Assignment 58

"The moving power of mathematical invention is not reasoning, but imagination." -- (Augustus De Morgan, 1806-1871)

Math History Tidbit:

Numbers have interesting historical backgrounds, and there are many interesting stories behind just about any counting number. Here are just a few very brief tidbits relating to numbers. (Some research will yield many interesting stories.)

1: This is the "unity" number, which serves as a base for all others. The concept of 1 is extremely important in the monotheistic religions of the world, including Christianity, Judaism, and Islam.

2. Regarded as a number representing many human characteristics (eg. male, female) and natural phenomena. It is the basis for the binary system and is the only even prime number. Interestingly, the number 2 also has many negative interpretations. Consider phrases like two-faced and double tongued.

3. Important in many religions (eg. Father, Son, and Holy Ghost) and directly associated with the triangle, which appears in many guises as a mystical symbol. Religions that put an emphasis on 3 include Hinduism, Islam, and Judaism.

4. Very common in the natural world (eg. north, south, east, west). The ancient Greeks saw "fourness" in things, such as the four basic elements of, air, fire, water. Many North American Indians, including the Dakota, Sioux, and Zuni, put special emphasis on the number 4.

Herkimer's Corner

When Herkimer was a pilot who was moonlighting as a male stripper, why was he arrested at the airport?

Answer: He attempted a takeoff on the runway.

Herky's friends:

CRYSTAL BALL...she made her living as a fortune teller.

AL E. GATOR...he studied water-dwelling reptiles.


Reading: Section 8.2, pages 474-476.

Written: Page 478/43-56 (Real problems!)

Items for reflection:

Mathematical word analysis:
SKEW: Two lines that cannot be contained in the same plane are called skew lines. The word skew comes from skiuhwan, a Middle English word meaning "escape." If two lines are skew lines, then each line has "escaped" any plane that contains the other line.

Exponential decay is very similar to exponential growth. Assume that a quantity of 10,000 decays at a rate of 2% a year. Then, in x years, the amount present could be represented by a function f, where

f(x) = y = 10000(1-.02)x = 10000(.98)x.

This should make sense. If a quantity loses 2% a year, then for any year, you should have 98% of what you had the previous year. We can ask questions like:

How much will be available in 50 years?

Answer is f(50) = 10000(.98)50 = 3642.

How long will it take before only 20% of the original amount remains?

Note that we don't yet have the MATH POWER to directly solve for x in the equation 2000 = 10000(.98)x. We lack the concept of a logarithm. (Don't fret! We will learn about logarithms soon.) Perhaps the best way to address the question is to use the TABLE feature on the TI-83. Set Y1 = 10000*.98^x and look at TABLE. If you scroll down the table and look for 2000 in the Y1 column, you will find it will take about 80 years.


The table below relates to finance (previous assignment):



10% (true rate).

Annual yield = 10%

10% compounded semiannually

(1+.10/2)2 = 1.1025. Annual yield = 10.25%

10% compounded quarterly

(1+.10/4)4 = 1.103812891. Annual yield = 10.38%

10% compounded monthly

(1+.10/12)12= 1.104713067. Annual yield = 10.47%

10% compounded daily

(1+.10/365)365= 1.05155782. Annual yield = 10.51%

10% compounded continuously

e.10 = 1.105170918. Annual yield = 10.51%

Note that the annual yields for the daily rate and the continuous rate are equal when expressed to two decimal places. However, the continuous yield is slightly greater. This could make a difference with the investments of large amounts of money.

Problem: A purchased item loses 11% of its value each year. The purchase price was $4,000.

(a) What will be the value of the item after 5 years?
(b) How many years will it take for the value to decrease to $1,000?

Solution (with communication):

(a) $4,000(1- 0.11)5 = $4,000(0.89)5 = $2,233.62.

(b) We need to solve $4,000(0.89x) = $1,000 ==> 0.89x = 0.25. By trial and error (possibly using the TABLE feature on your calculator) we find the 0.8911 = 0.27752 and 0.8912 = 0.24699. We conclude it will take 12 years for the value to decrease to $1,000.

Problem: An item valued at $1,000 increases in value by 6% a year. Another item valued at $5,000 decreases in value by 8% year. After how many years will the items have equal value, and what is that value?

Solution (with communication): If x is the number of years requested, then we must solve

$1,000(1.06)x = $5,000(0.92)x

Using the TABLE feature on the calculator, we find that at x = 11, the values (to the nearest dollar) are $1,898 and $1,998 respectively. At x = 12, the respective values are $2,012 and $1838. So, a reasonable response would be that both would have a value of approximately $1,940 after 11 years.

If we use the calculator for a very accurate graphics solution, we would find that both values are $1,938.77 when x = 11.36.