# The number of ants in Bob's kitchen increase at a rate that is proportional to the number of ants...

## Question:

The number of ants in Bob's kitchen increase at a rate that is proportional to the number of ants present each day. If there are 20 ants on day 0 and 60 ants on day 4, how many ants will there be on day 14? Round to the nearest ant.

## Separable Differential Equations:

Recall that a derivative is a rate of change, so whenever we see one, we are dealing with a derivative. In this problem, our rate of change (read: derivative) is a proportion. Recall that if {eq}y {/eq} is proportional to {eq}x {/eq} then it means

{eq}\begin{align*} y &\propto x \\ y &= kx \end{align*} {/eq}

where {eq}k {/eq} is a constant. When we can separate a differential equation so that one variable is on either side of the equals sign, then it is called separable. To solve such differential equations, we separate and integrate.

"I got ants in my pants!" says Bob. Or at least he will, because that guy's in trouble. We are told the rate at which the number of ants changes with time (measured in days) is proportional to the number of ants present each day. Let's call the number of ants {eq}N {/eq} and the number of days {eq}t {/eq} (for time). Then we can write

{eq}\begin{align*} \frac{dN}{dt} &= kN \end{align*} {/eq}

This is a separable differential equation. We separate and integrate to find

{eq}\begin{align*} \frac{dN}{dt} &= kN \\ \frac{dN}{N} &= k\ dt \\ \int \frac{dN}{N} &= \int k\ dt \\ \ln N &= kt + C \\ N &= Ae^{kt} \end{align*} {/eq}

where we wrote {eq}A = e^C {/eq} for the constant of integration. Now, since {eq}N(0) = 20 {/eq} we have

{eq}\begin{align*} Ae^{k(0)} &= 20 \\ A &= 20 \end{align*} {/eq}

And then since {eq}N(4) = 60 {/eq} we also have

{eq}\begin{align*} 20e^{k(4)} &= 60 \\ e^{4k} &= 3 \\ 4k &= \ln 3 \\ k &= \frac14 \ln 3 \end{align*} {/eq}

And so the number of ants on day {eq}t {/eq} is

{eq}\begin{align*} N (t) &= 20e^{(\ln3/4)t} \\ &= 20 (3)^{t/4} \end{align*} {/eq}

And so on day 14 the number of ants in Bob's kitchen will be

{eq}\begin{align*} N (14) &= 20 (3)^{14/4} \\ &\approx 935 \end{align*} {/eq}

If Bob's pants weren't filled with ants before, they are now. 