How do u determine exponential growth and deay?

Answer 1

A=P(1+r)^t
exponential growth
with P being principal(original amount)
r=the rate(in decimal form)
and t being the time
exponential decay
A=P(1-r)^t

Answer 2

If dy/dx = ky, then y = Ce^(kx), where C is the value of y when x=0

For example it can be shown that if the temperature is held constant, the rate of change of the atmospheric pressure p at any height h is proportional to the pressure at that height. Thus dp/dh = -kp. The minus sign is used because the p decreases as h increases. It follows that p=p0e^(-kh), where p0 is the pressure when h =0.

y=Ce^(kx) is the exponential law or the law of growth and decay. You get growth when k >0 and decay when k<0.

Answer 3

You mean how do you derive the equation?
If a quantity grows or decays at a rate directly proportional to its initial value, then

dN/dt = kN where k is a proportionality constant, N is initial value. Conversely, dN/dt = -kN if we are talking about decay.

In order to solve as a function of one variable we must rearrange and integrate both sides.

dN/N = k*dt
ln(N) = kt + c
N = e^(kt+c)
N = (e^kt)*(e^c)
e^c = another constant, "d"
Initial condition, N(0) = N_naught
N(0) = "d"*e^0
N_naught = "d"

Therefore our final equation is:

N(t) = N_naught * e^kt

Answer 4

you know how a straight line has constant slope, slope = change in y over change in x, change computed by subtraction? well, for exponential growth and decay, the log of change in y (computed as a ratio) is a constant times change in x. before graphing calculators, we used to use log-log and semi-log graph paper. data that showed a straight line on semi-log was exponential.

let y1 = ae^(kx1), y2 = ae^[k(x2)]
y2/y1 = e^[k(x2-x1)] [cancelling a and subtracting exponents]
ln (y2/y1) = ln e^[k(x2-x1)] = k(x2-x1)
ln (y2/y1) / (x2-x1) = k