Lecture 3 1D Newtons Law
The Chain Rule and the Natural Logarithm
The Chain Rule
Consider two functions:
( x ) as a function of ( t ): ( \frac{dx}{dt} = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} )
( y ) as a function of ( x ): ( \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} )
We can combine these to express ( y ) as a function of ( t ): ( y(x(t)) ) and calculate ( \frac{dy}{dt} )
The relationship is:
( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} ) (Chain Rule)
Visualize as the ( dx ) terms "cancelling" each other in the context of derivatives.
Example of the Chain Rule
Given:
( x(t) = t^2 - t )
( y(x) = 1 - x^2 )
Substitute:
( y(x(t)) = 1 - (t^2 - t)^2 = -t^4 + 2t^3 - t^2 + 1 )
Differentiate:
( \frac{dy}{dt} = -4t^3 + 6t^2 - 2t )
( \frac{dy}{dx} = -2x ) where ( x = t^2 - t )
( \frac{dx}{dt} = 2t - 1 )
Combine:
( \frac{dy}{dt} = -2x \cdot (2t - 1) = -2(t^2 - t)(2t - 1) = -4t^3 + 6t^2 - 2t )
Proof of the Chain Rule
Small change in ( t ) gives:
( \Delta x = x(t + \Delta t) - x(t) )
( \Delta y = y(x(t + \Delta t)) - y(x(t)) = y(x + \Delta x) - y(x) )
From the definitions:
( \frac{\Delta y}{\Delta t} = \frac{\Delta y}{\Delta x} \cdot \frac{\Delta x}{\Delta t} )
- As ( \Delta t \to 0 ): ( \frac{\Delta y}{\Delta x} \to \frac{dy}{dx} ) and ( \frac{\Delta x}{\Delta t} \to \frac{dx}{dt} )
Thus, the chain rule follows.
Application in Motion
For one-dimensional motion:
Show that ( \frac{dv}{dt} = \frac{d}{ds}\left(\frac{1}{2}v^2\right) )
Recall: ( a = \frac{dv}{dt} ), where ( v ) is a function of position ( s ): ( v = v(s(t)) )
With differentiation, we find:
( \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = \frac{dv}{ds} \cdot v )
Then, using the Leibniz rule gives:
( \frac{d}{ds}\left(\frac{1}{2}v^2\right) = v \frac{dv}{ds} )
If uniform acceleration, integrate:
( \int s = 0 ) to get: ( \frac{v^2 - u^2}{2} = as )
The Natural Logarithm
Integral definition:
For ( n
eq -1 ): ( \int t^n dt = \frac{1}{n+1}t^{n+1} + C )Special case when ( n = -1 ): define natural logarithm as ( \ln(t) = \int_{1}^{t} \frac{1}{t} dt )
Understand area under the curve of ( \frac{1}{t} ) between 1 and ( t \geq 1 )
Properties of the Natural Logarithm
Inequality:
( \ln(t) \geq 0 ) for ( t \geq 1 )
( \ln(t) \leq 0 ) for ( 0 < t < 1 )
Logarithmic properties:
( \ln(t_1 t_2) = \ln(t_1) + \ln(t_2) )
( \ln(t^a) = a \ln(t) ) for any ( a )
Notation:
Also noted as ( \log(t) ) or ( \log_e(t) ), while ( \log_{10}(t) ) denotes log base 10.
Proofs of Logarithm Properties
For ( t \geq 1 ): Area under curve gives ( \ln(t) \geq 0 )
For ( t < 1 ): ( \ln(t) = -\int_{t}^{1} \frac{1}{t} dt \leq 0 )
For product property, assume ( t_1, t_2 \geq 1 ):
Evaluate ( \ln(t_1 t_2) ) through area relationships.
For exponent property, evaluate ( \ln(t^{a}) ) via change of variable and integration.