Calculus Notes: Average Velocity, Instantaneous Velocity, and Introduction to Limits

Position, Average Velocity, and the Secant Line

• Definitions and Formulas:

  • Position Function: Let $s(t)$ represent the position of an object as a function of time $t$.
  • Time Interval: Consider a time interval bounded between two values, denoted as $[a, b]$.
  • Change in Position (Displacement): Measured by evaluating position at the ending time minus position at the beginning time:     $\Delta s = s(b) - s(a)$
  • Change in Time:     $\Delta t = b - a$
  • Average Velocity Formula:     $\text{Average Velocity} = \frac{\Delta s}{\Delta t} = \frac{s(b) - s(a)}{b - a}$

• Geometric Interpretation via Secant Line Slope:

  • Plot the position function $s(t)$ on a Cartesian coordinate system with time $t$ on the horizontal axis and position $s$ on the vertical axis.
  • Identify two points on the graph corresponding to the boundaries of time interval $[a, b]$:
  • First point: $(a, s(a))$
  • Second point: $(b, s(b))$
  • Applying the standard slope formula (change in $y$ over change in $x$) between these two ordered pairs yields:     $\text{Slope} = \frac{s(b) - s(a)}{b - a}$
  • The average velocity over time interval $[a, b]$ equals the slope of the secant line passing through points $(a, s(a))$ and $(b, s(b))$.

Instantaneous Velocity and Tangent Lines

• Distinction Between Velocity Types:

  • Average Velocity: Represents average rate of motion across a extended time interval $[a, b]$.
  • Instantaneous Velocity: Describes exact velocity at a single specific instant in time $t = a$

• Approximation via Interval Reduction:

  1. Select a small time interval containing time instant $a$.
  2. Calculate average velocity across this initial interval to obtain a rough estimate.
  3. Refine the approximation by choosing progressively smaller time intervals surrounding $a$.
  4. As interval width shrinks toward zero, the secant line rotates and approaches the tangent line.
  5. Instantaneous velocity at $t = a$ equals the slope of the tangent line to position function $s(t)$ at point $(a, s(a))$.

Questions and Discussion: Theoretical Concept of Instantaneous Change

• Theoretical Paradox:

  • Question: How can change occur at a single instant of time? Velocity implies movement, which requires elapsed time. At an isolated instant, how can change exist?

• Metaphor and Clarification:

  • Vehicle Speedometer Metaphor: When driving a car and viewing the speedometer at a specific moment, reading $60\,\text{miles per hour}$ measures speed at that exact instant.
  • If speed remains constant over the subsequent hour, distance traveled will be $60\,\text{miles}$.
  • Velocity continuously fluctuates, yet at every single instant, an object possesses an instantaneous speed due to ongoing motion.

Physics Application: Free-Falling Object (Galileo's Law)

• Galileo's Law of Falling Bodies:

  • If an object is dropped and falls freely under gravity (ignoring air resistance), distance fallen is directly proportional to time squared:     $s(t) = 4.9 t^2$
  • Units: Time $t$ in seconds ($\text{s}$), position $s(t)$ in meters ($\text{m}$).
  • Scenario: Object dropped from top of a $450\,\text{m}$ building.

• Problem Setup:

  • Determine instantaneous velocity at time $t = 5\,\text{s}$.
  • Direct derivative calculations are avoided; numerical approximation via shrinking intervals is applied.

• Sequential Numerical Approximations:

  • Interval 1: $[5, 5.1]$

  • Ending position: $s(5.1) = 4.9 \times 5.1^2 = 127.449\,\text{m}$

  • Beginning position: $s(5) = 4.9 \times 5^2 = 122.5\,\text{m}$

  • Average velocity:       $\text{Average Velocity} = \frac{s(5.1) - s(5)}{5.1 - 5} = \frac{4.9 \times 5.1^2 - 4.9 \times 5^2}{0.1} = \frac{127.449 - 122.5}{0.1} = 49.49\,\text{m/s}$

  • Interval 2: $[5, 5.05]$

  • Right endpoint moved closer to $5$

  • Average velocity:       $\text{Average Velocity} = \frac{s(5.05) - s(5)}{5.05 - 5} = \frac{4.9 \times 5.05^2 - 4.9 \times 5^2}{0.05} = \frac{124.99225 - 122.5}{0.05} = 49.245\,\text{m/s}$

  • Interval 3: $[5, 5.01]$

  • Right endpoint shifted to $5.01$

  • Average velocity = $49.049\,\text{m/s}$

  • Interval 4: $[5, 5.001]$

  • Right endpoint shifted to $5.001$

  • Average velocity = $49.0049\,\text{m/s}$

• Tabulated Trend Analysis:

  • Interval $[5, 5.1]$: $49.49\,\text{m/s}$
  • Interval $[5, 5.05]$: $49.245\,\text{m/s}$
  • Interval $[5, 5.01]$: $49.049\,\text{m/s}$
  • Interval $[5, 5.001]$: $49.0049\,\text{m/s}$
  • Convergence Target: As interval size approaches zero, average velocities converge toward $49\,\text{m/s}$.
  • Instantaneous velocity at $t = 5\,\text{s}$ equals $49\,\text{m/s}$.

Section 2.2: The Limit of a Function

• Theoretical Foundation:

  • Limit concept focuses on function behavior as input variable $x$ approaches target value $a$ without reaching $a$
  • Notation:     $\lim_{x \to a} f(x) = L$

• Example 1: Simple Continuous Linear Function

  • Function: $f(x) = 2x$
  • Target Input: $x \to 3$
  • Approaching $x = 3$ from the Left ($x < 3$):
  • $x = 2 \implies f(2) = 4$
  • $x = 2.5 \implies f(2.5) = 5$
  • $x = 2.9 \implies f(2.9) = 5.8$
  • $x = 2.99 \implies f(2.99) = 5.98$
  • $x = 2.999 \implies f(2.999) = 5.998$
  • Numerical trend approaches $6$
  • Approaching $x = 3$ from the Right ($x > 3$):
  • $x = 4 \implies f(4) = 8$
  • $x = 3.5 \implies f(3.5) = 7$
  • $x = 3.1 \implies f(3.1) = 6.2$
  • $x = 3.01 \implies f(3.01) = 6.02$
  • $x = 3.001 \implies f(3.001) = 6.002$
  • Numerical trend approaches $6$
  • Result:     $\lim_{x \to 3} (2x) = 6$
  • Geometric Visualization: Straight line graph containing points $(0,0)$, $(1,2)$, $(2,4)$, $(3,6)$, $(4,8)$, $(5,10)$. Moving along the line from either direction approaches point $(3,6)$.

• Example 2: Function Undefined at Target Point

  • Scenario: Function $f(x)$ where $f(2)$ does not exist (undefined due to division by zero or structural hole).
  • Evaluating limit as $x \to 2$ using numerical inputs:
  • $x = 2.5 \implies f(2.5) = 0.2222$
  • $x = 2.1 \implies f(2.1) = 0.244$
  • $x = 2.01 \implies f(2.01) = 0.249$
  • $x = 2.001 \implies f(2.001) = 0.2499$
  • Numerical Trend: Function values approach $0.25$
  • Limit Value:     $\lim_{x \to 2} f(x) = 0.25$
  • Key Principle: A limit can exist at $x = a$ even when $f(a)$ is undefined. Limits evaluate neighboring behavior, ignoring exact point values.

• Example 3: Discontinuous Function with Isolated Point

  • Scenario: Graph featuring an open circle at $(a, L)$ and a solid point at $(a, M)$.
  • Direct Function Evaluation:     $f(a) = M$
  • Limit Evaluation:     $\lim_{x \to a} f(x) = L$
  • Principle: Disconnections or isolated defined values do not affect limits. The limit remains $L$ as long as surrounding curves approach $L$

One-Sided Limits

• Right-Sided Limit:

  • Notation:     $\lim_{x \to a^+} f(x)$
  • Definition: Evaluates behavior of $f(x)$ as $x$ approaches $a$ exclusively from values strictly greater than $a$ (moving from right to left).

• Left-Sided Limit:

  • Notation:     $\lim_{x \to a^-} f(x)$
  • Definition: Evaluates behavior of $f(x)$ as $x$ approaches $a$ exclusively from values strictly less than $a$ (moving from left to right).

• Jump Discontinuities:

  • Functions containing vertical jumps at $x = a$ yield distinct left-sided and right-sided limits matching the respective boundaries of each branch.

Position Function: Let $s(t)$ show where an object is at any time $t$.

• Time Interval: Think of a time period between two points, $[a, b]$.
• Change in Position (Displacement): This tells us how much the position has changed from the start to the end of the time period: $riangle s = s(b) - s(a)$.
• Change in Time: $riangle t = b - a$.
• Average Velocity Formula: The average speed of the object during the time period is given by: $ext{Average Velocity} = \frac{ riangle s}{ riangle t} = \frac{s(b) - s(a)}{b - a}$.

Geometric Interpretation via Secant Line Slope:

• When we plot the position function $s(t)$ on a graph with time on the horizontal axis and position on the vertical axis, we can see how it moves.
• We mark the starting point at $(a, s(a))$ and the endpoint at $(b, s(b))$.
• The average velocity equals the slope (or steepness) of the line connecting these two points, which is calculated with: $ext{Slope} = \frac{s(b) - s(a)}{b - a}$.

Instantaneous Velocity and Tangent Lines:

• Average Velocity: This is how fast something moves over a period of time, $[a, b]$.
• Instantaneous Velocity: This is how fast something is moving at a specific moment in time, at $t = a$.
• To find instantaneous velocity, we look at smaller and smaller time intervals around $a$. As we make the intervals tiny, the average velocity over those intervals gives us the instantaneous speed at that moment.

Questions and Discussion:

• Theoretical Concept of Instantaneous Change:
  • How can something change at an exact moment? Typically, we think movement needs some time. Can there be change at a single instant?
  • Think of a car speedometer. When you check it, it may show $60 ext{ miles per hour}$, which tells you the speed right at that moment. If it stays that way, you'll travel $60 ext{ miles}$ in an hour. Although speed changes all the time, at every moment, the car has a speed due to its continuous motion.

Physics Application: Free-Falling Object (Galileo's Law):

• Galileo's Law of Falling Bodies: If an object falls freely without any air resistance, the distance it falls is related to the time squared: $s(t) = 4.9 t^2$.
• Example Problem: What is the speed of an object after it falls for $5 ext{ seconds}$? Instead of calculating directly, we can approximate by using smaller intervals.
  • Starting Interval: $[5, 5.1]$:
    • Find positions at these times:
    • $s(5.1) = 4.9 imes 5.1^2 = 127.449 ext{ m}$
    • $s(5) = 4.9 imes 5^2 = 122.5 ext{ m}$
    • Therefore, average velocity = $\frac{s(5.1) - s(5)}{5.1 - 5} = \frac{127.449 - 122.5}{0.1} = 49.49 ext{ m/s}$.
  • Repeat for smaller intervals, like $[5, 5.05]$, to get more precise velocity readings:
    • Average velocity = $49.245 ext{ m/s}$.
• As the intervals get very small, the average velocities all get closer to $49 ext{ m/s}$. That means the exact speed of the object at $t = 5 ext{ seconds}$ is approximately $49 ext{ m/s}$.