The derivative of y = ecx has taken time and effort. The result was y' = cecx, which means that y' = cy. That computation brought others with it, virtually for free—the derivatives of bx and xx and eu(x). But I want to stay with y' = cy —which is the most important differential equation in applied mathematics.
Compare y' = x with y' = y. The first only asks for an antiderivative of x. We quickly find y = ½x² + C. The second has dy/dx equal to y itself—which we rewrite as dy/y = dx. The integral is ln y = x + C. Then y itself is exec. Notice that the first solution is ½x² plus a constant, and the second solution is ex times a constant.
There is a way to graph slope x versus slope y. Figure 6.7 shows "tangent arrows," which give the slope at each x and y. For parabolas, the arrows grow steeper as x grows—because y' = slope = x. For exponentials, the arrows grow steeper as y grows—the equation is y' = slope = y. Now the arrows are connected by y = Aex. A differential equation gives afield of arrows (slopes). Its solution is a curve that stays tangent to the arrows—then the curve has the right slope.
A field of arrows can show many solutions at once (this comes in a differential equations course). Usually a single y0 is not sacred. To understand the equation we start from many y0—on the left the parabolas stay parallel, on the right the heights stay proportional. For y' = -y all solution curves go to zero.
From y' = y it is a short step to y' = cy. To make c appear in the derivative, put c into the exponent. The derivative of y = ecx is cecx, which is c times y. We have reached the key equation, which comes with an initial condition—a starting value y0:
dy/dt = cy with y = y0 at t = 0. (1)
A small change: x has switched to t. In most applications time is the natural variable, rather than space. The factor c becomes the "growth rate" or "decay rate"—and ecx converts to ect.
The last step is to match the initial condition. The problem requires y = y0 at t = 0. Our ect starts from ec0 = 1. The constant of integration is needed now—the solutions are y = Aect. By choosing A = y0, we match the initial condition and solve equation (1). The formula to remember is y0ect.
6l The exponential law y = y0ect solves y' = cy starting from y0.
The rate of growth or decay is c. May I call your attention to a basic fact? The formula y0ect contains three quantities y0, c, t. If two of them are given, plus one additional piece of information, the third is determined. Many applications have one of these three forms: find t, find c, find y0.
1. Find the doubling time T if c = 1/10. At that time y0ecT equals 2y0:
ecT = 2 yields cT = ln 2 so that T = ln 2 / c ≈ .7/.1. (2)
The question asks for an exponent T. The answer involves logarithms. If a cell grows at a continuous rate of c = 10% per day, it takes about .7/.1 = 7 days to double in size. (Note that .7 is close to ln 2.) If a savings account earns 10% continuous interest, it doubles in 7 years.
In this problem we knew c. In the next problem we know T.
2. Find the decay constant c for carbon-14 if y = ½y0 in T = 5568 years.
ecT = ½ yields cT = ln ½ so that c &asym; (ln ½)/5568. (3)
After the half-life T = 5568, the factor ecT equals ½. Now c is negative (ln ½ = -ln 2).
Question 1 was about growth. Question 2 was about decay. Both answers found ecT as the ratio y(T)/y(0). Then cT is its logarithm. Note how c sticks to T. T has the units of time, c has the units of "1/time."
Main point: The doubling time is (ln 2)/c, because cT = ln 2. The time to multiply by e is 1/c. The time to multiply by 10 is (ln 10)/c. The time to divide by e is -1/c, when a negative c brings decay.
3. Find the initial value y0 if c = 2 and y(1) = 5:
y(t) = y0ect yields y0 = y(t)e-ct = 5e-2.
All we do is run the process backward. Start from 5 and go back to y0. With time reversed, ect becomes e-ct. The product of e² and e-2 is 1—growth forward and decay backward.
Equally important is T + t. Go forward to time Tand go on to T + t:
y(T + t) is y0ec(T + t) which is (y0ecT)ect. (4)
Every step t, at the start or later, multiplies by the same ect. This uses the fundamental property of exponentials, that eT+t = eTet.
EXAMPLE 1 Population growth from birth rate b and death rate d (both constant):
dy/dt = by − dy = cy (the net rate is c = b − d).
The population in this model is y0ect = y0ebte-dt. It grows when b > d (which makes c > 0). One estimate of the growth rate is c = 0.02/year:
The earth's population doubles in about T = ln 2 / c ≈ .7 / .0.2 = 35 years.
First comment: We predict the future based on c. We count the past population to find c. Changes in c are a serious problem for this model.
Second comment: y0ect is not a whole number. You may prefer to think of bacteria instead of people. (This section begins a major application of mathematics to economics and the life sciences.) Malthus based his theory of human population on this equation y' = cy—and with large numbers a fraction of a person doesn't matter so much. To use calculus we go from discrete to continuous. The theory must fail when t is very large, since populations cannot grow exponentially forever. Section 6.5 introduces the logistic equation y' = cy − by², with a competition term -by² to slow the growth.
Third comment: The dimensions of b, c, d are "1/time." The dictionary gives birth rate = number of births per person in a unit of time. It is a relative rate—peopledivided by people and time. The product ct is dimensionless and ect makes sense (also dimensionless). Some texts replace c by λ(lambda). Then 1/λ is the growth time ordecay time or drug elimination time or diffusion time.
EXAMPLE 2 Radioactive dating A gram of charcoal from the cave paintings in France gives 0.97 disintegrations per minute. A gram of living wood gives 6.68 disintegrations per minute. Find the age of those Lascaux paintings.
The charcoal stopped adding radiocarbon when it was burned (at t = 0). The amount has decayed to y0ect. In living wood this amount is still y0, because cosmic rays maintain the balance. Their ratio is ect = 0.97/6.68. Knowing the decay rate c from Question 2 above, we know the present time t:
ct = ln(0.97/6.68) yields t = (5568/-.7)·ln(0.97/6.68) = 14,400 years.
Here is a related problem—the age of uranium. Right now there is 140 times as much U-238 as U-235. Nearly equal amounts were created, with half-lives of (4.5)109 and (0.7)109 years. Question: How long since uranium was created? Answer: Find t bysybstituting c = (ln ½)/(4.5)109 and C = (ln ½)/(0.7)109:
ect/eCt = 140 ⇒ ct − Ct = ln 140 ⇒ t = ln 140 / (c−C) = 6(109) years.
EXAMPLE 3 Calculus in Economics: price inflation and the value of money
We begin with two inflation rates—a continuous rate and an annual rate. For the price change Δy over a year, use the annual rate:
Δy = (annual rate) times (y) times (Δt). (5)
Calculus applies the continuous rate to each instant dt. The price change is dy:
dy = (continuous rate) times (y) times (dt). (6)
Dividing by dt, this is a differential equation for the price:
dy/dt = (continuous rate) times (y) = .05y.
The solution is y0e.05t. Set t = 1. Then e.05 ≈ 1.0513 and the annual rate is 5.13%.
When you ask a bank what interest they pay, they give both rates: 8% and 8.33%. The higher one they call the "effective rate." It comes from compounding (and depends how often they do it). If the compounding is continuous, every dt brings an increase of dy-and eeo8is near 1.0833.
Section 6.6 returns to compound interest. The interval drops from a month to a day to a second. That leads to (1 + lln)", and in the limit to e. Here we compute the effect of 5% continuous interest:
Future value A dollar now has the same value as esoSTdollars in T years.
Present value A dollar in T years has the same value as e--OSTdollars now.
Doubling time Prices double (emosT= 2) in T= In 21.05 x 14 years.
With no compounding, the doubling time is 20 years. Simple interest adds on 20 times 5% = 100%. With continuous compounding the time is reduced by the factor In 2 z -7, regardless of the interest rate.
EXAMPLE 4 In 1626 the Indians sold Manhattan for $24. Our calculations indicate that they knew what they were doing. Assuming 8% compound interest, the original $24 is multiplied by e.08'. After t = 365 years the multiplier is e29.2and the $24 has grown to 115 trillion dollars. With that much money they could buy back the land and pay off the national debt.
This seems farfetched. Possibly there is a big flaw in the model. It is absolutely true that Ben Franklin left money to Boston and Philadelphia, to be invested for 200 years. In 1990 it yielded millions (not trillions, that takes longer). Our next step is a new model.
