Lecture 4 1D Newtons Law

The Exponential Function

Definition and Properties

  • Let ( x = \ln(t) ) and consider the graph of ( x ) versus ( t ).

  • ( \ln(t) ) is defined for ( t > 0 ).

  • The derivative of ( \ln(t) ) is given by:

    • ( \frac{d}{dt}(\ln(t)) = \frac{1}{t} )

  • The function ( \ln(t) ) is increasing in ( t ) but the rate of increase decreases as ( t ) grows.

Graphical Insights

  • The function ( x = \ln(t) ) takes all values ( -\infty < x < \infty ) for ( 0 < t < \infty ).

  • For each ( x ), there exists a unique ( t > 0 ): ( t = \exp(x) ) and ( x = \ln(t) ).

  • Properties of the exponential function include:

    • ( \exp(\ln(t)) = t ) and ( \ln(\exp(x)) = x )

    • ( \exp(x) > 0 ) for all ( -\infty < x < \infty )

    • ( \exp(0) = 1 )

    • The number ( e = \exp(1) ) which is approximately ( 2.7182818 \ldots

  • ( e ) is regarded as more significant than ( \pi ) in mathematics.

Exponential Function Formalism

Exponential Definition

  • It is shown that ( \exp(x) = e^x ).

  • Proof:

    • Given property of ( \ln ): ( \ln(t^x) = x \ln(t) )

    • Set ( t = e ): ( \ln(e^x) = x \ln(e) = x )

    • Hence, ( \exp(\ln(e^x)) = \exp(x) ) leading to ( e^x = \exp(x) ).

  • The notation ( \ln(t) ) can also be written as ( \log_e(t) ).

Derivative of the Exponential Function

Finding the Derivative

  • Given that ( \frac{d}{dt}(\ln(t)) = \frac{1}{t} ), let ( t = e^x ):

    • ( \frac{dx}{dt} = \frac{1}{e^x} )

  • Chain Rule Application:

    • ( 1 = \frac{dt}{dt} = \frac{dt}{dx} \frac{dx}{dt} )

    • Rearranging gives: ( \frac{dt}{dx} = e^x ) hence, ( \frac{d}{dx}(e^x) = e^x )

  • For constant ( k ), ( \frac{d}{dx}(e^{kx}) = ke^{kx} ).

Exponential Growth and Decline

Behavior of Exponential Functions

  • As ( x \to +\infty ), ( \exp(x) ) increases rapidly (exponential growth).

  • As ( x \to -\infty ), ( \exp(x) \to 0 ) indicating exponential decay.

  • Numerical Examples:

    • ( \exp(10) \approx 22026.47 )

    • ( \exp(100) \approx 2.6881 \times 10^{43} )

Population Modeling

Exponential Growth in Population

  • Let ( P(t) ) denote population at time ( t ).

  • Assumed growth model: ( \frac{dP}{dt} = kP ) for a constant ( k > 0 ), with initial population ( P(0) = P_0 ).

  • Integrating:[ \int_{P_0}^{P} \frac{1}{P} dP = \int_0^{t} k dt ] leads to[ \ln(P) - \ln(P_0) = kt ]

  • Exponentiating provides:

    • ( P = P_0 \exp(kt) )

  • This describes exponential growth as ( t ) increases.

Example of Population Data

  • The population of Sweden (1750-1960) modeled with:

    • Initial population: ( P_0 = 1.68 \times 10^6 )

    • Growth constant: ( k = 0.0072/year ).

Decay Model Example

Mass in Frictional Medium

  • A mass ( m ) through a frictional medium experiences deceleration ( -2v ).

  • If initial velocity ( v = u ) leads to:

    • Governing Equation: ( \frac{dv}{dt} = -2v )

  • Solving gives:

    • ( v = ue^{-2t} )

    • Implying ( v \to 0 ) as ( t \to \infty ) (exponential decay).

Graphical Representation

  • Solutions plotted indicate behavior of ( v ) towards zero as time increases.