In equations whats the diff. btwn exponetial growth & exponetial decay?how do we use these principles in life?
406962
When something grows exponentially it describes the relationship between two values.  One value grows at an exponenet (x2 or x3 or x3, ...) while the other item grows linearly (a+a+a+a)\n\nA simple example is a bacterial culture.  One value is time and it is the linear value.  The number of bacteri is the exponential vlaue (x2).\n\nSo one cell in the culture divides, giving us 2 cells.  Then those 2 cells each divide giving us 4 cells (from 2 to 4) then 4 double to 8, 8 to 16, 16 to 32.  Now look at the number of cells.\n\n1\n2\n4 = 2 squared (2x2)\n8 = 2 cubed (2x2x2)\n16 = 2 to the (4th power) [2x2x2x2]\n32 = 2 to the (5th power) [2x2x2x2x2]\n\n\nExponential decay is basically just the opposite.\n\nA radio signal degrades by a factor of 2 for every unit distance traveled.  The half-life of a radioactive substance is described by exponential decay.
176116
A halflife is a form of time constant for characterizing exponential trends with a decay rate of 1/2. Another common time constant decay rate is 1/e.
223647
i have found that once the timescale of fermentation is reached it will result in the coeficiant expansion of the original quantity. bearing in mind the properties of  the 3 states of matter are of course intermingled with H20  its understandable that time is of the essence . You may refer to Von danikens theory of alien fermentation equations based on the helgar experiance this will show that in time is of no consequence. So to conclude if you wait a long time you have wasted your time because what is TIME.
1019110
The difference between exponential decay and inverse relationships between two variables is as follows.  Exponential decay refers primarily to rates of radioactive decay.  You have a rate of decay over time for some element, say 1/e^(.2x) to keep it simple, and you compare it with the amount of the the element over that same time period.  It's called exponential when its rate decreases at a rate proportional to its value.   An inverse relationship is easier to describe and conceive.  An inverse relationship is one where one variable increases while the other one decreases, in a simple one-to-one manner.
45317
The simplest model of population growth is based on the straightforward quantities of four properties of population change:\n\nB - Births or Natality \nD - Deaths or Mortality \nI - for Immigration and \nE - for Emigration\n\n\n\nNt+1 = Nt + B - D + I - E\n\n[Equation 3.1]\n\n\n\n\nThe net rate at which the population is increasing or decreasing is called the net rate of population growth - Ro. The net rate of population growth is simply the ratio of the size of the population at some time Nt+1 compared to the size of the population during some the previous moment in time Nt: \n\n\n\n\n[Equation 3.2]\n\n \n\nIf birth rates (b) and death rates (d) in a population are determined for a specific time interval we can calculate another measure of population growth call the intrinsic rate of increase - r from these two population parameters: \n\n\nr = b - d\n\n[Equation 3.3]\n\n\n\n\nThe rate of change of population size dN/dt is then given by the differential equation: \n\n\n\n\n[Equation 3.4]\n\n\n\n\nwhere N is the size of the population at a particular instant in time. To obtain the size of the population Nt at some time t, is integrated to give: \n\n\n\n\n[Equation 3.5]\n\n\n\n\nwhere No is the initial size of the population at some time designated zero, e is a constant = 2.718 (the base of natural logarithms), r is the intrinsic rate of increase and t is the time elapsed since zero.\n\nThe population growth described by equation [3.5] is geometric, which means the population size goes on doubling (or halving if d is greater than b) at a constant rate. This situation is clearly unrealistic, since the growth of all species populations must be limited by the availability of some finite resource.
546696
Ok, So you know the Time, the Starting mass we'll say Ao, and the Half life.  So by the equation:\nA/Ao = (1/2)^n,\n\nwhere A is the amount remaining, Ao is the starting amount and n is an integer, which needs to be determined.\n\nn = T/(t)\n\nWhere t is the half life and T is the time elapsed.  \n\nSo you first solve for n, and then derive the expression for A.  you should get the equation,\n\nA = ((1/2)^n)*(Ao)\n\nThat'll give you the new mass and from the mass you can use Avogadros number to determine the particles emitted.
109334
Think of the chemicals u consume every day....a disproportionate amount of some chemical will be harmful to you...arithmetic & algebra is used to determine what qntity of what to use...don't forget expiration dates are calculated based on the "life expectancy" of used products....ofcourse statistics comes in with the financial mgt (be it the food industry or what they have to account for their expenditure)!
814943
There is no balanced equation for that question.\n\nYou need to show your products going to reactants for the world to be able to balance the equation.\n\nGood Luck!
766444
time to the minus one.\n\nDo a dimensional analaysis on a rate equation to confirm this.
581102
You use derivatives to compute how values are changing. A good example is to compute velocities.\n\nIntegration is used to compute areas and volumes.
990985
Living things have growth and metabolic activities\nNon living does not have any growth.\n\nliving ex-> man,plants,fish,etc.\nnon living ->building,stone,car,etc.
526116
Basically, all you need to know is the initial amount (8.0 grams I assume), and the half life (46 days).  Then you can build an exponential model of the form:\n\nA(t)=A*e^(k*t)  \nA is initial amount\nk is decay constant\nt is time\n\nKnowing that after 46 days, there will be half as much present:\n\n4=8*e^(k*46)\n\nThen solve for k using logarithms.  \n\nln(0.5)=46k\nk=ln(0.5)/46=-0.01507\n\nSubstitute into equation to get time dependent equation:\n\nA(t)=8*e^(-0.01507*t)\n\nNow you want to know how much is left after t=184:\n\nA(184)=8*e^(-0.01507*184)\n\nA(184)=0.5 [grams]\n\nViola....
806438
The exponential formula for growth would be\n\nn(t) = n(0)* e^(t/tc), where n0 is the initial number of cells, and tc the "time constant" for growth.  This can also be written\n\nn(t)/n(0) = e^(t/tc)\n\nYou are given that when t = 6mo, n(6)/n(0) = 2\n\ntherefore 2 = e^(6/tc);   solve this for tc:\n\nln(2) = 6/tc, and tc  =  6/ln(2) =  8.656\n\nThe equation is now n(t)/n(0) = e^t*/8.656\n\nwhere t is in months.   For t = 3 years, that is 36 months, so n(36)/n(0) = e^(36/8.656) = 64.0   For t = 6 years, or 72 months n(72) = e^(72/8.656) = 4.10*10^3
493182
P(n+1) - P(n) is the change in population over one period. \n\nrP is the rate of growth of population, tempered by (1-P(n)/E). \n\nBasically, imagine P(n) is very small compared to the equilibrium population. P(n)/E would be small, and 1-P(n)/E would be close to one. Thus, the growth of population rP(n)*(1-P(n)/P) would be close to rP. \n\nHowever, the model takes into account other factors, e.g. environment, that limits population growth. As P(n) grows, P(n)/E gets closer to 1. If P(n) is almost exactly E, (1-P(n)/E) approaches 0. Thus rP(1-P(n)/E) approaches 0, and the population growth P(n+1) - P(n) approaches 0. \n\nAs a result, the population starts growing very fast, and then slows down as it approaches equilibium value. This relationship would also hold if the population started out too high, and decreased towards equilibrium.
134985
Here's an example of how e ties in with real life.  Put water in a can with a hole at the bottom.  Measure the rate of water flow with the can full, and calculate how long it would take to empty the can at that rate.  That's the "time constant".  Say it's a 32 ounce can and the initial flow is 1 oz/sec, then the TC is 32 sec.  But what really happens after 32 sec?  As the water goes down, so does the flow rate which is proportional to the amount of water remaining above the hole.  At 32 sec, the can is actually 1/e full.  At 64 sec the can is 1/e^2 full, and so on.  Any time the rate of approach to final value of some quantity is proportional to the value of the quantity relative to the final value, you get similar behavior.  This is called exponential decay, which is an asymptotic approach to the final value (in this case 0).  Acoustic echoes, radioactive decay, temperature change of an object subjected to heat or cooling, the amplitude of a damped oscillation and the charge or discharge of a capacitor through a resistance are other examples, and the principal remaining in an account that pays out a fixed percentage rate approximates this behavior (with TC ~ 1/payout rate).  At any time t, what remains from an initial X value of X0 that approaches a final value Xf is X = X0 + (Xf-X0)/e^(t/TC).\nAnother kind of e-dominated behavior is unbounded proportional change, called exponential growth.  There is no final value, and rate of change only depends on the size of X.  Thus at time t, X = X0 * (1 + e^(k*t)), where k is the growth rate constant.  Examples of this growth (which in real life, generally hit a limit caused by finite resources) are an account that accumulates a fixed interest, chain reactions, population growth, etc.
883446
Hi, it depends on particular organisms. Check the link below for basic mathematical formulas you may be able to use.\n\nA study is done in which the number of bacteria in a liquid culture is monitored over a period of several hours. Ten observations were made:\n\n\nTwo questions need to be answered:\n\nIs this data consistent with an exponential growth model: N = N0ekt, where \nN0 = the number of bacteria present initially (at t = 0)\nk = a growth rate constant\nN = the number of bacteria present at a given time t,\nand e is the fundamental mathematical constant forming the base for natural logarithms. \nIf the answer to the first question is "yes", then what are the values of N0, k, and t2 (the so-called doubling time) for this bacterial system?\nThe experimental measurements listed above indicate the number of bacteria are growing with time. If we plot the observed values of N (vertically) against the elapsed time, t, in hours, we get a graph (available on web page below).\n\nYou can draw a smooth curve through the experimental points as shown in this figure, and the points all seem to lie relatively close to this curve. Further, this curve looks quite a bit like the graph of an exponential function, but from what we see, it might just as well be part of a parabola or of one of a number of other curves that result when various algebraic forms are graphed. This graph suggests that the answer to the first question may be "yes", but we would usually want stronger evidence than this to draw a definite conclusion. The trouble here is that it is difficult to distinguish precisely and reliably between the curves are the graphs of quite different mathematical functions.\n\nThere is, of course, one curve that is easily recognized by everyone - the straight line! In this situation, the logarithmic function can be used to rewrite the exponential growth formula into a form that will allow us to create a plot that does give a straight line when data is the result of an exponential growth (or decay) process. Start with the exponential growth formula:\n\nN = N0 ekt\n\nSince this involves an exponential with base e, it is "natural" to use natural logarithms - those based on the number e. Using the more common notation ‘ln’ for ‘loge’, we simply take the logarithm of both sides of the previous equation:\n\nln N = ln(N0 ekt)\n\n= ln N0 + ln(ekt)\n\n= ln N0 + kt ln e\n\n= ln N0 + kt\n\nHere, we have used two of the three basic properties of logarithms to manipulate the right-hand side, namely that\n\nlog (AB) = log A + log B \n\nlog (Am) = m log A \n\nFurther, in the last step we took advantage of the fact that by choosing to work with natural logarithms, ln e = logee = 1. Thus, the exponential growth formula can be written in the logarithmic form\n\nln N = kt + ln N0\n\nIf the exponential growth formula is correct and we were to plot ln N along the vertical axis of our graph (rather than N itself), and t along the horizontal axis, then the equation of the plot we are preparing will have the generic form\n\nvertical quantity = constant x horizontal quantity + another constant \n\nBut this is just the equation of a straight line. The plot that results using the data given above is available on web site below.\n\nAlthough the values written on the vertical axis are values of N, this graph is ruled vertically so that the vertical distances are actually proportional to ln N. Since one scale is logarithmic (the vertical one) whereas the other graph scale, the horizontal one, is the ordinary uniform spacing type, this is called a ‘semilogarithmic plot’.\n\nAt any rate, the figure indicates that the points are scattered about a straight line, the scatter due to small random errors in measurement. Since the plot of ln N vs t gives a straight line, we have confirmed that\n\nln N = kt + ln N0 (equation *) \n\nis a valid formula for this data, and therefore, so is its equivalent, N = N0 ekt, valid for this data.
757820
The general exponential population growth is:\n\nA(t) = P(1+r)^t\n\nwhere\n\nP is the inital population\nr is the growth rate\nt is the time