# How do you graph exponential decay?

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 $\frac{\mathrm{dy}}{\mathrm{dt}} = - k y$, 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 $\frac{1}{y} \mathrm{dy} = - k \mathrm{dt}$. Now we can integrate:
$\ln \left(y\right) = - k t + t i l \mathrm{de} \left\{c\right\}$
where $t i l \mathrm{de} \left\{c\right\}$ is a constant. In the end: $y \left(t\right) = c {e}^{- k t}$, where $c = {e}^{t i l \mathrm{de} \left\{c\right]} > 0$.

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 ${y}_{0} = y \left(0\right) = c$. 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 $y \left(t\right)$ 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). To get a visual intuition of what the differential equation $\frac{\mathrm{dy}}{\mathrm{dt}} = - k y$ really means, let's consider the $\left(t , y\right)$-plane. We are indeed interested in the behavior of the quantity $y$ in time $t$.
If we fix a point $\left({t}_{p} , {y}_{p}\right)$ 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 $\left({t}_{p} , {y}_{p}\right)$ "decay" after a "bit" of time. This "bit" can be thought as infinitesimally small: we denote it by ${\left(\mathrm{dt}\right)}_{p}$. In this infinitesimal time, the quantity $y$ changes by an infinitesimally small amount ${\left(\mathrm{dy}\right)}_{p}$.
So, the point we are searching for is the point $\left({t}_{p} + {\left(\mathrm{dt}\right)}_{p} , {y}_{p} + {\left(\mathrm{dy}\right)}_{p}\right)$, 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 $\left({t}_{p} , {y}_{p}\right)$, pointing in the direction of $\left({t}_{p} + {\left(\mathrm{dt}\right)}_{p} , {y}_{p} + {\left(\mathrm{dy}\right)}_{p}\right)$. The vector's direction is given by the difference
$\left({t}_{p} + {\left(\mathrm{dt}\right)}_{p} , {y}_{p} + {\left(\mathrm{dy}\right)}_{p}\right) - \left({t}_{p} , {y}_{p}\right) = \left({\left(\mathrm{dt}\right)}_{p} , {\left(\mathrm{dy}\right)}_{p}\right)$
From the differential equation we get that $\left({\left(\mathrm{dt}\right)}_{p} , {\left(\mathrm{dy}\right)}_{p}\right) = \left({\left(\mathrm{dt}\right)}_{p} , - k {y}_{p} {\left(\mathrm{dt}\right)}_{p}\right)$. Now we can choose the size of the arrow by setting ${\left(\mathrm{dt}\right)}_{p}$ 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 $\left({t}_{p} , {y}_{p}\right)$, 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 $\left({t}_{p} , {y}_{p}\right) = \left(2 , 6\right)$ and $k = 0.2$ as in the previous examples. 