This problem aims to familiarize us with the **law** of **natural growth** and **decay.** The concept behind this problem is **exponential growth formulas** and their **derivates.** We have seen that **numerous** entities **grow** or **decay** according to their **size.**

For **instance,** a group of **viruses** may **triple every hour.** After some time $(t)$, if the extent of the **group**Â is given by $y (t)$, then we can **illustrate** this knowledge in **mathematical** terms in the form of an equation:

\[ \dfrac{dy}{dt} = 2y \]

So if an **entity** $y$ **grows** or **wears proportional** to its size with some **constant** $k$, then it can be expressed as:

\[ \dfrac{dy}{dt} = ky \]

If $k > 0$, the expression is known as the **law of natural growth**,

If $k < 0$, then the expression is known as **the law of natural decay.**

## Expert Answer

As we have seen the **formula** for **growth** and **decay:**

\[ \dfrac{dy}{dt} =ky \]

You might have also seen the **exponential function** of the form:

\[ f(t) = Ce^{kt} \]

This **function satisfies** the **equation** $\dfrac{dy}{dt} = ky$, such that:

\[ \dfrac{dC\cdot e^{kt}}{dt} = C\cdot k\cdot e^{kt} \]

So it seems that it is one of the **possible solutions** to the above **differential** equation.

So we’ll be using this **equation** to get the value of $k$:

\[ P[t] = Ce^{kt} \]

Consider that the **initial population** is set as $P[t] = 1$, when the time $t = 0$, so the **equation** becomes:

\[ 1 = Ce^{k|0|} \]

\[1 = Ce^{0} \]

\[1 = C\cdot 1 \]

Hence, we get $C = 1$.

So if the **population double** after every **decade** then, we can rewrite the **equation** as:

\[2 = 1\cdot e^{10k} \]

Taking **natural log** to remove the **exponential:**

\[\ln 2 = \ln [e^{10k}] \]

\[\ln 2 = 10k \]

So $k$ **comes** out to be:

\[k = \dfrac{\ln 2}{10} \]

**OR,**

\[k = 0.0693 \]

As you can see that $k > 0$, indicates that the **population** is growing **exponentially.**

## Numerical Result

$k$ comes out to be $0.0693$, which **states**Â that $k > 0$, indicating the **population** growing **exponentially.**

## Example

A pack of **wolves** has $1000$ wolves in it, and they are **increasing** in number **exponentially.** After $4$ year the **pack** has $2000$ wolves. **Derive** the **formula** for the **number** of **wolves** at **random** time $t$.

The **phrase growing exponentially** gives us an **indication** of the situation that is:

\[f(t)=Ce^{kt} \]

Where $f(t)$ is the **number** of **wolves** at time $t$.

Given in the **statement,** initially means at $t = 0$ there were $1000$ **wolves** and at **time$** t=4$ there are **doubles** $2000$.

The **formula** to find $k$ given two **different time lapses** is:

\[k= \dfrac{\ln f(t_1)-\ln f(t_2)}{t_1 -t_2} \]

**Plugging** in the values gives us:

\[k= \dfrac{\ln 1000-\ln 2000}{0 -4} \]

\[k= \ln \dfrac{1000}{2000}-4 \]

\[k= \dfrac{\ln{\dfrac{1}{2}}}{-4} \]

\[k= \dfrac{\ln 2}{4} \]

**Therefore:**

\[f(t) = 1000\cdot e^{\dfrac{\ln 2}{4}t}\]

\[f(t) = 1000\cdot 2^{\dfrac{t}{4}}\]

Hence, the **preferred formula** for the **number** of **wolves** at any time $t$.