Question 487642: In 2000, there were about 202 million vehicles and about 283 million people in a certain country. The number of vehicles has been growing at 4.5% a year, while the population has been growing at 1% a year.
(a) Write a formula for the number of vehicles (in millions) as a function of t, the number of years since 2000.
V(t) =
(b) Write a formula for the number of people (in millions) as a function of t, the number of years since 2000.
P(t) =
(c) If the growth rates remain constant, when is there, on average, one vehicle per person? Give your answer in exact form and decimal form.
Exact form:
______years since 2000
Decimal form (nearest tenth):
_______years since 2000
Thank you so much and God bless you.
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Let's see if we can't figure this problem out.
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Since it's not very well defined by the problem, I'm going to assume that the number of vehicles for the year 2000 represents the number of vehicles on December 31st of that year.
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First for the number of vehicles. Each year that goes by, the number of vehicles increases by 4.5% (that is by the decimal 0.045). So that on December 31st of the year 2001 the increase in the number of cars is 202M times 0.045. That means that at the end of the first year (the end of 2001) the number of cars is the 202M at the end of 2000 plus the increase of 202M times 0.045. In algebraic form this is:
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202M + (202M*0.045)
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Factor out the 202M and this expression for the number of cars at the end of the first year (that is at the end of 2001) becomes:
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202M*(1 + 0.045) = 202M*1.045
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Next what happens during the second year? You start the second year with the number of vehicles at the end of the year 2001. We just determined it to be:
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202M*(1.045)
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So the increase by the end of the second year (the end of 2002) will be 0.045 times 202M*(1.045) and at the end of the second year the total number of cars will be what you had at the end of 2001 plus the increase of 0.045 times the number number of cars at the end of 2001. In algebraic form this is:
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202M*(1.045) + (202M*(1.045))*(0.045)
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Factor out (202M*1.045) and you have:
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(202M*1.045)*(1 + 0.045) = (202M*1.045)*(1.045) = 202M*(1.045)^2
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If you try this analysis for another year or two, it will become apparent that for each passing year, the number of cars increases by a factor of 1.045. This means that the number of cars at the end of a given year (call it V(t)) can be determined from the equation:
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V(t) = 202M*(1.045)^(t)
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Where t is the number of years after the year 2000. So for example in the year 2003, t would be 2003 minus 2000 or t would equal 3.
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So on December 31 of the year 2005 the number of cars would be:
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V(t) = 202M*(1.045)^t = 202M*(1.045)^(2005 - 2000) = 202M*(1.045)^5
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You can use a calculator to determine that 1.045 raised to the exponent 5 is 1.246181938 and when you multiply this by 202M the answer becomes 251.728751M.
At the end of the year 2005 the number of cars will be 251,728,751
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The same type of analysis can be done for the population. (Again assume that the population each year is for the last day of that year.) The difference is that each year the increase is 1% or 0.01. And the population in the year 2000 is 258M on December 31st 2000. So by introducing these changes into the equation for V(t) we can say that the equation for the population at the end of a given year is:
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P(t) = 258M*(1.01^t)
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where t is again defined as the year of interest minus 2000. If you want to find the population on December 31st of 2004, t would be 2004 - 2000 or t would be 4.
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Now to find the year when the number of vehicles equals the population so that there is one vehicle per person on average. This is done by setting the right side of the equation for V(t) equal to the right side of the equation for P(t) and then solving for t. In other words:
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202M*(1.045^t) = 258M*(1.01^t)
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Our goal is to get terms containing t on the left side of the equation, and all other terms on the right side. Begin by dividing both sides of this equation by 202M and you get:
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1.045^t = (258M/202M)*(1.01^t)
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Divide the 258M by 202M and you get 1.277227723. Substitute this into the equation and it becomes:
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1.045^t = (1.277227723)*(1.01^t)
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Divide both sides by 1.01^t and the equation becomes:
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(1.045^t)/(1.01^t) = 1.277227723
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Note on the left side that the exponent in the numerator is the same as the exponent in the denominator. Therefore, by the rules of exponents we can say:
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(1.045/1.01)^t = 1.277227723
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to simplify this a little, divide 1.045 by 1.01 and you have 1.034653465. Substitute this and the equation becomes:
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(1.034653465)^t = 1.277227723
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Any time you have a variable in an exponent, you should consider taking the log of both sides so that the variable can be brought out as a multiplier of the log. Let's use log base 10 since we can readily use a scientific calculator to determine the logarithm. Take log base 10 of both sides:
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log((1.034653465)^t) = log(1.277227723)
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bring the t out as the multiplier of the log:
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t*log(1.034653465)= log(1.277227723)
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Use a calculator to find the two logs:
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t*(0.014794916) = 0.106268336
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Solve for t by dividing both sides of the equation by 0.014794916:
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t = 0.106268336/0.014794916 = 7.182760136
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Since t is the year of interest minus 2000, we know that t occurs exactly 7.182760136 years after 2000 or to the nearest tenth, 7.2 years after 2000 which would be 2 tenths of the way into the year 2008. (Since a tenth of a year is 1.2 months, 2 tenths of the year should be 2.4 months into 2008 which would be around the middle of March 2008).
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Check my work to ensure that I didn't make some dumb error or a "fat finger" calculator mistake.
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Hope this helps you to understand the problem.
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