Deriving Exponential Decay Equation

[Jimmy87]

Hi,

I was trying to see where the equation N = N_o e^-λt came from and it is derived from dN/dt = -λN which is discussed very well in this thread in post #2 (https://www.physicsforums.com/threads/derivations-of-the-decay-constant-equation.213312/). I understand the steps except for the reason why both sides are integrated. I always struggle with this concept during derivations. If we rearrange it into the form dN/N = - λdt, I really don't see how the next logical step is to integrate both sides. I know it gets us the correct outcome but what is the reason for integrating both sides as appose to some other operation? Integrating means summing/adding - I don't see how summing/adding is the next logical step - I only understand it retrospectively from the correct outcome it gives.

Any help is much appreciated!

 

[BvU]

Decay (or growth) that goes proportional to what is present can be described with a differential equation of the form$${df\over dx} = \alpha \,f(x)$$the solution is found by integrating (summing up) the infinitesimal contributions> Since $f$ is to be found you treat this as any algebraic equation: bring the knowns ($\alpha$ and the independent variable $x$) to one side and the unknowns to the other:$${df\over f} = \alpha \, dx\ .$$The only way to find $f$ from an equation in $df$ is to integrate.

Perhaps it helps to treat this as a difference equation: when you know $f(x_0)$ at some initial point, you can find $f(x_0+\Delta x)$ by addition:$$f(x_0+\Delta x) = f(x_0) + f(x_0) * \alpha * \Delta x $$ (think of the definition of the derivative). Letting $\Delta \downarrow 0$ gets you an integral.

(note: physicists are casual with infinitesimals -- they know a mathematician can back them up with thorough analysis for decently behaving functions )

 

[Orodruin]

note: physicists are casual with infinitesimals

For some things even physicists call foul ...

A less hand-wavy way would be to put everything with $f$ on one side and then integrate with respect to $x$. You would find
$$
\int_{x_0}^{x_1} \alpha\, dx = \int_{x_0}^{x_1} \frac{f'(x)}{f(x)} dx,
$$
which is a change of variables from $x$ to $y = f(x)$ away from
$$
\int_{f(x_0)}^{f(x_1)} \frac{dy}{y}.
$$

 

[Jimmy87]

Decay (or growth) that goes proportional to what is present can be described with a differential equation of the form$${df\over dx} = \alpha \,f(x)$$the solution is found by integrating (summing up) the infinitesimal contributions> Since $f$ is to be found you treat this as any algebraic equation: bring the knowns ($\alpha$ and the independent variable $x$) to one side and the unknowns to the other:$${df\over f} = \alpha \, dx\ .$$The only way to find $f$ from an equation in $df$ is to integrate.

Perhaps it helps to treat this as a difference equation: when you know $f(x_0)$ at some initial point, you can find $f(x_0+\Delta x)$ by addition:$$f(x_0+\Delta x) = f(x_0) + f(x_0) * \alpha * \Delta x $$ (think of the definition of the derivative). Letting $\Delta \downarrow 0$ gets you an integral.

(note: physicists are casual with infinitesimals -- they know a mathematician can back them up with thorough analysis for decently behaving functions )

Thanks. I think I have understood what you mean. So we integrate because our end goal is to find the total number of nuclei (dN integral on the left) after some total time period (dt integral on the right) hence we get an equation at the end that can find the total nuclei after some time period.