This comprehensive guide explores the natural exponential function, its foundational properties, common mathematical modeling forms, and quantitative applications across biology, physics, and finance.
Euler Constant e and Natural Exponential Function
The natural exponential function \( f \) is an exponential function with a base equal to Euler's Constant e, defined of the form:
A table of values for \( e^x \) followed by the graph of \( f \) are shown below.
Properties of the Natural Exponential Function \( f(x) = e^x \)
- Domain: \( (-\infty, +\infty) \)
- Range: \( (0, +\infty) \)
- y-intercept: \( (0, 1) \)
- Horizontal Asymptote: \( y = 0 \)
- Monotonicity: Strictly increasing on \( (-\infty, +\infty) \)
- Continuity: Continuous on \( (-\infty, +\infty) \)
- Differentiability: Differentiable on \( (-\infty, +\infty) \)
- One-to-One Function: It is one-to-one and therefore possesses an inverse.
- Inverse Function: The inverse of \( f(x) = e^x \) is the natural logarithm \( \ln(x) \).
- Composition Identities: \( e^{\ln(x)} = x \) for \( x > 0 \) and \( \ln(e^x) = x \) for all real \( x \).
- Derivative: \( \dfrac{d}{dx}(e^x) = e^x \)
- Indefinite Integral: \( \int e^x \, dx = e^x + C \), where \( C \) is the constant of integration.
Mathematical Models Using Exponential Functions
Many mathematical models in physics, engineering, chemistry, and economics are described by natural exponential functions:
- Exponential Decay: \( f(x) = A e^{-kx} \), where \( k > 0 \) (Example: \( y = e^{-0.5x} \)).
- Exponential Growth: \( f(x) = A e^{kx} \), where \( k > 0 \) (Example: \( y = e^{2x} \)).
- Gaussian Model: \( f(x) = A e^{-\left(\frac{x-a}{b}\right)^2} \) (Example: \( f(x) = 3 e^{-\left(\frac{x-1}{2}\right)^2} \)).
Applications of the Natural Exponential Function – Examples with Solutions
Example 1: Population Growth Modeling
The population \( P \) of a city increases continuously according to the formula:
\[ P = 120000 e^{0.01 t} \]where \( t \) is the number of years after the year 2000. What will be the population of this city in the year 2030?
Solution:
- The number of years \( t \) between 2000 and 2030 is \( 30 \).
- Substitute \( t = 30 \) into the formula: \[ P = 120000 e^{0.01 \times 30} = 120000 e^{0.3} \approx 161983 \]
Example 2: Radioactive Material Decay Modeling
The amount \( A \) of a decaying radioactive material after \( t \) days is given by:
\[ A = A_0 e^{-0.005 t} \]If the initial amount at \( t = 0 \) is 20 milligrams:
a) What is the amount of radioactive material left after 30 days? Round your answer to the nearest hundredth.
b) How many days does it take for the material to decay to 10 milligrams?
Solution:
- Part a: Since \( A = 20 \) at \( t = 0 \), \( A_0 = 20 \). The formula becomes \( A = 20 e^{-0.005 t} \). At \( t = 30 \): \[ A = 20 e^{-0.005 \times 30} = 20 e^{-0.15} \approx 17.21 \text{ mg} \]
- Part b: Set \( A = 10 \): \[ 10 = 20 e^{-0.005 t} \implies e^{-0.005 t} = 0.5 \] Taking the natural logarithm of both sides: \[ -0.005 t = \ln(0.5) \implies t = \frac{\ln(0.5)}{-0.005} \approx 138.63 \text{ days} \]
Example 3: R-C Circuit Modeling
The voltage \( V_c \) across a charging capacitor in an R-C circuit is given by:
\[ V_c = V \left(1 - e^{-t/rc}\right) \]where \( t \) is in seconds, \( V \) in volts, \( r \) in Ohms, and \( c \) in Farads.
a) Let \( V = 10 \) volts. Make a table of values of \( V_c \) for \( t = 0, rc, 2rc, 3rc, 4rc, 5rc \) (rounded to the nearest hundredth).
b) Graph \( V_c \) as a function of \( t \).
c) Find the value of \( t \) in terms of \( rc \) for which \( V_c \) equals 9.99 volts.
Solution:
- Part a: Table values for \( V_c = 10(1 - e^{-t/rc}) \):
\( t \) \( 0 \) \( rc \) \( 2rc \) \( 3rc \) \( 4rc \) \( 5rc \) \( V_c \) \( 0 \) \( 6.32 \) \( 8.65 \) \( 9.50 \) \( 9.82 \) \( 9.93 \) - Part b: Graph of voltage across a charging capacitor:
- Part c: Solve \( 9.99 = 10 \left(1 - e^{-t/rc}\right) \): \[ 0.999 = 1 - e^{-t/rc} \implies e^{-t/rc} = 0.001 \] Taking natural logarithms: \[ -\frac{t}{rc} = \ln(0.001) \approx -6.9078 \implies t \approx 6.91 rc \]
Example 4: Continuous Compounding Interest
The balance \( B \) in a savings account after \( t \) years under continuous compounding at rate \( r \) is given by:
\[ B = P e^{rt} \]An amount of \( \$10000 \) is invested at \( r = 6.25\% \) and compounded continuously. How many years will it take for the investment to double?
Solution:
- Initial deposit \( P = 10000 \), so \( B = 10000 e^{0.0625 t} \).
- We want the balance to double to \( B = 20000 \): \[ 20000 = 10000 e^{0.0625 t} \implies 2 = e^{0.0625 t} \]
- Taking natural logarithms: \[ \ln(2) = 0.0625 t \implies t = \frac{\ln(2)}{0.0625} \approx 11.09 \text{ years} \]