## What is Exponential Function?

The relation that represents an exponential function is:

$y=A{e}^{kx}$, where k and A are constants.

The exponential function has the following properties:

1. Consider that P (t) is a given function of time t. The representation of the rate of change of P with respect to t is given by the model $\frac{dP}{dt}={P}^{\prime }\left(t\right)$.

2. For a function P (t) to get growth over an interval I, the condition $\frac{dP}{dt}={P}^{\prime }\left(t\right)>0$must be satisfied for each $t\in I$.

3. A function P(t) is said to decrease over an interval if $\frac{dP}{dt}={P}^{\prime }\left(t\right)<0$ for each $t\in I$.

4. A function P(t) is neither increasing nor decreasing over an interval if $\frac{dP}{dt}=P\left(t\right)=0$ for each $t\in I$.

## What is an Exponential Growth Function?

An exponential model is described to show the condition where the same number is multiplied repeatedly. This model has numerous applications, especially in the existing sciences and in financial aspects.

The process in which a quantity increases over time such that the rate of increase of the quantity at any time is proportional to the value of the quantity at that time is known as exponential growth.

The model of population growth each year could be an example of an exponential model.

For the population increase model, the growth rate is proportional to the population size at time t. Let us assume the variable P(t) (sometimes we use the only P) represents the population at any time t. Additionally, let P0 be the initial population at time t = 0.

P (0) = P0.

If the growth of population is exponential, then:

The rate of change of population at time t = k (Population at time t)

Mathematically, it can be represented as:

$\frac{dP}{dt}=kP\cdots \cdots \left(1\right)$

The equation can be simplified as:

$K=\frac{\frac{dP}{dt}}{P}$

Where k denotes the relative growth rate and is a constant.

From the equation (1),

Using the method of separation of variable in the model we have,

$\frac{dP}{P}=kdt$

Integrating both the sides we get the model as:

$\begin{array}{l}\int \left(\frac{1}{P}\right)dP=\int kdt\\ \mathrm{ln}\left|P\right|=kt+c\cdots \cdots \left(2\right)\end{array}$

Raising both sides of equation (2) to the exponent e, we get:

${e}^{\mathrm{ln}\left|P\right|}={e}^{kt+c}$

Using the inverse property ${e}^{\mathrm{ln}k}=k$ and law of exponents ${b}^{x+y}={b}^{x}{b}^{y}$, we get,

$\left|P\right|={e}^{kt}{e}^{c}$

Using absolute value definition for the model, we get:

$P=±{e}^{c}{e}^{kt}$

Replace the constant term with A:

$P\left(t\right)=A{e}^{kt}\cdots \cdots \left(3\right)$

Equation (3) shows the general solution for the differential equation.

Now, use the initial condition $P\left(0\right)={P}_{o}$, substitute t = 0 in equation (3), and equate:

${P}_{o}=P\left(0\right)=A{e}^{k\left(0\right)}\cdots \cdots \left(4\right)$ 

We know that ${x}^{0}=1$.

Hence, equation (4) becomes,

${P}_{o}=A\left(1\right)$

So, the value of A is:

${P}_{o}=A$

Hence, equation (3) becomes:

$P\left(t\right)={P}_{0}{e}^{kt}\cdots \cdots \left(5\right)$

Equation (5) is the particular solution of the exponential model, where:

${P}_{o}$ = Initial population at time t = 0.

k = The relative growth (increase) rate. It is a constant.

t = The time taken for the growth of the population.

P(t) = The population at time t.

### Notes

1. For an exponential growth model, if there is a case where the growth rate, k, is unknown, then it should be determined. This value of growth is determined using the known value of population for the two particular times.
2. Exponential growth is a good indicator for little populace in an enormous populace with bountiful assets. For short time periods, exponential growth is used.

3. The graph of the exponential growth model $P\left(t\right)={P}_{o}{e}^{kt}$ is given by,

## What is Exponential Decay?

For an exponential decay such as population, when the rate of decrement is proportional to the size at time t, the exponential decay model is as follows:

$\begin{array}{l}\frac{dP}{dt}=-kP\\ P\left(t\right)={P}_{o}\\ k=\frac{-1}{P}\left(\frac{dP}{dt}\right)\end{array}$

Where k is called the relative decay constant.

The population will decrease for the value of k>0. That is,

$\begin{array}{l}\frac{dP}{dt}<0\\ k=\frac{-1}{P}\left(\frac{dP}{dt}\right)>0\end{array}$

The initial value problem using the separation of variables method is

$P\left(t\right)={P}_{o}{e}^{-kt}$

Where the variables denote the following:

• At time t=0, P0 represents the initial population
• Constant relative decay rate is k. Note that k > 0.
• The time in which the population decays is denoted by t.
• The population that is left after time t is P(t).

### Note

Most of the time, the decay rate is given as half-life, where the half-life represents the decay time for half of the substance, i.e., when only half the original substance is left.

Example

The half-life for Bismuth-210 is 5 days.

1. Consider a sample with 800 mg mass. Determine the formula for mass remaining after t days.
2. What will be the mass remaining after 30 days?
3. Find the time when the mass is reduced to 1 mg.

Sol.: (a) The condition of exponential decay is:

$P\left(t\right)={P}_{o}{e}^{-kt}$

It is given that the initial mass is 800 mg.

Thus, the formula becomes:

$P\left(t\right)=800{e}^{-kt}$

Now, to complete the equation, we must find the relative decay rate using the half-life of the substance.

For Bismuth-210, the half-life is five days.

Thus, after t = 5, the initial value of the population will get decayed to half of its real value.

$\begin{array}{l}\frac{1}{2}\left(800\right)=400\\ P\left(5\right)=400\end{array}$

Mathematically, we have:

P(5) = 400.

From the decay equation:

$400=P\left(5\right)=800{e}^{-k\left(5\right)}$

We get:

$800{e}^{-5k}=400$

Now, we find the value of k.

$\begin{array}{l}{e}^{-5k}=\frac{400}{800}\\ {e}^{-5k}=0.5\cdots \cdots \left(1\right)\\ \mathrm{ln}\left({e}^{-5k}\right)=\mathrm{ln}\left(0.5\right)\\ -5k\left(lne\right)=ln\left(0.5\right)\end{array}$

We know that ln e = 1. So,

$\begin{array}{l}-5k\left(1\right)=\mathrm{ln}\left(0.5\right)\\ k=\frac{-\mathrm{ln}\left(0.5\right)}{5}\\ k=0.1386\end{array}$

Substituting k = 0.1386 and ${P}_{o}=800$gives a formula for finding the remaining mass as follows:

$P\left(t\right)=800{e}^{-0.1386t}$

(b) From the above formula, we get that the mass remaining after t = 30 is:

$\begin{array}{l}P\left(30\right)=800{e}^{-0.1386\left(30\right)}\\ P\left(30\right)=800{e}^{-4.158}\\ P\left(30\right)\approx 12.5grams\end{array}$

(c)  Here, we need to find the time needed for the sample to get reduced to 1 mg.

When P(t) = 1, solve for t using the decay equation as follows:

$\begin{array}{l}P\left(t\right)=800{e}^{-0.1386t}\\ P\left(t\right)=1\\ 1=P\left(t\right)=800{e}^{-0.1386t}\\ 1=800{e}^{-0.1386t}\\ {e}^{-0.1386t}=\frac{1}{800}\end{array}$

Taking log on both sides, the equation becomes:

$\begin{array}{l}-0.1386t\mathrm{ln}\left(e\right)=\mathrm{ln}\frac{1}{800}\\ t=\frac{\mathrm{ln}\frac{1}{800}}{-0.1386}\\ t\approx 48.2\end{array}$

Thus, it will take approximately 48.2 days to decay to the value of 1mg.

## Newton's Law of Cooling

According to Newton’s law of cooling, the cooling rate for an object is proportional to the difference between the temperature of the object and its surroundings.

At time t, consider T(t) is the temperature of an object, and the temperature of the surrounding environment is Ts. Assume it to be constant.

Mathematically, for Newton’s law of cooling:

$\begin{array}{l}\frac{dT}{dt}=k\left(T-{T}_{s}\right)\\ y=T-{T}_{s}\end{array}$

Differentiating both sides with respect to t, we get:

$\begin{array}{l}\frac{dy}{dt}=\frac{dT}{dt}-0\\ \frac{dy}{dt}=\frac{dT}{dt}\\ \frac{dy}{dt}=k\left(T-{T}_{s}\right)\\ \frac{dy}{dt}=ky\cdots \cdots \left(1\right)\end{array}$

This is just the basic exponential growth.

The general solution is: $y\left(t\right)={y}_{o}{e}^{kt}$

Recall that, $y\left(t\right)=T\left(t\right)-{T}_{s}$

From this equation, we have: ${y}_{o}=y\left(0\right)=T\left(0\right)-{T}_{s}={T}_{o}-{T}_{s}$

Here, T= initial temperature at t = 0.

Now, substituting,

$\begin{array}{l}y\left(t\right)=T\left(t\right)-{T}_{s}\\ {y}_{o}={T}_{o}-{T}_{s}\\ y\left(t\right)={y}_{o}{e}^{kt}\end{array}$

We get:

$\begin{array}{l}T\left(t\right)-{T}_{s}=\left({T}_{o}-{T}_{s}\right){e}^{kt}\\ T\left(t\right)={T}_{s}+\left({T}_{o}-{T}_{s}\right){e}^{kt}\cdots \cdots \left(2\right)\end{array}$

Equation (2) is the main equation for Newton's law of cooling, where:

• The temperature of the surrounding environment is given by Ts.
• The initial temperature of the object at time t = 0 is denoted by T0.
• The proportionality constant is k.
• The temperature of the object after time t is T(t).

## Common Mistakes

• Population per capita is independent of the population size for exponential growth.
• In nature, the growth of populations may be exponential for some period.
• Exponential growth produces a j-shaped curve.

## Formulas

• Newton's Law of Cooling
$T\left(t\right)={T}_{s}+\left({T}_{o}-{T}_{s}\right){e}^{kt}$
• The solution of the initial value problem $\frac{dP}{dt}=kP;\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}P\left(0\right)={P}_{0}$, using the separation of variables method is,
$P\left(t\right)={P}_{o}{e}^{-kt}$

## Context and Applications

This topic is significant in the professional exams for undergraduate and postgraduate courses, especially for:

• Bachelor of Science in Mathematics
• Bachelor of Science in Mathematics

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