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 [t1,t2]
  • \text{AV} = \dfrac{\Delta s}{\Delta t}=\dfrac{s(t2)-s(t1)}{t2-t1}
  • 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.