4.1 Growth and Decay

Since the applications in this section deal with functions of time, we’ll denote the independent variable by $t$. If $Q$ is a function of $t$, $Q'$ will denote the derivative of $Q$ with respect to $t$; thus,

\[ Q'={dQ\over dt}. \]

Exponential Growth and Decay

One of the most common mathematical models for a physical process is the exponential model, where it’s assumed that the rate of change of a quantity $Q$ is proportional to $Q$; thus

\begin{equation} Q'=aQ, \tag{4.1.1}\end{equation}

where $a$ is the constant of proportionality.

From Example 3, the general solution of (4.1.1) is

\[ Q=ce^{at} \]

and the solution of the initial value problem

\[ Q'=aQ, \quad Q(t_0)=Q_0 \]

\begin{equation} Q=Q_0e^{a(t-t_0)}. \tag{4.1.2}\end{equation}

Since the solutions of $Q'=aQ$ are exponential functions, we say that a quantity $Q$ that satisfies this equation grows exponentially if $a > 0$, or decays exponentially if $a < 0$ (Figure 4.1.1).

Figure 4.1.1. Exponential growth and decay

Radioactive Decay

Experimental evidence shows that radioactive material decays at a rate proportional to the mass of the material present. According to this model the mass $Q(t)$ of a radioactive material present at time $t$ satisfies (4.1.1), where $a$ is a negative constant whose value for any given material must be determined by experimental observation. For simplicity, we’ll replace the negative constant $a$ by $-k$, where $k$ is a positive number that we’ll call the decay constant of the material. Thus, (4.1.1) becomes

\[ Q'=-kQ. \]

If the mass of the material present at $t=t_0$ is $Q_0$, the mass present at time $t$ is the solution of

\[ Q'=-kQ,\quad Q(t_0)=Q_0. \]

From (4.1.2) with $a=-k$, the solution of this initial value problem is

\begin{equation} Q=Q_0e^{-k(t-t_0)}. \tag{4.1.3}\end{equation}

The half–life $\tau$ of a radioactive material is defined to be the time required for half of its mass to decay; that is, if $Q(t_0)=Q_0$, then

\begin{equation} Q(\tau+t_0)={Q_0\over 2}. \tag{4.1.4}\end{equation}

From (4.1.3) with $t=\tau+t_0$, (4.1.4) is equivalent to

\[ Q_0e^{-k\tau}={Q_0\over 2}, \]

\[ e^{-k\tau}={1\over 2}. \]

Taking logarithms yields

\[ -k\tau=\ln{1\over 2}=-\ln2, \]

so the half-life is

\begin{equation} \tau={1\over k}\ln2. \tag{4.1.5}\end{equation}

(Figure 4.1.2). The half-life is independent of $t_0$ and $Q_0$, since it’s determined by the properties of material, not by the amount of the material present at any particular time.

Figure 4.1.2. Half-life of a radioactive substance

Example 4.1.1

A radioactive substance has a half-life of 1620 years.

If its mass is now 4 g (grams), how much will be left 810 years from now?
Find the time $t_1$ when 1.5 g of the substance remain.

Solution (a) From (4.1.3) with $t_0=0$ and $Q_0=4$,

\begin{equation} Q=4e^{-kt}, \tag{4.1.6}\end{equation}

where we determine $k$ from (4.1.5), with $\tau$= 1620 years:

\[ k={\ln2\over\tau}={\ln2\over 1620}. \]

Substituting this in (4.1.6) yields

\begin{equation} Q=4e^{-(t\ln2)/1620}. \tag{4.1.7}\end{equation}

Therefore the mass left after 810 years will be

\[\begin{array}{rl} Q(810) &=4e^{-(810\ln2)/1620}=4e^{-(\ln2)/2} \\ &=2\sqrt{2} \mbox{ g}. \end{array}\]

Solution (b) Setting $t=t_1$ in (4.1.7) and requiring that $Q(t_1)=1.5$ yields

\[ {3\over2}=4e^{(-t_1\ln2)/1620}. \]

Dividing by 4 and taking logarithms yields

\[ \ln{3\over8}=-{t_1\ln2\over1620}. \]

Since $\ln3/8=-\ln8/3$,

\[ t_1=1620{\ln8/3\over\ln2}\approx 2292.4\;\mbox{ years}. \]

Interest Compounded Continuously

Suppose we deposit an amount of money $Q_0$ in an interest-bearing account and make no further deposits or withdrawals for $t$ years, during which the account bears interest at a constant annual rate $r$. To calculate the value of the account at the end of $t$ years, we need one more piece of information: how the interest is added to the account, or—as the bankers say—how it is compounded. If the interest is compounded annually, the value of the account is multiplied by $1+r$ at the end of each year. This means that after $t$ years the value of the account is

\[ Q(t)=Q_0(1+r)^t. \]

If interest is compounded semiannually, the value of the account is multiplied by $(1+r/2)$ every 6 months. Since this occurs twice annually, the value of the account after $t$ years is

\[ Q(t)=Q_0\left(1+{r\over 2}\right)^{2t}. \]

In general, if interest is compounded $n$ times per year, the value of the account is multiplied $n$ times per year by $(1+r/n)$; therefore, the value of the account after $t$ years is

\begin{equation} Q(t)=Q_0\left(1+{r\over n}\right)^{nt}. \tag{4.1.8}\end{equation}

Thus, increasing the frequency of compounding increases the value of the account after a fixed period of time. Table 4.1.1 shows the effect of increasing the number of compoundings over $t=5$ years on an initial deposit of $Q_0=100$ (dollars), at an annual interest rate of 6%.

Table 4.1.1. Table The effect of compound interest

\[\begin{array}{|c|c|}\hline n & \dst\$100 \left(1+{.06\over n}\right)^{5n}\\ \hbox{(number of compoundings} & \hbox{(value in dollars}\\ \hbox{per year)} &\hbox{after 5 years)}\\ \hline 1 & \$133.82 \\ 2 & \$134.39 \\ 4 & \$134.68 \\ 8 & \$134.83 \\ 364 & \$134.98 \\\hline \end{array}\]

You can see from Table 4.1.1 that the value of the account after 5 years is an increasing function of $n$. Now suppose the maximum allowable rate of interest on savings accounts is restricted by law, but the time intervals between successive compoundings isn’t ; then competing banks can attract savers by compounding often. The ultimate step in this direction is to compound continuously, by which we mean that $n\to\infty$ in (4.1.8). Since we know from calculus that

\[ \lim_{n\to\infty} \left(1+{r\over n}\right)^n=e^r, \]

this yields

\[\begin{array}{rl} Q(t) & \dst=\lim_{n\to\infty} Q_0\left(1+{r\over n}\right)^{nt}=Q_0 \left[ \lim_{n\to\infty} \left(1+{r\over n}\right)^n\right]^t \\[12pt] &=Q_0e^{rt}. \end{array}\]

Observe that $Q=Q_0e^{rt}$ is the solution of the initial value problem

\[ Q'=rQ, \quad Q(0)=Q_0; \]

that is, with continuous compounding the value of the account grows exponentially.

Example 4.1.2

If $150 is deposited in a bank that pays $5{1\over2}$% annual interest compounded continuously, the value of the account after $t$ years is

\[ Q(t)=150e^{.055t} \]

dollars. (Note that it’s necessary to write the interest rate as a decimal; thus, $r=.055$.) Therefore, after $t=10$ years the value of the account is

\[ Q(10)=150e^{.55} \approx \$259.99. \]

Example 4.1.3

We wish to accumulate $10,000 in 10 years by making a single deposit in a savings account bearing $5{1\over2}$% annual interest compounded continuously. How much must we deposit in the account?

Solution The value of the account at time $t$ is

\begin{equation} Q(t)=Q_0e^{.055t}. \tag{4.1.9}\end{equation}

Since we want $Q(10)$ to be $10,000, the initial deposit $Q_0$ must satisfy the equation

\begin{equation} 10000=Q_0e^{.55}, \tag{4.1.10}\end{equation}

obtained by setting $t=10$ and $Q(10)=10000$ in (4.1.9). Solving (4.1.10) for $Q_0$ yields

\[ Q_0=10000e^{-.55} \approx \$5769.50. \]

Mixed Growth and Decay

Example 4.1.4

A radioactive substance with decay constant $k$ is produced at a constant rate of $a$ units of mass per unit time.

Assuming that $Q(0)=Q_0$, find the mass $Q(t)$ of the substance present at time $t$.
Find $\lim_{t\to\infty} Q(t)$.

Solution (a) Here

\[ Q'=\mbox{ rate of increase of } Q - \mbox{ rate of decrease of } Q. \]

The rate of increase is the constant $a$. Since $Q$ is radioactive with decay constant $k$, the rate of decrease is $kQ$. Therefore

\[ Q'=a-kQ. \]

This is a linear first order differential equation. Rewriting it and imposing the initial condition shows that $Q$ is the solution of the initial value problem

\begin{equation} Q'+kQ=a, \quad Q(0)=Q_0. \tag{4.1.11}\end{equation}

Since $e^{-kt}$ is a solution of the complementary equation, the solutions of (4.1.11) are of the form $Q=ue^{-kt}$, where $u'e^{-kt}=a$, so $u'=ae^{kt}$. Hence,

\[ u={a\over k}e^{kt}+c \]

and

\[ Q=ue^{-kt}={a\over k}+ce^{-kt}. \]

Since $Q(0)=Q_0$, setting $t=0$ here yields

\[ Q_0={a\over k}+c \mbox{\quad or \quad} c=Q_0-{a\over k}. \]

Therefore

\begin{equation} Q={a\over k}+\left(Q_0-{a\over k}\right)e^{-kt}. \tag{4.1.12}\end{equation}

Solution (b) Since $k > 0$, $\lim_{t\to\infty} e^{-kt}=0$, so from (4.1.12)

\[ \lim_{t\to\infty} Q(t)={a\over k}. \]

This limit depends only on $a$ and $k$, and not on $Q_0$. We say that $a/k$ is the steady state value of $Q$. From (4.1.12) we also see that $Q$ approaches its steady state value from above if $Q_0 > a/k$, or from below if $Q_0 < a/k$. If $Q_0=a/k$, then $Q$ remains constant (Figure 4.1.3).

Q(t) approaches the steady state value ak as t — Figure 4.1.3. $Q(t)$ approaches the steady state value $\dst{a\over k}$ as $t\to\infty$

Carbon Dating

The fact that $Q$ approaches a steady state value in the situation discussed in Example 4 underlies the method of carbon dating, devised by the American chemist and Nobel Prize Winner W.S. Libby.

Carbon 12 is stable, but carbon-14, which is produced by cosmic bombardment of nitrogen in the upper atmosphere, is radioactive with a half-life of about 5570 years. Libby assumed that the quantity of carbon-12 in the atmosphere has been constant throughout time, and that the quantity of radioactive carbon-14 achieved its steady state value long ago as a result of its creation and decomposition over millions of years. These assumptions led Libby to conclude that the ratio of carbon-14 to carbon-12 has been nearly constant for a long time. This constant, which we denote by $R$, has been determined experimentally.

Living cells absorb both carbon-12 and carbon-14 in the proportion in which they are present in the environment. Therefore the ratio of carbon-14 to carbon-12 in a living cell is always $R$. However, when the cell dies it ceases to absorb carbon, and the ratio of carbon-14 to carbon-12 decreases exponentially as the radioactive carbon-14 decays. This is the basis for the method of carbon dating, as illustrated in the next example.

Example 4.1.5

An archaeologist investigating the site of an ancient village finds a burial ground where the amount of carbon-14 present in individual remains is between 42 and 44% of the amount present in live individuals. Estimate the age of the village and the length of time for which it survived.

Solution Let $Q=Q(t)$ be the quantity of carbon-14 in an individual set of remains $t$ years after death, and let $Q_0$ be the quantity that would be present in live individuals. Since carbon-14 decays exponentially with half-life 5570 years, its decay constant is

\[ k={\ln2\over 5570}. \]

Therefore

\[ Q=Q_0e^{-t(\ln2)/5570} \]

if we choose our time scale so that $t_0=0$ is the time of death. If we know the present value of $Q$ we can solve this equation for $t$, the number of years since death occurred. This yields

\[ t=-5570 {\ln Q/Q_0\over\ln2}. \]

It is given that $Q=.42Q_0$ in the remains of individuals who died first. Therefore these deaths occurred about

\[ t_1=-5570 {\ln .42\over\ln2} \approx 6971 \]

years ago. For the most recent deaths, $Q=.44 Q_0$; hence, these deaths occurred about

\[ t_2=-5570 {\ln .44\over\ln2} \approx 6597 \]

years ago. Therefore it’s reasonable to conclude that the village was founded about 7000 years ago, and lasted for about 400 years.

A Savings Program

Example 4.1.6

A person opens a savings account with an initial deposit of $1000 and subsequently deposits $50 per week. Find the value $Q(t)$ of the account at time $t > 0$, assuming that the bank pays 6% interest compounded continuously.

Solution Observe that $Q$ isn’t continuous, since there are 52 discrete deposits per year of $50 each. To construct a mathematical model for this problem in the form of a differential equation, we make the simplifying assumption that the deposits are made continuously at a rate of $2600 per year. This is essential, since solutions of differential equations are continuous functions. With this assumption, $Q$ increases continuously at the rate

\[ Q'=2600+.06 Q \]

and therefore $Q$ satisfies the differential equation

\begin{equation} Q'-.06Q=2600. \tag{4.1.13}\end{equation}

(Of course, we must recognize that the solution of this equation is an approximation to the true value of $Q$ at any given time. We’ll discuss this further below.) Since $e^{.06t}$ is a solution of the complementary equation, the solutions of (4.1.13) are of the form $Q=ue^{.06t}$, where $u'e^{.06t}=2600$. Hence, $u'=2600e^{-.06t}$,

\[ u=- {2600\over .06}e^{-.06t}+c \]

and

\begin{equation} Q=ue^{.06t}=-{2600\over .06}+ce^{.06t}. \tag{4.1.14}\end{equation}

Setting $t=0$ and $Q=1000$ here yields

\[ c=1000+{2600\over .06}, \]

and substituting this into (4.1.14) yields

\begin{equation} Q=1000e^{.06t}+{2600\over .06}(e^{.06t}-1), \tag{4.1.15}\end{equation}

where the first term is the value due to the initial deposit and the second is due to the subsequent weekly deposits. ∎

Mathematical models must be tested for validity by comparing predictions based on them with the actual outcome of experiments. Example 6 is unusual in that we can compute the exact value of the account at any specified time and compare it with the approximate value predicted by (4.1.15) (See Exercise 21.). The follwing table gives a comparison for a ten year period. Each exact answer corresponds to the time of the year-end deposit, and each year is assumed to have exactly 52 weeks.

\[ \vbox{\tabskip=0pt \offinterlineskip \def\tablerule{\noalign{\hrule}} \halign{\hfill#\hfill&\qquad\hfill#\hfill&\qquad\hfill#\hfill &\qquad\hfill#\hfill&\qquad#\hfill\cr \noalign{\vskip12pt} \noalign{\vskip22pt} Year & Approximate Value of $Q$ & Exact Value of $P$ & Error & Percentage Error \cr & (Example~\ref{example:4.1.6}) & (Exercise~\ref{exer:4.1.21}) & $Q-P$ &\ \; $(Q-P)/P$ \cr \noalign{\vskip16pt} \ 1 & \$ 3741.42 & \$ 3739.87 & \$ 1.55 &\ \; $.0413 $\% \cr \noalign{\vskip7pt} \ 2 &\ \, 6652.36 &\ \; 6649.17 &\ \; 3.19 &\ \; $.0479$ \cr \noalign{\vskip7pt} \ 3 &\ \, 9743.30 &\ \; 9738.37 &\ \; 4.93 &\ \; $.0506$ \cr \noalign{\vskip7pt} \ 4 & \, 13,025.38 & \, 13,018.60 &\ \; 6.78 &\ \; $.0521$ \cr \noalign{\vskip7pt} \ 5 & \, 16,510.41 & \, 16,501.66 &\ \; 8.75 &\ \; $.0530 $ \cr \noalign{\vskip7pt} \ 6 & \, 20,210.94 & \, 20,200.11 & 10.83 &\ \; $.0536$ \cr \noalign{\vskip7pt} \ 7 & \, 24,140.30 & \, 24,127.25 & 13.05 &\ \; $.0541 $ \cr \noalign{\vskip7pt} \ 8 & \, 28,312.63 & \, 28,297.23 & 15.40 &\ \; $.0544 $ \cr \noalign{\vskip7pt} \ 9 & \, 32,742.97 & \, 32,725.07 & 17.90 &\ \; $.0547 $ \cr \noalign{\vskip7pt} 10 & \, 37,447.27 & \, 37,426.72 & 20.55 &\ \; $.0549 $ \cr}} \]

4.1 Exercises

The half-life of a radioactive substance is 3200 years. Find the quantity $Q(t)$ of the substance left at time $t > 0$ if $Q(0)=20$ g.

Show answer

$Q=20e^{-(t\ln2)/3200}$ g
The half-life of a radioactive substance is 2 days. Find the time required for a given amount of the material to decay to 1/10 of its original mass.

Show answer

${2\ln10\over\ln2}$ days
A radioactive material loses 25% of its mass in 10 minutes. What is its half-life?

Show answer

$\dst{\tau=10{\ln2\over\ln4/3}}$ minutes
A tree contains a known percentage $p_0$ of a radioactive substance with half-life $\tau$. When the tree dies the substance decays and isn’t replaced. If the percentage of the substance in the fossilized remains of such a tree is found to be $p_1$, how long has the tree been dead?

Show answer

$\dst{\tau {\ln(p_0/p_1)\over\ln2}}$
If $t_p$ and $t_q$ are the times required for a radioactive material to decay to $1/p$ and $1/q$ times its original mass (respectively), how are $t_p$ and $t_q$ related?

Show answer

$\dst{{t_p\over t_q}={\ln p\over\ln q}}$
Find the decay constant $k$ for a radioactive substance, given that the mass of the substance is $Q_1$ at time $t_1$ and $Q_2$ at time $t_2$.

Show answer

$\dst{k={1\over t_2-t_1}\ln {Q_1\over Q_2}}$
A process creates a radioactive substance at the rate of 2 g/hr and the substance decays at a rate proportional to its mass, with constant of proportionality $k=.1 (\mbox{hr})^{-1}$. If $Q(t)$ is the mass of the substance at time $t$, find $\lim_{t\to\infty}Q(t)$.

Show answer

20 g
A bank pays interest continuously at the rate of 6%. How long does it take for a deposit of $Q_0$ to grow in value to $2Q_0$?

Show answer

$\dst{{50 \ln2\over 3}}$ yrs
At what rate of interest, compounded continuously, will a bank deposit double in value in 8 years?

Show answer

$\dst{{25\over 2}\ln2}$%
A savings account pays 5% per annum interest compounded continuously. The initial deposit is $Q_0$ dollars. Assume that there are no subsequent withdrawals or deposits.
1. How long will it take for the value of the account to triple?
2. What is $Q_0$ if the value of the account after 10 years is $100,000 dollars?
Show answer

(a) $=20\ln3$ yr (b). $Q_0=100000e^{-.5}$
A candymaker makes 500 pounds of candy per week, while his large family eats the candy at a rate equal to $Q(t)/10$ pounds per week, where $Q(t)$ is the amount of candy present at time $t$.
1. Find $Q(t)$ for $t > 0$ if the candymaker has 250 pounds of candy at $t=0$.
2. Find $\lim_{t\to\infty} Q(t)$.
Show answer

(a) $Q(t)=5000-4750e^{-t/10}$ (b) 5000 lbs
Suppose a substance decays at a yearly rate equal to half the square of the mass of the substance present. If we start with 50 g of the substance, how long will it be until only 25 g remain?

Show answer

$\dst{1\over 25}$ yrs;
A super bread dough increases in volume at a rate proportional to the volume $V$ present. If $V$ increases by a factor of 10 in 2 hours and $V(0)=V_0$, find $V$ at any time $t$. How long will it take for $V$ to increase to $100 V_0$?

Show answer

$V=V_0e^{t\ln10/2}\; 4$ hours
A radioactive substance decays at a rate proportional to the amount present, and half the original quantity $Q_0$ is left after 1500 years. In how many years would the original amount be reduced to $3Q_0/4$? How much will be left after 2000 years?

Show answer

$\dst{{1500\ln {4\over 3}\over\ln2}} \mbox{ yrs}; \; 2^{-4/3}Q_0$
A wizard creates gold continuously at the rate of 1 ounce per hour, but an assistant steals it continuously at the rate of 5% of however much is there per hour. Let $W(t)$ be the number of ounces that the wizard has at time $t$. Find $W(t)$ and $\lim_{t\to\infty}W(t)$ if $W(0)=1$.

Show answer

$W(t)=20-19e^{-t/20}$; $\lim_{t\to\infty}W(t)=20$ ounces
A process creates a radioactive substance at the rate of 1 g/hr, and the substance decays at an hourly rate equal to 1/10 of the mass present (expressed in grams). Assuming that there are initially 20 g, find the mass $S(t)$ of the substance present at time $t$, and find $\lim_{t\to\infty} S(t)$.

Show answer

$S(t)=10(1+e^{-t/10}); \; \lim_{t\to\infty}S(t)=10$ g
A tank is empty at $t=0$. Water is added to the tank at the rate of 10 gal/min, but it leaks out at a rate (in gallons per minute) equal to the number of gallons in the tank. What is the smallest capacity the tank can have if this process is to continue forever?

Show answer

10 gallons
A person deposits $25,000 in a bank that pays 5% per year interest, compounded continuously. The person continuously withdraws from the account at the rate of $750 per year. Find $V(t)$, the value of the account at time $t$ after the initial deposit.

Show answer

$V(t)=15000+10000e^{t/20}$
A person has a fortune that grows at rate proportional to the square root of its worth. Find the worth $W$ of the fortune as a function of $t$ if it was $1 million 6 months ago and is $4 million today.

Show answer

$W(t)=4\times 10^6(t+1)^2$ dollars $t$ years from now
Let $p=p(t)$ be the quantity of a product present at time $t$. The product is manufactured continuously at a rate proportional to $p$, with proportionality constant 1/2, and it’s consumed continuously at a rate proportional to $p^2$, with proportionality constant 1/8. Find $p(t)$ if $p(0)=100$.

Show answer

$\dst{p={100\over 25-24e^{-t/2}}}$
1. In the situation of Example 4.1.6 find the exact value $P(t)$ of the person’s account after $t$ years, where $t$ is an integer. Assume that each year has exactly 52 weeks, and include the year-end deposit in the computation.
  
  Hint
  
  At time $t$ the initial $\$1000$ has been on deposit for $t$ years. There have been $52t$ deposits of $\$50$ each. The first $\$50$ has been on deposit for $t-1/52$ years, the second for $t-2/52$ years $\cdots$ in general, the $j$th $\$50$ has been on deposit for $t-j/52$ years $(1 \le j \le 52t)$. Find the present value of each $\$50$ deposit assuming $6$% interest compounded continuously, and use the formula
  
  \[ 1+x+x^2+\cdots+x^n={1-x^{n+1}\over 1-x}\ (x \ne 1) \]
  
  to find their total value.
2. Let
  
  \[ p(t)={Q(t)-P(t)\over P(t)} \]
  
  be the relative error after $t$ years. Find
  
  \[ p(\infty)=\lim_{t\to\infty}p(t). \]
Show answer

(a) $\dst{P(t)=1000e^{.06t}+50{e^{.06t}-1\over e^{.06/52}-1}}$ (b) $5.64 \times 10^{-4}$
A homebuyer borrows $P_0$ dollars at an annual interest rate $r$, agreeing to repay the loan with equal monthly payments of $M$ dollars per month over $N$ years.
1. Derive a differential equation for the loan principal (amount that the homebuyer owes) $P(t)$ at time $t>0$, making the simplifying assumption that the homebuyer repays the loan continuously rather than in discrete steps. (See Example 4.1.6 .)
2. Solve the equation derived in (a).
3. Use the result of (b) to determine an approximate value for $M$ assuming that each year has exactly 12 months of equal length.
4. It can be shown that the exact value of $M$ is given by
  
  \[ M={rP_0\over 12}\left(1-(1+r/12)^{-12N}\right)^{-1}. \]
  
  Compare the value of $M$ obtained from the answer in (c) to the exact value if (i) $P_0=\$50,000$, $r=7{1\over2}$%, $N=20$ (ii) $P_0=\$150,000$, $r=9.0$%, $N=30$.
Show answer

(a) $P'=rP-12M$ (b) $\dst{P={12M\over r}(1-e^{rt})+P_0e^{rt}}$ (c) $\dst{M\approx{rP_0\over 12(1-e^{-rN})}}$

(d) For (i) approximate $M=\$402.25$, exact $M=\$402.80$

for (ii) approximate $M=\$1206.05$, exact $M=\$1206.93$.
Assume that the homebuyer of Exercise 22 elects to repay the loan continuously at the rate of $\alpha M$ dollars per month, where $\alpha$ is a constant greater than 1. (This is called accelerated payment.)
1. Determine the time $T(\alpha)$ when the loan will be paid off and the amount $S(\alpha)$ that the homebuyer will save.
2. Suppose $P_0=\$50,000$, $r=8$%, and $N=15$. Compute the savings realized by accelerated payments with $\alpha=1.05,1.10$, and $1.15$.
Show answer

(a) $T(\alpha)=\dst{-{1\over r}\ln\left(1-\left(1-e^{-rN})/\alpha\right)\right)}\mbox{ years}$

$S(\alpha)=\dst{{P_0\over(1-e^{-rN})}\left[rN+\alpha\ln \left(1-(1-e^{-rN})/\alpha\right)\right]}$

(b) $T(1.05)=13.69$ yrs, $S(1.05)=\$3579.94$ $T(1.10)=12.61$ yrs,

$S(1.10)=\$6476.63$ $T(1.15)=11.70$ yrs, $S(1.15)=\$8874.98$.
A benefactor wishes to establish a trust fund to pay a researcher’s salary for $T$ years. The salary is to start at $S_0$ dollars per year and increase at a fractional rate of $a$ per year. Find the amount of money $P_0$ that the benefactor must deposit in a trust fund paying interest at a rate $r$ per year. Assume that the researcher’s salary is paid continuously, the interest is compounded continuously, and the salary increases are granted continuously.

Show answer

$P_0=\left\{\begin{array}{cl}\dst{S_0(1-e^{(a-r)T})\over r-a} \mbox{ if }a\ne r,\\[6pt] S_0T \mbox{ if }a=r.\end{array}\right. $
L A radioactive substance with decay constant $k$ is produced at the rate of

\[ {at\over1+btQ(t)} \]

units of mass per unit time, where $a$ and $b$ are positive constants and $Q(t)$ is the mass of the substance present at time $t$; thus, the rate of production is small at the start and tends to slow when $Q$ is large.
1. Set up a differential equation for $Q$.
2. Choose your own positive values for $a$, $b$, $k$, and $Q_0=Q(0)$. Use a numerical method to discover what happens to $Q(t)$ as $t\to\infty$. (Be precise, expressing your conclusions in terms of $a$, $b$, $k$. However, no proof is required.)
L Follow the instructions of Exercise 25, assuming that the substance is produced at the rate of $at/(1+bt(Q(t))^2)$ units of mass per unit of time.
L Follow the instructions of Exercise 25, assuming that the substance is produced at the rate of $at/(1+bt)$ units of mass per unit of time.