# Exponential Decay

##### 530 words — categories: calculus

In the equation of population growth, \(dP/dt = kP\), the constant \(k\) is given by \(k = (\text{birth rate}) - (\text{death rate})\). Thus, a population grows only when the birth rate is greater than the death rate. When the death rate is greater than the birth rate we have a decrease or “decay”. Our equation for population decay becomes \(dP/dt = -kP\), where \(k >0\). The function that satisfies that rate of return is then

\[P(t) = P_0e^{-kt}\]

where \(P_0\) is the amount at \(t = 0\). And, for such decaying functions

\[P_0 = \frac{P(t)}{e^{-kt}}\]

## Radioactive Decay

Strontium-90 has a decay rate of 2.8% per year. The rate of change of an amount \(N\) of this radioactive isotope is given by

\[\frac{dN}{dt} = -0.028N\]

- Find the function that satisfies the equation. Let \(N_0\) represent the amount present at \(t = 0\).
- Suppose that 1000 grams of strontium-90 is present at \(t = 0\). How much will remain after 70 years?
- After how long will half of the 1000 g remain?

```
from sympy import *
from math import e
import numpy as np
from sympy.abc import N, t
N = Function("N")(t)
C1 = symbols("C1")
dNdt = Eq(N.diff(t), -0.028 * N)
f = dsolve(dNdt, N)
f
```

`## Eq(N(t), C1*exp(-0.028*t))`

```
ff = lambdify((t, C1), f.rhs, 'numpy')
ff(70, 1000)
```

`## 140.858420921045`

```
def half_life(p0, k):
return np.log(2) / k
half_life(1000, 0.028)
```

`## 24.755256448569472`

## Half-Life

The half-life is the time required for a quantity to reduce half of its initial value. The **decay rate**, \(k\), and the **half-life**, \(T\), are related by

\[kT = \ln 2 = 0.693147\]

therefore

\[k = \frac{\ln 2}{T}\]

and

\[T = \frac{\ln 2}{k}\]

## Present Value

You make an initial investment of \(P_0\) that will grow to $10.000 in 20 years. Interest is compounded continuously at 6%. What should the initial investment be?

```
def P(p, k, t):
"""first method"""
return p / e**(k*t)
P(10000, 0.06, 20)
```

`## 3011.9421191220213`

```
def P(p, k, t):
"""second method"""
return p * e**(-(k*t))
P(10000, 0.06, 20)
```

`## 3011.9421191220213`

Economists call this value the **present value**. The process of computing present values is called **discounting**. The computation of present value can be interpreted as exponential decay from the future back to the present.

The present value \(P_0\) of an amount \(P\) due \(t\) years alter is found by solving the equation

\[P_0e^{kt} = P\] Therefore

\[P_0 = Pe^{-kt}\]

## US Farms

The number \(N\) of farms in the United States has declined continually since 1950. In 1950, there were 5,650,000 farms, and in 2005, that number had decreased to 2,100,990. Assuming the number of farms decreased according to the exponential decay model:

- Find the value of \(k\), and write the exponential function that describes the number of farms after time \(t\), where \(t\) is the number of years since 1950.
- Estimate the number of farms in 2009 and in 2015.
- At this decay rate, when will only 1,000,000 farms remain?

```
yrs = 2005 - 1950
k = np.log(2) / yrs
k
```

`## 0.012602676010180823`

\[N(t) = 5,650,000e^{-0.013t}\]

```
def N(t, k):
return 5650000 * e**(-(k*t))
n1 = 2009 - 1950
n2 = 2015 - 1950
N(n1, k)
```

`## 2686119.697425508`

`N(n2, k)`

`## 2490495.29536376`

`solve(5650000 * e*(k*t) - 1000000, t)`

`## [5.16647460146324]`