3.6.2 Derivatives as Rates of Change

Growth Models – Using Derivatives as Rates of Change

Key Ideas

  • “Growth model” ⇢ any function that represents how a quantity (population, revenue, users, etc.) increases or decreases with time.
  • Derivative of the model gives an instantaneous growth rate (slope of the tangent).
  • Slope of the secant line between two points gives an average growth rate over an interval.

Internet-User Example (2000–2015)

  • Data set: Worldwide Internet users, measured every February.
  • Best-fit polynomial provided by instructor (exact constants not written in transcript).
    – Let tt = years after February 2000 (so 2005 ⇢ t=5t = 5, 2010 ⇢ t=10t = 10, 2011 ⇢ t=11t = 11, 2015 ⇢ t=15t = 15, 2020 ⇢ t=20t = 20).
    – Model: P(t)P(t) (millions of users).
    – Instructor’s computed derivative: dPdt=12t+98\frac{dP}{dt}=12t+98.
A. Average growth rate (2005 → 2010)
  • Formula: Average rate=P(10)−P(5)10−5.\text{Average rate}=\frac{P(10)-P(5)}{10-5}.
  • Given values:
    P(10)=2,011.2 million,P(5)=1,071.2 million.P(10)=2{,}011.2\text{ million},\qquad P(5)=1{,}071.2\text{ million}.
  • Calculation:
    2,011.2−1,071.25=188  million users⋅year−1.\frac{2{,}011.2-1{,}071.2}{5}=188\;\text{million users·year}^{-1}.
  • Interpretation: During 2005-2010, world Internet adoption grew by ≈188 million new users each year (on average).
B. Instantaneous growth rate (February 2011)
  • Need the derivative value at t=11t=11:
    P′(11)=12(11)+98=230.P'(11)=12(11)+98 = 230.
  • Units: million users per year.
    → In Feb 2011, the user base was expanding at roughly 230 million per year at that instant.
C. Shape of the growth-rate graph (P′(t)P'(t))
  • Graph of P′(t)=12t+98P'(t)=12t+98 is a straight line with positive slope 12.
    – Always positive ⇒ adoption never shrinks in 2000-2015 window.
    – Increasing slope ⇒ rate of adoption is itself accelerating each year.
D. Extrapolation beyond 2015 (to 2020)
  • Instructor assumes same quadratic model valid to t=20t=20.
    – “Plug in 20” gives P(20)≈7,700 millionP(20)\approx7{,}700\text{ million} users (≈ worldwide population).
    – Caveat: Extrapolating a fitted quadratic many years past data can be unreliable, but this demonstrates the method.

Cost Analysis – Average vs. Marginal Cost

Setup

  • Cost function C(x)C(x) = total dollars to manufacture the first xx items.
    (Transcript gives) C(x)=−0.2x2+50x+100.C(x) = -0.2x^{2}+50x+100.
Definitions
  • Average cost per item: Cˉ(x)=C(x)x.\bar{C}(x)=\frac{C(x)}{x}.
  • Marginal cost (approximate cost of one additional item after the first xx): C′(x).C'(x).
Derivations
  1. Average cost function:
    Cˉ(x)=−0.2x2+50x+100x=−0.2x+50+100x.\bar{C}(x)=\frac{-0.2x^{2}+50x+100}{x}= -0.2x + 50 + \frac{100}{x}.
  2. Marginal cost function (take the derivative):
    C′(x)=ddx(−0.2x2+50x+100)=−0.4x+50.C'(x)=\frac{d}{dx}\bigl(-0.2x^{2}+50x+100\bigr)=-0.4x+50.

Numerical Evaluations & Interpretations

Production levelAverage cost Cˉ(x)\bar{C}(x)Marginal cost C′(x)C'(x)Interpretation
x=100x=100Cˉ(100)=49  $/item\bar{C}(100)=49\;\$\text{/item}C′(100)=46  $/itemC'(100)=46\;\$\text{/item}On the 100-item batch, each unit has cost $49 on average; making the 101st unit would cost ≈$46 more.
x=900x=900Cˉ(900)=32  $/item\bar{C}(900)=32\;\$\text{/item}C′(900)=14  $/itemC'(900)=14\;\$\text{/item}At 900 units, economies of scale drop average cost to $32; the 901st unit would cost only about $14 additional.
Observations
  • Cˉ(x)\bar{C}(x) decreases as xx grows ⇒ strong economies of scale.
  • C′(x)C'(x) is linear & decreasing; eventually marginal cost could become negative (mathematical artifact of the quadratic model, not realistic). Practical takeaway: producing more lowers incremental expense up to some physical capacity limit.

Conceptual Connections & Practical Significance

  • Derivatives link change to value: population growth, production cost, spread of technology, etc.
  • Secant-line slopes (average rates) guide broad planning; tangent slopes (instantaneous rates) guide fine-tuned decisions and forecasting.
  • For businesses, marginal-cost curves inform pricing, capacity expansion, and profit-maximizing output.
  • In demographic studies, a positive and increasing P′(t)P'(t) warns of accelerating demand on infrastructure, market size, and policy needs.

Equations & Formulas Recap

  • Average growth rate over [t<em>1,t</em>2][t<em>1,t</em>2]:
    AGR=P(t<em>2)−P(t</em>1)t<em>2−t</em>1.\text{AGR}=\frac{P(t<em>2)-P(t</em>1)}{t<em>2-t</em>1}.
  • Instantaneous growth rate: P′(t).P'(t).
  • Average cost: Cˉ(x)=C(x)x.\bar{C}(x)=\frac{C(x)}{x}.
  • Marginal cost: C′(x)=dCdx.C'(x)=\frac{dC}{dx}.
  • Quadratic marginal-cost example: C′(x)=−0.4x+50.C'(x)=-0.4x+50. (Slope −0.4 represents $0.40 cost reduction per extra unit produced.)

Ethical & Real-World Implications Discussed

  • Technological diffusion: rapid, accelerating Internet growth implies issues of digital equity, infrastructure, and global access.
  • Economic scaling: decreasing marginal cost encourages mass production, which can boost accessibility but also raises sustainability and labor-practice concerns if not managed responsibly.

Quick Study Checklist

  • Can you compute both average & instantaneous rates of change from a numeric table or formula?
  • Given C(x)C(x), can you derive Cˉ(x)\bar{C}(x) and C′(x)C'(x), then interpret them economically?
  • Do you recognize the limitations of extrapolating a fitted model far beyond its data range?