AP Calculus BC 2.1 Defining Average and Instantaneous Rates of Change at a Point Study Notes - New Syllabus
AP Calculus BC 2.1 Defining Average and Instantaneous Rates of Change at a Point Study Notes- New syllabus
AP Calculus BC 2.1 Defining Average and Instantaneous Rates of Change at a Point Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Derivatives allow us to determine rates of change at an instant by applying limits to knowledge about rates of change over intervals.
Key Concepts:
- Average Rates of Change Using Difference Quotients
- Representing the Derivative as a Limit
Average Rates of Change Using Difference Quotients
Average Rates of Change Using Difference Quotients
The difference quotient measures the average rate of change of a function and forms the foundation of the derivative.
1. Symmetric form centered at \( a \):
\( \dfrac{f(a + h) – f(a)}{h} \)
This gives the average rate of change from \( a \) to \( a + h \).
2. General form over an interval from \( a \) to \( x \):
\( \dfrac{f(x) – f(a)}{x – a} \)
This measures the average rate of change between the points \( (a, f(a)) \) and \( (x, f(x)) \).
Use Cases:
- These are used in evaluating slopes of secant lines and approximating derivatives.
- As \( h \to 0 \), the first form becomes the derivative \( f'(a) \).
- As \( x \to a \), the second form also approaches \( f'(a) \).
Example :
Let \( f(x) = x^2 + 3x \). Use the difference quotient formula
\( \dfrac{f(a + h) – f(a)}{h} \)
to find the average rate of change near \( a = 2 \).
▶️Answer/Explanation
Compute \( f(a + h) \) for \( f(x) = x^2 + 3x \)
\( f(2 + h) = (2 + h)^2 + 3(2 + h) \)
\( = 4 + 4h + h^2 + 6 + 3h = h^2 + 7h + 10 \)
Compute \( f(2) = 2^2 + 3(2) = 4 + 6 = 10 \)
$ \dfrac{f(2 + h) – f(2)}{h} = \dfrac{(h^2 + 7h + 10) – 10}{h} = \dfrac{h^2 + 7h}{h} $
\( = h + 7 \)
Conclusion: The difference quotient simplifies to \( h + 7 \), which approaches 7 as \( h \to 0 \).
Representing the Derivative as a Limit
Representing the Derivative as a Limit
The derivative of a function at a point is defined as the limit of the average rate of change (difference quotient) as the interval approaches zero.
This is the formal definition of the derivative:
\( f'(a) = \lim_{h \to 0} \dfrac{f(a + h) – f(a)}{h} \)
Alternate (but equivalent) form:
\( f'(a) = \lim_{x \to a} \dfrac{f(x) – f(a)}{x – a} \)
Meaning:
- This limit gives the instantaneous rate of change of the function at the point \( x = a \).
- It also represents the slope of the tangent line to the graph of the function at that point.
- Both definitions are used interchangeably depending on the situation or the question format.
Notation: The derivative of \( f(x) \) can be written in several ways:
- \( f'(x) \)
- \( \dfrac{dy}{dx} \)
- \( \dfrac{d}{dx}[f(x)] \)
Example :
Finding the Derivative Using the Limit Definition
Let \( f(x) = 3x^2 + 2x \). Find \( f'(x) \) using the limit definition of the derivative:
\( f'(x) = \lim_{h \to 0} \dfrac{f(x + h) – f(x)}{h} \)
▶️Answer/Explanation
\( f(x + h) = 3(x + h)^2 + 2(x + h) \)
\( = 3(x^2 + 2xh + h^2) + 2x + 2h \)
\( = 3x^2 + 6xh + 3h^2 + 2x + 2h \)
Compute \( f(x + h) – f(x) \)
\( f(x + h) – f(x) = [3x^2 + 6xh + 3h^2 + 2x + 2h] – [3x^2 + 2x] \)
\( = 6xh + 3h^2 + 2h \)
Divide by \( h \)
\( \dfrac{f(x + h) – f(x)}{h} = \dfrac{6xh + 3h^2 + 2h}{h} = 6x + 3h + 2 \)
Take the limit as \( h \to 0 \)
\( f'(x) = \lim_{h \to 0} (6x + 3h + 2) = 6x + 2 \)
Example:
Use the definition of derivative to find \( f'(x) \) for
\( f(x) = \sqrt{x + 1} \)
Use the limit definition:
\( f'(x) = \lim_{h \to 0} \dfrac{f(x + h) – f(x)}{h} \)
▶️Answer/Explanation
\( f'(x) = \lim_{h \to 0} \dfrac{\sqrt{x + h + 1} – \sqrt{x + 1}}{h} \)
Multiply numerator and denominator by the conjugate
\( = \lim_{h \to 0} \dfrac{[\sqrt{x + h + 1} – \sqrt{x + 1}] \cdot [\sqrt{x + h + 1} + \sqrt{x + 1}]}{h \cdot [\sqrt{x + h + 1} + \sqrt{x + 1}]} \)
2Simplify the numerator using identity \( (a – b)(a + b) = a^2 – b^2 \)
\( = \lim_{h \to 0} \dfrac{(x + h + 1) – (x + 1)}{h[\sqrt{x + h + 1} + \sqrt{x + 1}]} \)
\( = \lim_{h \to 0} \dfrac{h}{h[\sqrt{x + h + 1} + \sqrt{x + 1}]} \)
\( = \lim_{h \to 0} \dfrac{1}{\sqrt{x + h + 1} + \sqrt{x + 1}} \)
Take the limit as \( h \to 0 \)
\( f'(x) = \dfrac{1}{2\sqrt{x + 1}} \)