Section10.2Alternate Forms for Exponential Models Activities
Activity10.2.1.
In Section 10.1, we investigated the growth of a colony of the bacteria lactobacillus acidophilus, part of turning milk into yogurt. We noted that based on experiments a new generation of bacteria are formed every 70 minutes, that is, the population doubles every 70 minutes, and assumed that when we start tracking the data we have 3 g of cells. If \(a\) is the number of 70-minute periods since we started tracking and \(M(a)\) is the mass of the colony, we found that
Our current formula’s input is \(a\text{,}\) the number of 70-minute periods. That’s an awkward unit of time! It would be nicer to have a formula whose input was \(t\text{,}\) the number of minutes. Find a formula that expresses the relationship between \(a\) and \(t\text{.}\) That is, write an equation
\begin{equation*}
t = \text{a formula with } a \text{ in it} \text{.}
\end{equation*}
Suppose that a quantity \(Q\) is changing by \(r\%\) every \(n\)-minutes. Write a formula that shows the relationship between \(Q\) and \(t\text{,}\) where \(t\) is measured in minutes.
A new cat video is posted and 12 people view it the first day. Every 4 days afterward the number of people who see it triples. Let \(V(d)\) be the number of views the video has after \(d\) days.
Find the percent change in the number of views the video gets each day. Use this information to find an alternate form for the formula. Hint: Start by finding the number of views the video gets on the second day. Use this number and the fact that it got 12 views on the first day to find the percent change.
Barium-133 has a half-life of 10.551 years. This means that after 10.551 years only half of the original amount will remain. This is the result of the radioactive isotope breaking down into other substances. Suppose that we obtain 7.000 grams of barium-133.
In April 1986, a flawed reactor design played a part in the Chernobyl nuclear meltdown. Approximately 14252 becqurels (Bqs), units of radioactivity, were initially released into the environment. Only areas with less than 800 Bqs are considered safe for human habitation. The function \(f(x) = 14252(0.5)^{x/32}\) describes the amount, \(f(x)\text{,}\) in becqurels, of a radioactive element remaining in the area \(x\) years after 1986.
What is the half-life of the radioactivity in Chernobyl? That is, how long does it take for there to be half as much radioactivity as there was at the start? Explain in words how to answer this question by looking at the formula, without calculating anything.
The number \(e \approx 2.71828\) is often used as the base for exponential models in science, possibly because exponential models with base \(e\) play especially nicely with calculus. The number \(e\) is also used in finance to model continuously compounded interest, a useful theoretical tool. The amount, \(A(t)\text{,}\) that an initial deposit of \(P\) has grown to after \(t\) years in an account that is compounding continuously at a rate of \(r\%\) is
Suppose Parker deposits $1000 into an account that earns interest at a rate of 2% per year. Parker doesn’t remember how many times a year interest is compounded. Use the continuous compounding formula to find a model for the most money they could possibly have after \(t\) years (assuming they don’t deposit any more money in the account).
To solve this problem algebraically, you’ll need \(\log_e\text{.}\) The base \(e\) is so common in scientific applications that \(\log_e\) has been given a special name and notation: it’s called the natural logarithm, and denoted \(\ln\) (read as “el-en”). Many calculators have an \(\ln\) button. It might seem strange to call \(\log_e\) the “natural” logarithm - there doesn’t seem to be much about it that’s natural! This name makes more sense in the context of calculus, where \(\ln\) is the best behaved logarithm.
If Parker’s money is compounded continuously, what is the percent change in the amount in their account each year? Use this information to find an alternate form for the formula that looks more like the exponential models we’ve been working with.
