# Exponential Functions Business Applications

##### 839 words — categories: calculus

We go over some textbook exercises for exponential functions.

## Franchise Expansion

A franchise business is selling franchises for pizza shops throughout the country. The marketing manager estimates that the number of franchises, \(N\), will increase at the rate of 10% per year, that is,

\[\frac{dN}{dt} = 0.10N\]

- Find the function that satisfies this equation, assuming that the number of franchises at \(t = 0\) is 50.
- How many franchises will be there in 20 years?
- In what period of time will the initial number of 50 franchises double?

```
from sympy import *
from math import e
from sympy.abc import N, t
N = Function("N")(t)
dNdt = Eq(N.diff(t), 0.10 * N)
f = dsolve(dNdt, N)
f
```

`## Eq(N(t), C1*exp(0.1*t))`

```
def f(t, c):
return c * e**(0.1 * t)
f(20, 50)
```

`## 369.45280494653247`

`solve(50 * e**(0.1 * t) - 100, t)`

`## [6.93147180559944]`

## Compound Interest

Suppose that \(P_0\) is invested in a fund for which interest is compounded continuously at 5.9% per year. That is, the balance \(P\) grows at the rate given by

\[\frac{dP}{dt} = 0.059P\]

- Find the function that satisfies the equation.
- Suppose that $1000 is invested, what is the balance after 1 year and 2 years?
- When will a $1000 investment double itself?

```
from sympy.abc import P
P = Function("P")(t)
dPdt = Eq(P.diff(t), 0.059 * P)
dPdt
```

`## Eq(Derivative(P(t), t), 0.059*P(t))`

`dsolve(dPdt, P)`

`## Eq(P(t), C1*exp(0.059*t))`

```
def f(t, c):
return c * e**(0.059 * t)
f(1, 1000)
```

`## 1060.775240740159`

`f(2, 1000)`

`## 1125.2441113673424`

`solve(1000 * e**(0.059 * t) - 2000, t)`

`## [11.7482572976262]`

## Value of Manhattan Island

Peter Minuit o the Dutch West India Company purchased Manhattan Island from the natives living there in 1626 for $24 worth of merchandise. Assuming an exponential rate of inflation of 5%, how much will Manhattan be worth in 2020?

Let’s assume the growth model for the value of the island is given by

\[\frac{dV}{dt} = 0.05V\]

The function that satisfies this rate of change is

\[V(t) = ce^{0.05t}\]

```
from sympy.abc import V
V = Function("V")(t)
dVdt = Eq(V.diff(t), 0.05 * V)
dVdt
```

`## Eq(Derivative(V(t), t), 0.05*V(t))`

```
f = dsolve(dVdt, V)
f
```

`## Eq(V(t), C1*exp(0.05*t))`

```
years = 2020 - 1626
def f(t, c):
return c * e**(0.05*t)
f(years, 24)
```

`## 8626061203.204306`

## Cost of Hershey Bar

The cost of a Hershey bar was $0.05 in 1962 and $0.75 in 2010.

- Find an exponential function that fits the data.
- Predict the cost of a Hershey bar in 2015 and 2025.

The assumtion is that the value of a bar has grown exponentially

\[\frac{dV}{dt} = kV\]

Because the value at \(t = 0\) is $0.05, we can say that the exponential function is

\[V(t) = 0.05e^{kt}\]

We know that 48 years after 1962, the value is $0.75, therefire

\[0.75 = 0.05e^{k(48)}\]

and we need to find \(k\).

```
from sympy.abc import k
years = 2010 - 1962
equation = Eq(0.75, 0.05 * e**(k*years))
k = solve(equation, k)[0]
k
```

`## 0.0564177125229626`

```
def f(t, c):
return 0.05 * e**(c*t)
f(2015 - 1962, k)
```

`## 0.994422104712232`

`f(2025 - 1962, k)`

`## 1.74819462993714`

## Diffusion of Information

Pharmaceutical firms invest significantly in testing new medications. After a drug is approved by the FDA, it still takes time for physicians to fully accept and start prescribing the medication. The acceptance by physicians approaches a *limit value* of 100%, or 1, after time \(t\), in months. Suppose that the percentage \(P\) of physicians prescribing a new cancer medication after \(t\) months is approximated by

\[P(t) = 100(1 - e^{-0.4t})\]

- What percentage of doctors are prescribing the medication after 0, 1, 2, 3, 5, 12, and 16 months?
- Find \(P^\prime(7)\) and interpret its meaning.
- Sketch a graph of the function.

```
def p(t):
return 100 * (1 - e**(-0.4 *t))
percentages = [round(p(t), 2) for t in [1, 2, 3, 5, 12, 16]]
percentages
```

`## [32.97, 55.07, 69.88, 86.47, 99.18, 99.83]`

```
expr = 100 * (1 - e**(-0.4 *t))
expr.diff()
```

`## 40.0*2.71828182845905**(-0.4*t)`

The equation

\[p^\prime(t) = 40e^{-0.4t}\] tells us the rate of change of the percentage of phyicians that accept the new medication at any given time (in months).

```
import numpy as np
import matplotlib.pyplot as plt
x_values = np.linspace(0, 20)
plt.plot(x_values, p(x_values))
plt.title("% physicians accepting and prescribing new medication")
```

## Hullian Learning Model

The Hullian learning model asserts that the probability \(p\) of mastering a task after \(t\) learning trials is approximated by

\[p(t) = 1 - e^{-kt}\]

where \(k\) is a constant that depends on the task to be learned. Suppose a new dance is taught to an aerobics class. For this particular dance, the constant \(k = 0.28\).

- What is the probability of mastering the dance’s steps in 1, 2, 5, 11, 16, 20 and trials?
- Find the rate of change, \(p^\prime(t)\).
- Sketch a graph of the function.

```
def p(t, k):
return 1 - e**(-k*t)
constant = 0.28
probabilities = [round(p(t, constant), 2) for t in [1, 2, 5, 11, 16, 20]]
probabilities
```

`## [0.24, 0.43, 0.75, 0.95, 0.99, 1.0]`

```
expr = 1 - e**(-k*t)
expr.diff()
```

`## 0.0564177125229626*2.71828182845905**(-0.0564177125229626*t)`

```
x_values = np.linspace(0, 25)
plt.plot(x_values, p(x_values, constant))
plt.title("$p(t)$ of mastering dance steps after $t$ trials")
```