How do you graph exponential decay?

1 Answer

A quantity decades exponentially if it decreases in time at a rate that is linearly proportional to its value. In terms of equations, if y decays exponentially, then dy/dt = - k y, where k>0 is the exponential decay constant and it characterizes the decay.

The variables involved in this differential equation are separable. If we separate them, we get 1/y dy = -k dt. Now we can integrate:
ln(y)=-kt+tilde{c}
where tilde{c} is a constant. In the end: y(t)=ce^{-kt}, where c=e^{tilde{c]}>0.

This means that graphing the exponential decay turns into graphing an exponential function with negative exponent. For t=0 (the instant in which the decay starts), we get that y_0=y(0)=c. So the parameter c is the y-intercept of the function: it represents the value of our quantity y at the instant in which the decay starts.
We can plot y(t) for some values of c. In the following plot k=0.2 and different colors represent graphs for c=10, c=5, c=2, c=1 and c=0.5 (from the top to the bottom).
enter image source here


To get a visual intuition of what the differential equation dy/dt=-ky really means, let's consider the (t,y)-plane. We are indeed interested in the behavior of the quantity y in time t.
If we fix a point (t_p,y_p) on the plane, we are stating that the quantity y has the value y=y_p at time t=t_p. We want to represent the decay, so we ask ourselves where would the point (t_p,y_p) "decay" after a "bit" of time. This "bit" can be thought as infinitesimally small: we denote it by (dt)_p. In this infinitesimal time, the quantity y changes by an infinitesimally small amount (dy)_p.
So, the point we are searching for is the point (t_p+(dt)_p,y_p+(dy)_p), which is the point that is going to describe the decay an infinitesimal amount of time after t_p. To represent this information we draw an arrow (namely a vector) in (t_p,y_p), pointing in the direction of (t_p+(dt)_p,y_p+(dy)_p). The vector's direction is given by the difference
(t_p+(dt)_p,y_p+(dy)_p)-(t_p,y_p)=((dt)_p,(dy)_p)
From the differential equation we get that ((dt)_p,(dy)_p)=((dt)_p,-ky_p(dt)_p). Now we can choose the size of the arrow by setting (dt)_p as small as we like.
[This argument is not rigorous: we should speak about finite differences and how they are related to differentials, but the core idea emerges anyway. Also notation is invented for the purpose of the argument.]

If we repeat this operation for some points, we get the following picture. Note that vectors are normalized (i.e. made unitary dividing by their norm) and this particular plot is made fixing k=0.2 (other positive values of k don't change the qualitative behavior).

Now, given any point (t_p,y_p), we are "forced" to follow the arrows and we get the plot of the decay when the quantity y has value y_p at time t=t_p. In the following picture I chose (t_p,y_p)=(2,6) and k=0.2 as in the previous examples.