73. Savings Bonds U.S. Savings Bonds, Series I, pay interest at a rate of 3% com

Question

73. Savings Bonds U.S. Savings Bonds, Series I, pay interest at a rate of 3% compounded quarterly. How much would a bond purchase for $1000 be worth in 10 years? These bonds stop paying interest after 30 years. Why do you think this is so? (Hint: Think about how much this bond would be worth in 80 years.)

75. Pharmacology When a drug is administered orally, the amount of the drug present in the bloodstream

of the patient can be modeled by a function of the form

C (t) = atebt

where C(t) is the concentration of the drug (in milligrams per liter), t is the number of hours since the drug was administered, and a and b are positive constants. For a 300 mg dose of the asthma drug aminophylline, this function is

C (t) = 4.5te 0.275t

(Source: Merck Manual of Diagnosis and Therapy)

a. How much of this drug is present in the bloodstream at t = 0? b. How much of this drug is present in the bloodstream after 1 hour? c. Sketch the graph of this function, either by hand or with a graphing utility, with t ranging from 0 to 20. d. What happens to the value of the function as t ? Does this make sense in the context of this

problem? Why?

e. Use a graphing utility to find the time at which the concentration of this drug reaches its maximum.

f. Use a graphing utility to determine when the concentration of this drug reaches 3 mg/L for the second

time. (This will occur after the concentration peaks.)

77. In the definition of the exponential function, why is a = 1 excluded?

79. The graph of the function f (x) = Ca x passes through the points (0, 12) and (2, 3). a. Use f (0) to find C. b. Is this function increasing or decreasing? Explain. c. Now that you know C, use f (2) to find a. Does your value of a confirm your answer to partb?

Explanation / Answer

Principle, P = $1000

Rate, r = 3% = 0.03

Time t = 10 years

Compounding frequency n = quarterly = 4 (4 compounding in a year)

Final Amount after 10 years = P(1+r/n)nt

= 1000*(1+0.03/4)10*4

= 1000*(1.0075)40

= 1000*1.3484

= $1348.4

After 10 years this bond will be worth = $1348.4

Interest gained = 1348.4 - 1000 = 348.4

average interest per year = 348.4/10 = $34.84

At 30 years the value of this bond will be = P(1+r/n)nt

= 1000*(1+0.03/4)30*4

= 1000*(1.0075)120

= 1000*2.4514

= $2451.4

Similarly the value of this bond if interest is continued at 31 years = 1000*(1+0.03/4)31*4

= 1000*(1.0075)124

= 1000*2.5257

= 2525.7

Interest in 1 year from 30 to 31 years = 2525.7 - 2451.4 = $74.3

Now, if we compare this to average interest during 10 year period at $34.84 we find that it is almost double at 30 years at $74.3 hence these bonds stop paying interest after 30 years.

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