

Method-I:
S=P(1+i)^n
2P=P(1+0.02)^n
2 = 1.02^n
n=log 2 / log 1.02
n=35 years.

Method - II:

S=Pe^in
2P=Pe^0.02n
n=34.65

The correct answer is 34.65..... The problem is why METHOD - I is not working (even its close) ?

 

[stapel]

Q. The population of a country is growing exponentially at a constant rate of 2% per year. How much time this population will take to double itself?

Solution:

Method-I:
S=P(1+i)^n
2P=P(1+0.02)^n
2 = 1.02^n
n=log 2 / log 1.02
n=35 years.

Method - II:

S=Pe^in
2P=Pe^0.02n
n=34.65

The correct answer is 34.65..... The problem is why METHOD - I is not working?

Did the exercise say that the population was growing at two percent, compounded annually; or exponentially at a constant two percent?

 

[Ishuda]

Q. The population of a country is growing exponentially at a constant rate of 2% per year. How much time this population will take to double itself?

Solution:

Method-I:
S=P(1+i)^n
2P=P(1+0.02)^n
2 = 1.02^n
n=log 2 / log 1.02
n=35 years.

Method - II:

S=Pe^in
2P=Pe^0.02n
n=34.65

The correct answer is 34.65..... The problem is why METHOD - I is not working (even its close) ?

First the growth is exponential means
S = e^f(x)
We could actually use any base which is convenient since a^f= e^{f ln(a)} = e^g. Now if a rate of exponential growth is constant, that actually means that the function f(x) is growing at a constant rate. That is
dS/dx = f'(x) e^f(x) = f'(x) S
and the rate of exponential growth = (dS/dx)/S = f'(x) is constant [=0.02=2% in this case]. Since f'(x) is constant, f(x) is linear and f(x) =bx+c where b=0.02 and e^c=S(0) [=P in your notation], and x is in years.




To look at it another way, let's use a base of 1.02 for convenience. IF we have
S(n) = S(0) 1.02ⁿ [ = P 1.02ⁿ]
then we would have
S'(n)/S(n) = ln(1.02)

 

[naveed_786110]

First the growth is exponential means
S = e^f(x)
We could actually use any base which is convenient since a^f= e^{f ln(a)} = e^g. Now if a rate of exponential growth is constant, that actually means that the function f(x) is growing at a constant rate. That is
dS/dx = f'(x) e^f(x) = f'(x) S
and the rate of exponential growth = (dS/dx)/S = f'(x) is constant [=0.02=2% in this case]. Since f'(x) is constant, f(x) is linear and f(x) =bx+c where b=0.02 and e^c=S(0) [=P in your notation], and x is in years.




To look at it another way, let's use a base of 1.02 for convenience. IF we have
S(n) = S(0) 1.02ⁿ [ = P 1.02ⁿ]
then we would have
1.02 = S(1)/S(0) = S(n)/S(n-1)
and the ratio would be constant, not the rate.

What actually Exponential rate means?

 

[Ishuda]

What actually Exponential rate means?

Exponential increase, in its simplest form, means that how fast something increases is proportional to the size of that something [the bigger/smaller it is the faster/slower it grows]. How fast something increases is the rate of increase (that is the derivative of the function of amount). That's what you have in the problem: You have a population S which is exponential in growth with a rate of increase which is a constant 2%. So the derivative of S [how fast S is growing] divided by S is equal to the constant 0.02 [2%]. If you do not know the calculus for this, you should have been given it as a definition.

 

[Steven G]

What actually Exponential rate means?

What Ishuda said is much better than what I will say but I will say it anyways. Exponential rate is a continuous increase (decrease). The amount at any time changes constantly, not 4 times per year or every minute or every second but all the time.