3.6.1 Derivatives as Rates of Change

Review: Derivative as a Rate of Change

• The derivative measures the instantaneous rate at which a quantity changes.
• Geometric meaning:
  • Slope of the tangent line to a curve at a specific point.
• Practical meaning in kinematics:
  • Instantaneous velocity of an object whose position is given as a function of time.
• Reminder from Chapter 3: average rate of change ≈ slope of a secant; instantaneous rate of change = slope of the tangent → the derivative.

Average vs. Instantaneous Velocity

• Average velocity on $[t<em>1,t</em>2]$
  • $\text{AV} = \dfrac{\Delta s}{\Delta t}=\dfrac{s(t<em>2)-s(t</em>1)}{t<em>2-t</em>1}$
  • Represents overall change over an interval (slope of secant).
• Instantaneous velocity at $t=a$
  • $v(a)=s'(a)=\lim_{\Delta t\to0}\dfrac{s(a+\Delta t)-s(a)}{\Delta t}$
  • Slope of tangent; gives speed and direction at that instant.

Example 1 – Patrol Car on an East–West Freeway

Scenario & Qualitative Graph Analysis

• Time ($t$) in hours after noon on the $x$-axis; position $s$ (miles from station) on the $y$-axis.
• Motion inferred from graph:
  1. $0\le t \le1.5$ h — position increasing, car drives east, away from station.
  2. 1.5< t \le2 h — position flat, car is stopped.
  3. 2< t \le3.5 h — position decreases past $s=0$, car returns, crosses station, continues west.

Displacement & Average Velocity on $[2,3.5]$ ( 2 : 00 pm → 3 : 30 pm )

• Positions read from graph:
  • $s(2)=80$ mi, $s(3.5)=-20$ mi.
• Displacement: $\Delta s=s(3.5)-s(2)=-20-80=-100\text{ mi}$ (negative ⇒ west).
• Average velocity:
  $\dfrac{-100}{3.5-2}=\dfrac{-100}{1.5}\approx-66.7\text{ mi/h}$ (66.7 mph toward west).

Greatest Instantaneous Velocity while Traveling East

• Consider only the segment where $s$ is increasing (east-bound): 0<t<1.5 h.
• Steepest portion (largest positive slope) visually occurs about 0.5<t<1 h (≈ 12 : 30 pm – 1 : 00 pm).

Velocity, Speed, Acceleration – Formal Definitions

• Velocity: $v(t)=s'(t)$ (signed; indicates direction).
• Speed: $|v(t)|$ (non-negative; ignores direction).
• Acceleration: $a(t)=v'(t)=s''(t)$ (derivative of velocity; second derivative of position).

Example 2 – Horizontal Motion on a Line

Given position $s(t)=t^2-5t$ (ft), $0\le t\le5$ s.

Velocity Function and Direction of Motion

• $v(t)=s'(t)=2t-5$.
• Stationary ( $v=0$ ): $t=\dfrac{5}{2}=2.5\,\text{s}$ (plus endpoints 0 & 5 assumed by context).
• Moving left (negative $v$): 0< t<2.5.
• Moving right (positive $v$): 2.5< t<5.

Acceleration

• $a(t)=v'(t)=s''(t)=2\,\text{ft/s}^2$ (constant, always positive → rightward acceleration).
• Graph: horizontal line $a=2$ from $t=0$ to $5$.

Overall Description of Motion

• $t=0$ s: starts at origin.
• 0<t<2.5 s: slides left (negative $s$), slowing down (speed ↓ because $v$ approaches 0).
• $t=2.5$ s: momentarily at rest.
• 2.5<t<5 s: moves right, speeding up, returns to $s=0$ at $t=5$ s.
• Constant rightward acceleration explains decreasing then increasing speed.

Example 3 – Stone Thrown Vertically Upward

Initial conditions:

• Initial velocity $v_0=64\,\text{ft/s}$ upward.
• Launch point $96\,\text{ft}$ above water.
• Position model (Newton’s laws):
  $s(t)=-16t^2+64t+96$.

Velocity & Acceleration

• $v(t)=s'(t)=-32t+64$.
• $a(t)=v'(t)=s''(t)=-32\,\text{ft/s}^2$ (constant downward gravitational acceleration).

Highest Point

• Occurs when $v(t)=0$ ⇒ $-32t+64=0 \Rightarrow t=2\,\text{s}$.
• Height at that time:
  $s(2)=-16(2)^2+64(2)+96=-64+128+96=160\,\text{ft}.$
  → The stone peaks $160\,\text{ft}$ above water at $t=2\,\text{s}$.

Velocity on Impact with Water

1. Find impact time by solving $s(t)=0$:
   $-16t^2+64t+96=0$
   $t=\dfrac{-64\pm\sqrt{64^2-4(-16)(96)}}{2(-16)}$
   Positive root ⇒ $t\approx5.16\,\text{s}$.
2. Impact velocity:
   $v(5.16)= -32(5.16)+64\approx-101.12\,\text{ft/s}$ (negative ⇒ downward).

Connections, Significance & Real-World Context

• Police-car problem shows how secant/tangent slopes translate to traffic speed and direction analysis.
• Horizontal‐motion example highlights how sign changes in $v(t)$ correspond to direction reversals and how a constant, non-zero acceleration produces non-uniform velocity.
• Vertical‐throw scenario applies Newton’s laws: constant gravity leads to quadratic position; derivatives quickly yield apex time & impact velocity without graphing.
• Ethical / practical aspects:
  • Accurate velocity estimation critical for law enforcement & safety (e.g.
    speed enforcement, pursuit tactics).
  • Understanding projectile motion under gravity informs bridge design, ballistics, and safety measures near heights.
• Mathematical takeaway: Derivatives turn geometric slope ideas into precise tools for predicting and explaining motion.