# Have I concluded a good formula

• October 1st 2011, 07:17 AM
David Green
Have I concluded a good formula
The population of animal in a country Park increases in number by 11 percent each year. A survey counted the number of animal in the park at the start of a particular year. If the number of animals counted was K, is the following formula connecting the current number, N, with time t in years since the initial count a correct formula to use?

N = K x 11 x t
• October 1st 2011, 07:41 AM
earboth
Re: Have I concluded a good formula
Quote:

Originally Posted by David Green
The population of animal in a country Park increases in number by 11 percent each year. A survey counted the number of animal in the park at the start of a particular year. If the number of animals counted was K, is the following formula connecting the current number, N, with time t in years since the initial count a correct formula to use?

N = K x 11 x t

1. Your formula is not correct.

2. In the beginning, that means t = 0, you have K animals.

After one year, t = 1, you have $N(1)= K + K \cdot \underbrace{0.11}_{11\%} = K(1+0.11)$

Next year, t = 2, you have $N(2)= K(1+0.11) + K (1+0.11)\cdot 0.11 = K(1+0.11)(1+0.11)=K(1+0.11)^2$

After x years, t = x, you have $N(x)=K (1+0.11)^x$
• October 1st 2011, 07:47 AM
Plato
Re: Have I concluded a good formula
Quote:

Originally Posted by David Green
The population of animal in a country Park increases in number by 11 percent each year. A survey counted the number of animal in the park at the start of a particular year. If the number of animals counted was K, is the following formula connecting the current number, N, with time t in years since the initial count a correct formula to use?
N = K x 11 x t

At the start $N_0=K$.
Then after one year $N_1=N_0+0.11N_0=1.11K$
After two years $N_2=N_1+0.11N_1+=(1.11)^2K$
So after t years $N_t=~?$
• October 1st 2011, 08:01 AM
David Green
Re: Have I concluded a good formula
Quote:

Originally Posted by earboth
1. Your formula is not correct.

2. In the beginning, that means t = 0, you have K animals.

After one year, t = 1, you have $N(1)= K + K \cdot \underbrace{0.11}_{11\%} = K(1+0.11)$

Next year, t = 2, you have $N(2)= K(1+0.11) + K (1+0.11)\cdot 0.11 = K(1+0.11)(1+0.11)=K(1+0.11)^2$

After x years, t = x, you have $N(x)=K (1+0.11)^x$

Thanks for your help, I read it as; N(x) = K x 1.11^x