According to United Nations data, the world population at the beginning of 1990

Question

According to United Nations data, the world population at the beginning of 1990 (which means t=0) was approximately 5.3 left parenthesis w h i c h space m e a n s space P subscript 0 equals 5.3 right parenthesis billion and growing at a rate of about 2% open parentheses w h i c h space m e a n s space k equals 2 % equals 0.02 close parentheses per year. This population growth can be modeled by the differential equation fraction numerator d P over denominator d t end fraction equals k P space given on Page 357 of the text book, where P is the population function, and k is the constant of proportionality, and in our this problem k equals 2 % equals 0.02 .

The particular solution of fraction numerator d P over denominator d t end fraction equals k P space is given by the textbook as P equals P subscript 0 e to the power of k t end exponent . Use this given particular solution formula to estimate the world population at the beginning of the year 2015 left parenthesis t equals 25 right parenthesis in the unit of billion. Show your result with two digits on the right side of decimal point. For example, if your results is 7.93 billion, just show 7.93 as the result. (You also use 5.3 as the value of P subscript 0 in which the unit 'billion' is implied.)

Explanation / Answer

P0 = 5.3 ---> yes correct

dP/dt = 2% = 0.02*P

dP/dt = 0.02P

dP/P = 0.02dt

Integrating :

ln|p| = 0.02t + C

P = e^(0.02t + C)

P = De^(0.02t)

When t = 0, P = 5.3

5.3 = De^(0.02*0)
5.3 = De^0
D = 5.3

So, P = 5.3e^(0.02t)

For 2015, t = 25 ---> correct

Plug in t = 25 :

P = 5.3 * e^(0.02 * 25)

P = 8.7382227347106795

So, 8.74 million ----> ANSWER

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