A proportionality statement involving the rate of change of a quantity y, the quantity y itself and the independent variable x can be interpreted as a differential equation. We look at some examples of such situations.
To find the value of an inverse function g^-1(x) at a given value of x we can look at the graph of g(x). Locating the specified x on the y axis, we project over to the graph of g(x) and the up or down to the x axis; the value we obtain on this axis is the value of g^-1(x). Alternatively we can evaluate g(x) at different values of the independent variable until our result is sufficiently close to the specified x.
We construct the graph of the natural log function y = ln(x) by first constructing the graph of the exponential function y = e^x, then reflecting through the y = x line to get the graph of the inverse function, which is y = ln(x).
The forms y = A b^t and y = A e^(kt) are equivalent, with b = e^k or equivalently with k = ln(b). For exponential growth k is positive and b is greater than 1; for exponential decay k is negative and b is less than 1.
Most of the functions we use in calculus and in modeling the real world are composite functions of the form f(g(x)), with f and g usually being the simple power functions, exponential functions, logarithmic functions, polynomial functions, etc.. The ability to decompose a given composite function f(g(x)) into its constituent functions f and g is essential in later applications. It is to learn to do this now so that the skill is available when it is needed later. Don't wait to develop the skill until you need to apply to learn something else.
Given a differential equation of the form dT / dt = k (T - Troom), and given a value of T at a clock time t, we can determine the approximate value of T at clock time t + `dt by using the fact that `dT = dT / dt * `dt. If `dt is small enough that dT / dt doesn't change by much between t and t + `dt, the approximation will be a good one.
The process can be continued for successive intervals to determine approximations that t + 2 `dt, t + 3 `dt, etc.. The accuracy of the approximation decreases more and more rapidly with succeeding intervals.
If the rate at which the Fahrenheit temperature T of a hot potato immersed in ice water changes is proportional to the excess temperature of the potato over the 32 degree Fahrenheit temperature of the water, then we represent the rate which the potato's temperature changes with respect to clock time t as dT / dt and temperature excess as T - 32.
The proportionality equation is therefore written dT / dt = k (T - 32).
For water flowing from a uniform cylinder, the rate at which the water depth y, as measured relative to the outflow hole, changes with respect clock time is proportional to the square root of the depth of the water. The time rate at which the water depth changes is designated by dy / dt, so the proportionality will read dy / dt = k `sqrt(y).
If the rate which the illumination I provided by a point source of light changes with respect to clock time, as observed by a detector moving away from the source and constant speed, is inversely proportional to the clock time, then the rate which illumination changes is dI / dt, and the inverse proportionality is expressed by dI / dt = k (1/t).
Note in the two equations expressed in the figure above, that the right-hand side of the first contains the dependent variable y, while the right hand side of the second contains the independent variable t. The right hand side of such an equation can contain either the dependent variable, the independent variable, or both. The right hand side of an equation involving, saying, y and t and the derivative dy / dt, is therefore generally a function of both y and t and we can express this equation as
dy / dt = f(y,t).
We could for example write dy / dt = (y^2 + cos(t)) / (t - y/t). This equation probably doesn't have a very significant meaning, and it probably can't even be solved in the sense of expressing the function y(t) in closed algebraic form, but from a mathematical point of view it is still a perfectly good differential equation. And as we will see below it can be solved as accurately as we wished using numerical approximations.
A question was raised regarding the text question about whether the inverse of the function g(x) = x^3 + 3^x exists and if so, what the value of the inverse would be when x = 20.
To answer the question of whether the function has an inverse, we first sketch graphs of y = x^3 and y = 3^x. At this point we could easily note that both functions are strictly increasing, so their sum will be strictly increasing and must therefore have an inverse (any strictly increasing or decreasing function will certainly passed the horizontal-line test, having a unique x value for every y value). For the purposes of illustration we nevertheless indicate the sum function by the red dotted line. We indicate for this function an arbitrary point y on the y axis, its horizontal projection to a unique graph point, and a projection of this point down to the x axis. The x axis value is the inverse value g^-1(y).
The question of finding the value of the inverse when x = 20 is somewhat ambiguous, since it isn't clear whether the x value corresponds to the original function or to the inverse. A better way of stating the question would have requested the value of g^-1(x) when x = 20.
Our first attempt at a solution should be to find the formula for g^-1(x). We could do this by solving the equation y = x^3 + 3^x for x, then reversing the roles of x and y. At least that is what we would like to do. Unfortunately the equation simply cannot be solved for the exact value of x. It's not that we are too stupid to figure out; we have figured out perfectly well, and what have figured out is that the solution is in fact impossible. So that leaves us with the alternative of approximate in the solution.
In this case we can try different values of x, until we find one that gives us something close enough to y = 20. (To see why we use y = 20 and sulfur acts, think of locating y = 20 on the graph and projecting over to the graph and down to the x axis). We could start by letting x = 1, which when substituted would give us y = 4. That isn't very close to 20, so we try x = 2. We get y = 17, which is getting much closer. Our y value will therefore probably be between 2 and 3, and probably, since the function keeps increasing at an increasing rate, closer to 2. Our next attempt at an x value might be x = 2.2. We can keep narrowing down the range of the x value until we feel we are close enough.
To construct the graph of the natural log function ln(x) we start by constructing the graph of y = e^x.
The number e, which we will later see is the limiting value of the expression (1 + 1/n) ^ n as n becomes larger and larger, has a value of approximately 2.7. So e lies between 2 and 3, somewhat closer to 3. The graph of y = e^x will therefore lie between the graph of y = 2^x and y = 3^x, and probably somewhat closer to the graph of y = 3 ^ x. This is indicated in the figure below. You should be able to construct this graph quickly and easily.
Just as the function y = log(x) is the inverse of the function y = 10 ^ x, the natural logarithms function y = ln(x) is defined as the inverse of the function y = e ^ x. The laws of logarithms are the same for the natural log as for the log function. The graph of the natural log is obtained by inverting the graph of the function y = e ^ x.
The form of an exponential function might be P(t) = P0 b ^ t, or it might be P(t) = P0 e^(kt). Recall that b is regarded as the growth factor for the exponential function, and is obtained as 1 + r, where r is the growth rate.
The two forms express the same function as long as b ^ t = e ^ (kt). We can see how the value of k is related to the growth factor b by solving this equation for b. Solution is simple enough if we know the laws of logarithms. We simply take the natural log both sides of the equation, apply the laws of logarithms, and we obtain k = ln(b).
We note that when the growth rate r is positive, b = 1 + r is greater than 1. Noting that the natural log of 1 is 0, we see from the graph of the natural log that the natural log of a number > 1 must be positive. We therefore conclude that, for exponential growth, k = ln(b) must be a positive number.
We can similarly conclude that when the growth rate is negative, corresponding to an exponential decay, the growth factor b will be less than 1, so that its natural log will be negative. We therefore conclude that for exponential decay, k = ln(b) must be a negative number.
Your text expresses these ideas somewhat differently, saying that k is always positive and separating the ideas of exponential growth and decay. Using this convention an exponential growth will be expressed by P0 e ^ (kt), and an exponential decay by P0 e ^ (-kt). You should understand both conventions, because you are likely to encounter both. If you understand why exponential decay requires a negative exponent while exponential growth requires a positive exponent, you have the main idea.
The idea of composite functions is extremely important in calculus. You need to be able to look at a composite function and tell what simple functions it is composed of, and how it is put together from those functions. Be sure to practice this skill now so you don't have to develop it later when your attention will need to be directed toward applying the skill.
Consider the function f(x) = e ^ (x^2). If we let g(x) = e ^ x, we can ask what should be substituted for the ? in the expression g(?) = e ^ x^2. The obvious answer it is that we substitute x^2 for the ? in order to obtain e^(x^2): g(x^2) = e ^ (x^2).
In this sense we see that f(x) = e ^ (x^2) is built up from the functions g(x) = e^x and h(x) = x^2. The specific way we build the function up is to form the composite function g(h(x)) = e^(h(x) = e ^ (x^2).
Numerical Solution of a Differential Equation
Recall from the preceding class that when the rate in which temperature changes is proportional to the difference between the temperature and the 80 degree room temperature, the knowledge that when the temperature is 150 degrees the rate of temperature changes -2 degrees/second yields the proportionality equation dT / dt = -.028 deg / s (T - 80). This equation allows us to find the rate which temperature changes for any given temperature T.
It follows that if we start at clock time t = 0 with a temperature of 150 degrees, the initial rate at which temperature changes will be -2 degrees/second. If we wish to predict the temperature five seconds later, corresponding to a time interval `dt = 5 seconds, we would easily reason out the result: at -2 degrees/second, in five seconds to temperature will change by -10 degrees.
These calculations are summarized in the first row of the table below.
It follows that clock time t = 5 seconds the temperature will be approximately 140 degrees. We say 'approximately' because the -2 degrees/second rate will actually decrease during the first five seconds as the temperature T decreases, and we have made no provision for correcting for that decrease in rate.
We can proceed to determine the rate at which temperature changes when T = 140 degrees. We simply substitute this temperature into the equation for the rate dT / dt, and we obtain a rate of -1.68 degrees/second. Over a five second interval this will apply a change of -8.4 degrees. These calculations are summarized in the second row of the table.
We see that at clock time t = 10 seconds we have an approximate temperature of 131.6 degrees, which by the same process used before results in a rate of -1.44 degrees/second. Over another five second interval this will result in a temperature change of -7.2 degrees, and we could proceed to calculate another row of the table corresponding to clock time t = 15 seconds.
As an additional homework problem, continue this procedure for a few more steps, until you obtain a temperature T which is less than 100 degrees.
Furthermore, starting with temperature T = 150 degrees at clock time t = 0, follow the same process for five steps but using a time increment `dt of 1 second rather than five seconds. Compare the t = 5 second value of the temperature T with the 140 degrees predicted by a single five-second increment.
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