9.3 Applications of the Exponential Function 321
i. Approximate the -values where the graph of intersects the horizontal line . Equivalently, trace along the graph until the -coordinate is .
ii. Approximate the zeros of .
LetÕs use the second method to approximate the solution to 2 1.04. Estimate the zeros of the function 1.04 2. You should come up with approximately 17.67; the money doubles approximately every 17.67 years.9
Suppose we put "0 dollars in a bank at interest rate per year compounded annually. (If theinterestrateis5%, then 0.05.)Assumethemoneyisputinthebankatthebeginning of the year and interest is compounded at the end of the year. Then each year the balance is multiplied by 1 . The balance after years is given by the function
""01 . (9.1)
Inparticular,if$5000isputinabankpaying4%interestperyearcompoundedannually,this formula says that " 5000 1.04, which agrees with our answer from Example 9.6. Notice that is 0.04 and not 4; if we used 4 instead, then at the end of one year, the account would have grown from $5000 to $5000 1 4 $25,000! (This is a nice deal if you can get it, but not a reasonable answer to this problem.)
COMMENT When we write the equation "5000 1.04 and let the domain be 0, we are modeling a discontinuous phenomenon with a continuous model. This equation will only mirror reality if is an integer, i.e., if the bank has just paid the annual interest to the account. For instance, if 0, 1, then the interest has not yet been compounded so the balance should be exactly $5000 over this interval. At 1 it should jump to $5200. However, it is convenient to model this discrete process with a continuous function, as we have seen done in examples in previous chapters.
actual function (not continuous)
0 1 2 3 t
EXAMPLE 9.7 Suppose we put the $5000 in a bank paying interest at an interest rate of 4% per year but this time with interest compounded quarterly.
This does not mean that you get 4% interest each quarter.
9 Thisillustratesaveryusefulruleofthumbforestimatingdoublingtime:theRuleof70.Thisruleofthumbsaysthattoestimate the number of years it will take something growing at a rate of % per year to double, we calculate 70. In Example 9.6, we calculate 704 17.5, and see that this is a very close estimate. Keep in mind that this is not a hard and fast rule but rather only an approximation. When we study logarithms we will see why this rule works, and for what values of it works best.
322 CHAPTER 9 Exponential Functions
Rather, your money earns 1%, 4% , interest each quarter of a year. The advantage is that
interest is computed on the interest more frequently (four times per year, as opposed to just once per year in Example 9.6). The balance is growing by 1% every 1/4 of a year, so it grows according to
(Refer to Example 9.3 if you need help with this.) Notice that the balance increases by 1% (is multiplied by 1.01) each quarter, so when 1, interest has been compounded four times, as expected. "150001.014 50001.040604 , so each year the interest is effectively 4.0604 %. This is called the effective annual interest rate.
4% is called a nominal annual interest rate because the interest is 4% Òin name only.Ó
EXERCISE 9.6 Compare the amount of money you would have after 10 years in Example 9.6 with the amount in Example 9.7.
Answers are provided at the end of the chapter.
Suppose you put $"0 in a bank at interest rate per year compounded times per year. Then the interest rate per compounding period is . Each year there are compounding
periods, so in years there are compounding periods. (The growth is every th of a year.) Then
"the amount of money in the account at the end of years,
"0 the original amount of money,
the annual interest rate (If the rate is 5%, then 0.05., the number of compounding periods per year.
(If you have any trouble following this, make a table as was done in Example 9.2. Also, verify that Equation (9.1) is just a special case of the formula presented above.)
EXERCISE 9.7 Bank A gives a nominal interest rate of 6% per year compounded monthly. Bank B gives 6.1% interest per year compounded annually. Which is the preferable arrangement for an investor?
Throughexperimentationscientistshavefoundthatunstableradioactiveisotopesrevertback to nonradioactive form at a rate proportional to the amount of the isotope present. The rate of decay of radioactive substances is typically indicated by identifying the half-life of the substance; this is the amount of time it takes for half of the original amount of radioactive substance to revert (decay) to its nonradioactive form.
9.3 Applications of the Exponential Function 323
EXAMPLE 9.8 Tritium, the radioactive hydrogen isotope H3, has a half-life of 12.3 years. (Tritium can be used for determining the age of wines in a process similar to carbon-dating, which we will
look at in Exercise 9.8.) Let #0 denote the amount of tritium at time 0. Find a formula for #, the amount of tritium years later.
Use the ideas of Example 9.3. This idea of radioactive half-life is very similar to the idea of doubling time we discussed when working with E. coli. When given information about doubling time we used 2 as the base; in the case of tritium it makes sense to use 1/2 as our base.
Make a table of values.
0 #0 12.3 0.5 #0
24.6 (0.5)(0.5) #0 36.9 0.53 #0
Radioactive material decays at a rate proportional to itself, so an exponential function models the process. Therefore, # for some constants and .
The Òinitial condition,Ó the value of # when 0, helps us Þnd . At 0, # #0. Therefore, #0 0 and ##0.
Now we must Þnd by using the fact that #12.3 2#0.
0.5#0 #012.3, so 0.5 12.3.
How do we solve for ?
Recall: If 3 7, we can solve for by taking the cube root of both sides. That is,
313 713, so 713.
We can apply the same approach to solving for in the equation 0.5 12.3.
0.5112.3 12.3112.3 0.5112.3
##0[0.5112.3] or #00.512.3
324 CHAPTER 9 Exponential Functions
A person using Approach 3 could decide that 0.5112.3 is a clumsy way to express the value of and therefore replace the expression by a numerical approximation. 0.5112.3 0.945.Theanswercanbeexpressedas:##00.945.#00.945 isapproximatelythe same as #00.512.3. Very large values of will yield greater discrepancies than smaller ones. It is advisable to work with exact values and save any rounding off until the end of
the problem. If you donÕt do this, then youÕll need to think hard about error propagation. We have two different forms in which we can express # as a function of , in years:
12.3 ##00.945 and # #0 2 .
From the former, we can quickly determine that every year there is 94.5% of the amount of tritium there was the previous year. (Another way of putting this is that each year the amount of tritium decreases by 5.5%, because 100% 94.5% 5.5%.) From the latter, we can read the information that every 12.3 years half the tritium remains.
We can take advantage of the fact that there are many (inÞnitely many, in fact) ways to write any given number. For example, 64 26 409612 0.56 43, and so on.
EXERCISE 9.8 Carbon Dating Cosmic rays in the upper atmosphere convert some of the stable form of carbon, 12, to the unstable radioactive isotope 14, a form with two extra electrons. Throughphotosynthesis,plantstakeincarbondioxidefromtheatmosphereandincorporate the carbon atoms; therefore plants contain the same ratio of 14 to 12 atoms as exists in the atmosphere. Animals eat plants and so have the same ratio of 14 to 12 as do the plants and the atmosphere. Thus, during the lifetime of a living organism the ratio of 14 to 12 in the organism is Þxed; it is the same as the ratio in the earthÕs atmosphere.
When an organism dies it stops incorporating carbon atoms. The 14 that is in the organism decays exponentially (reverting to the stable 12 form), and the 14 supply is not replenished.Assumingthattheratioof12 to14 hasremainedconstantintheatmosphere, scientists are able to use the unstable 14 isotope to date organisms.
The half-life of carbon-14 is approximately 5730 years.
(a) Let 0 denote the amount of radioactive carbon in an organism right before it died. Find a formula for , the amount of radioactive carbon in this organism years after it has died.
(b) A sample from a tree that perished in a volcanic eruption has been analyzed and found to have 70% of the 14 it would have had were it alive. Approximately how long ago did the volcanic eruption occur?
Answers are provided at the end of the chapter.
Writing and Rewriting Exponential Functions; Cheap Information, Yours for the Taking
Doubling Time in Minutes; Doubling Time in Hours. We know that the number of organisms in a colony of E. coli under ideal reproductive conditions doubles in size every 20 minutes. The number of E. coli after minutes is given by
0 220, where 0 0 and is given in minutes.
9.3 Applications of the Exponential Function 325
Doubling every 20 minutes is equivalent to doubling every 1/3 hour (or doubling three times each hour) so we can write
0 23, where is given in hours.
Percentage Change Per Minute, Percentage Change Per Hour. If we rewrite 0 220 in the form 0, we can readily read off the percent change in population each minute.
0 220 02120 01.0353 01 0.0353,
so every minute the E. coli population increases by about 3.53%.10 For measured in hours,
023 023 08 01 7 0 1 100 ,
so every hour the E. coli population increases by 700%.
Nominal Annual Interest and Effective Annual Interest. Suppose a bank has a nominal interest rate of 5.4% per year compounded monthly. If "0 dollars are deposited in the account at time 0 and left for years, the balance in the account will be
""0 1 0.05412 .
Effectively the annual interest will be slightly more than 5.4% due to the compounding. Rewriting the function above lets us read off the effective annual interest rate:
""0 1 0.05412 "01.004512 "0[1.004512] "01.05536.
We see that a nominal annual interest rate of 5.4% compounded monthly is equivalent to an effective annual growth rate of about 5.536%.
The idea behind the last few examples is that an exponential equation can be written in many different forms. Different forms make different bits of information (annual growth rate, or quarterly growth rate, or doubling time, or tripling time, etc.) available at our Þngertips.
EXERCISE 9.9 A Þltering system is used to purify water used in industrial processes. The longer the water remains in the Þltering system the less contaminants remain in the water. The system removes contaminants at a rate proportional to the amount of contaminants in the water. Let be the number of milligrams of contaminants in the water hours after it enters the system. Suppose that when the water enters the system it contains 0 milligrams of contaminant.
(a) Find if
i. the system removes 20% of the contaminants every hour.
10 Alternatively, we could have calculated 1 and computed the percent change from 0 to 1: