3.6.1 Derivatives as Rates of Change

Conceptual Foundations

• Derivative as Rate of Change
  • The derivative measures how a quantity changes instantaneously.
  • Slope of the tangent line at a point on a graph.
• Average vs. Instantaneous Velocity
  • Average velocity: slope of the secant line between two points.
  • \text{Average velocity}=\dfrac{\Delta s}{\Delta t}=\dfrac{f(t2)-f(t1)}{t2-t1}
  • Instantaneous velocity: slope of the tangent line; numerically, v(t)=f'(t).

Example 1 – Patrol Car on an East–West Freeway

Scenario & Graph Interpretation

• Time axis: t in hours past noon.
• Position axis: s in miles (east = positive, west = negative).
• Motion description from the graph:
  • 0\le t\le1.5 h: position increases → car drives east, away from station.
  • 1.5\le t\le2.0 h: position constant → car stops.
  • 2.0\le t\le3.5 h: position decreases toward, then past, station → car heads west.

(B) Displacement & Average Velocity, 2:00 PM – 3:30 PM

• Positions: f(2)=80\text{ mi},\;f(3.5)=-20\text{ mi}.
• Displacement: \Delta s=f(3.5)-f(2)=(-20)-80=-100\text{ mi}
  • Negative sign ⇒ 100 mi toward the west.
• Average velocity: \bar v=\dfrac{-100}{3.5-2}=\dfrac{-100}{1.5}\approx-66.7\text{ mph}
  • Negative indicates westbound travel.

(C) Greatest Eastbound Instantaneous Velocity

• Relevant only while heading east (0\le t\le1.5 h).
• Steepest tangent occurs on steepest segment of graph: roughly 0.5\text{ h}\le t\le1.0\text{ h} (12:30 – 1:00 PM).

Formal Definitions

• Velocity: v(t)=\dfrac{ds}{dt}=s'(t) (direction-sensitive).
• Speed: |v(t)| (direction ignored).
• Acceleration: a(t)=\dfrac{d^2s}{dt^2}=v'(t).

Example 2 – Horizontal Motion Along a Line

Position function (feet): s(t)=t^2-5t,\;0\le t\le5, with rightward motion positive.

(A) Velocity Analysis

• Velocity: v(t)=\dfrac{d}{dt}(t^2-5t)=2t-5.
• Stationary points: solve v(t)=0\Rightarrow2t-5=0\Rightarrow t=2.5\text{ s}.
• Sign chart or graph shows:
  • 0<t<2.5: v(t)<0 → moving left.
  • t=0,\,2.5,\,5 s: stationary.
  • 2.5

(B) Acceleration Graph

• Acceleration: a(t)=\dfrac{d}{dt}(2t-5)=2 (constant, positive).
• Graph: horizontal line at a=2 from t=0 to t=5 s.

(C) Narrative of Motion

• t=0: object at origin.
• 0\rightarrow2.5 s: moves left, slows (speed ↓ because v(t)\to0), constant rightward acceleration.
• t=2.5 s: momentarily at rest, furthest left.
• 2.5\rightarrow5 s: moves right, speed ↑ (accel = 2 ft/s²), returns to origin at t=5 s.

Example 3 – Stone Thrown Vertically Upward

• Initial conditions: launched from a bridge 96 ft above river with initial velocity 64 ft/s.
• Position (feet) via Newton’s laws: s(t)=-16t^{2}+64t+96.

(A) Velocity & Acceleration Functions

• Velocity: v(t)=\dfrac{d}{dt}(-16t^{2}+64t+96)=-32t+64.
• Acceleration: a(t)=\dfrac{d}{dt}(-32t+64)=-32\text{ ft/s}^2 (constant downward, Earth’s gravity with sign convention).

(B) Highest Point Above River

1. Set velocity to zero (top of flight):
   -32t+64=0\Rightarrow t=2\text{ s}.
2. Height at t=2:
   s(2)=-16(2)^2+64(2)+96=-16(4)+128+96=160\text{ ft}.

• Conclusion: max height 160 ft, reached at 2 s.

(C) Impact Velocity at River Level

1. Determine impact time by solving s(t)=0:
   -16t^{2}+64t+96=0
   \Rightarrow-16(t^{2}-4t-6)=0
   Quadratic formula ⇒ positive root t\approx5.16\text{ s}.
2. Velocity at that time:
   v(5.16)=-32(5.16)+64\approx-101.12\text{ ft/s}.

• Negative sign: downward motion toward river.

Overarching Insights & Connections

• Using signs is crucial: positive/negative distinguish directions (east/west, left/right, up/down).
• Average quantities summarize over intervals; derivatives give instant snapshots.
• Constant acceleration (gravity or linear motion example) yields quadratic position functions.
• Turning points (max/min height or furthest displacement) occur where velocity = 0.
• Speed is magnitude only: useful for questions ignoring direction (e.g.
  “how fast?” vs. “which way?”).
• Graphical analysis (slopes, regions above/below axis) complements algebraic derivatives.