Solving equations and setting up formulas for solving them.?
March 9th, 2010Examples, sources, would help.
1) Read the problem and understand it.
2) "What is the question?"
3) Write down "Let x denote (what you are looking for)." If there are two unknown values (variables) add in "Let y denote ...", etc.
4) Read the sentences and develop one equation if there is one variable, two equations if there are two variables, etc.
5) Solve the equations for the variables.
6) Translate the solutions in 5) to answer the question.
Let's look at two examples ...
I. Suppose that Alice is driving on an interstate at a constant speed. If she is one hundred and sixty miles from her destination at 1:15 p.m. and is one hundred and thirty miles from her destination at 1:45 p.m., then at what time will Alice reach her destination?
Let t denote the number of minutes after 1:15 p.m. and let f(t) denote the number of miles Alice is from her destination at time t described above. The question is to find the value of t for which f(t) = 0.
Here, f(0) = 160 and f(30) = 130 (with 1:45 being 30 minutes after 1:15).
As the speed is constant, f(t) is a linear equation and f(t) = mt + b for two constants m and b.
Therefore, f(0) = m(0) + b = b = 160 and
f(30) = m(30) + b = 30m + 160 = 130 which is true when m = - 1.
Therefore, f(t) = - t + 160.
f(t) = 0 when t = 160. What time is it 160 minutes after 1:15 p.m.?
Calculate that and you have the answer.
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II. Suppose that, on July 1, 2011, you make one investment that can provide an annual interest of eight percent until 2030. On what day will your investment have doubled? On what day will it have tripled?
First, let us assume that you invest $1000 and let t denote the number of years after July 1, 2011, and let f(t) denote the value of your investment t years after that day.
Then f(t) = 1000 (1 + .08)^t .
The investment will have doubled when f(t) = 2000 and
(1.08)^t = 2. Solving for t: ln[(1.08) ^ t] = ln(2),
t ln(1.08) = ln(2) and t = ln(2) / ln(1.08) = 0.6931 / 0.0764 = 9.02
.02 is 1/50 and 1/50th of a year is one week.
Nine years and one week after July 1, 2011 is July 8, 2020.
(I did not use a calculator, so I may be a few days off.)
Find the day on which it has tripled.
One other comment on II:
It is good to know the "72 rule" (an appoximation). If you invest one dollar, the interest each year is x percent and it is kept there for y years, then the value will be close to two dollars (doubling) if x times y equals 72. In II. above, the value doubled with the product equaling about 8 times 9.02. This approximation is close when the interest ranges from 4 percent to 12 percent.
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