  Assignment 63

 "Were it not for number and its nature, nothing that exists would be clear to anybody either in itself or in its relation to other things. You can observe the power of number exercising itself in all the acts and the thoughts of men."- (Philolaus, a Pythagorean, 5th century B.C.) Grace Chisholm Young (1864-1944): The English-born Young was one of the first to demonstrate applications of set theory to problems in mathematical analysis. She overcame considerable prejudice against women to achieve a Ph.D. in mathematics from Gottingen University in Germany in 1895. Her most distinguished work appears in a group of papers published from 1914 to 1916 in which she presented and developed theories and concepts in differential calculus. She worked with her husband, English mathematician William Young (1863-1942) on more than 200 articles and books. In 1905, she published a geometry book that included many paper-folding patterns for three-dimensional models. In 1906, her publication The Theory of Sets of Points, demonstrated applications of set theory to problems in mathematical analysis. Young was a well-loved, generous woman with a variety of talents and tremendous energy. Why did Herkimer raise his son to be nasty in a very quiet way? Answer: He was told that children should be obscene, but not heard. Herky wants to know: What should you assume when a sign on a tailor shop reads "Closed for alterations?" When you read an autobiography, why do you learn nothing about the history of cars? ASSIGNMENT #63 Reading: Review Chapter 8, as necessary. Written: Page 507/69,70 Page 508 (Self-Test)/1-17. NOTE: Consider this to be a review sheet for the next exercise. Mathematical word analysis:SECANT: From the Latin secare ("to cut"). A geometric secant is a line that "cuts through" a circle..
If \$1000 is invested now, here is the accumulation after 10 years at

• 8% (a true rate): \$1000(1.08)10 = \$2,158.92.
• 8% compounded monthly: \$1000(1 + .08/12)120 = \$2,219.64.
• 8% compounded daily: \$1000(1 + .08/365)3650 = \$2,225.25.
• 8% compounded continuously: \$1000e(.08)10 = \$1000e.8 = \$2,225.54.

Suppose you want to have \$100,000 ten years from now. Let x = the amount that must be invested now in order to accumulate to \$100,000. What would you have to invest if the interest rate was (a) a true rate of 8%? (b) 8% compounded monthly? (c) 8% compounded continuously? Here are the solutions.

(a) x(1.08)10 = \$100,000 ==> x = \$100,000/(1.08)10 = \$46,319.35.

(b) x(1 + .08/12)120 = \$100,000 ==> x = \$100,000/(1 + .08/12)120 = \$45,052.35.

(c) xe(.08)10 = \$100,000 ==> x = \$100,000/e.8 = \$44,932.90.

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The law of uninhibited growth is exactly similar to money that grows under continuous compounding.

Here is an example:

If the number N of bacteria present in a culture at time t (in hours) obeys the equation N = 1000e0.01t, then we can ask questions like "How long will it take the culture to increase to 10,000?" We need to solve the following equation for t: 10,000 = 1000e0.01t ==> e0.01t = 10 ==> 0.01t = ln(10) ==> t = ln(10)/.01 = 230.265, or about 230 hours.

Here is an example similar to problem #4 on page 419.

It is known that Iodine-131 decays according to the equation A = A0e-0.087t, where A0 is the initial amount present at time t (in days). We can ask "How long will it take an amount of Iodine-131 to decrease to 10% of its original amount?" We need to solve the equation A0/10 = A0e-0.087t ==> 0.1 = e-0.087t ==> -0.087t = ln(0.1) ==> t = ln(0.1)/(-0.087) = 26.466, or about 26.5 days.

 Problem: Solve for x: 5e3x -log(1000) = ln e57. Solution (with communication): 5e3x -log(1000) = ln e57 ==> 5e3x - 3 = 57 ==> 5e3x = 60 ==> e3x = 12 ==> 3x = ln 12 ==> x = (1/3)ln 12 = 0.8283. Problem: If w = -2000ln(y/27). what is the value of y when w = 12,000? Solution (with communication): w = 12,000 ==> 12,000 = -2000ln(y/982) ==> ln(y/982) = -6 ==> y/982 = e-6 ==> y = 982e-6 = 2.434.