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Understanding Instantaneous Rates of Change
Aug 16, 2024
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Review flashcards
Lesson 4.2: Introduction to Instantaneous Rates of Change
Key Objectives
Understand and practice calculating difference quotients.
Recognize the difference quotient as an average rate of change on an interval ( x ) to ( x + h ).
Introduction to instantaneous rates of change.
Difference Quotient
Definition
: Difference quotient is the average rate of change of a function on the interval ( x ) to ( x + h ).
Formula
: [ \frac{f(x+h) - f(x)}{h} ]
This represents the slope of the secant line between the points ( (x, f(x)) ) and ( (x+h, f(x+h)) ).
Examples
Example 1: Linear Function
Function: ( f(x) = -10x + 7 )
Find the average rate of change from ( x ) to ( x + h ).
Steps
:
Calculate ( f(x+h) - f(x) ):
( = -10(x+h) + 7 - (-10x + 7) )
Simplifies to ( -10h )
Divide by ( h ):
( \frac{-10h}{h} = -10 )
Result: Consistent with the slope of the line.
Example 2: Quadratic Function
Function: ( f(x) = 3x^2 - 2 )
Calculate from ( x = 2 ) to ( x = 2 + h ).
Steps
:
Evaluate ( f(2+h) - f(2) ):
( = 3(2+h)^2 - 2 - (3 \times 2^2 - 2) )
Simplifies to ( 11h + 3h^2 )
Divide by ( h ):
( \frac{11h + 3h^2}{h} = 11 + 3h )
Example 3: Rational Function
Function: ( f(x) = \frac{2}{x} )
Calculate from ( x = 3 ) to ( x = 3 + h ).
Steps
:
Evaluate ( f(3+h) - f(3) ):
( = \frac{2}{3+h} - \frac{2}{3} )
Simplifies to ( \frac{-2h}{3(3+h)} )
Divide by ( h ):
( \frac{-2}{3(3+h)} )
Instantaneous Rate of Change
Conceptualized as the rate of change at a precise moment.
Calculation
: Let ( h ) approach 0 in the difference quotient.
Caution
: Direct substitution of ( h = 0 ) leads to an indeterminate form (( \frac{0}{0} )).
Use limits to resolve.
Example: Velocity
Function: ( p(t) = -16t^2 + 50t + 5 )
Average velocity
from ( t=0 ) to ( t=1 ):
( \frac{p(1) - p(0)}{1} = 34 ) feet per second.
Instantaneous velocity
at ( t=1 ):
Evaluate ( \frac{p(1+h) - p(1)}{h} ) and let ( h \to 0 ):
Result: ( 18 ) feet per second.
Derivative
Definition
: Derivative ( f'(x) ) of a function is the instantaneous rate of change.
Calculated using the difference quotient approach.
Final Example: Rational Function
Function: ( g(x) = \frac{1}{2x + 3} )
Calculate instantaneous rate of change at ( x=1 ).
Steps
:
Evaluate ( g(1+h) - g(1) ) over ( h ).
Simplify and evaluate as ( h \to 0 ):
Result: (-\frac{2}{25} ).
Conclusion
Mastery of the difference quotient is essential for understanding rates of change in calculus.
Practice with various functions to gain familiarity.
Understand that the derivative represents the instantaneous rate of change.
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