Exponential Growth, Rule of 70, and Power Functions - Notes

  • Policy and context for homework and technical issues

    • Over the semester you can skip up to 20% of problems. Reason: flexibility and to account for technical issues in WileyPLUS.
    • Early on there were many issues with WileyPLUS (e.g., wrong minus signs like m-dash vs n-dash causing input mismatches).
    • If you encounter a technical issue with a submission, there is leeway in grading; you can skip a question if you’re confident your answer is correct and the entry issue is the cause.
    • The instructor emphasizes this is about practice, not punishment; it’s not about judging you for missing HW.
    • Some questions in the course are short answer; others require reading graphs. In some cases, you may not be able to tell what’s going on from the screen alone.
    • For questions that require exact answers (not approximations), you should avoid rounding too early because of error tolerances in grading.
    • The course content for the current material centers on linear and exponential functions; not all sections from Chapter 1 are covered in every assignment.
    • When solving homework problems, treat them like written homework: work on paper with a pencil to practice the process of solving, not just to get a right numeric result.
    • The instructor recommends solving problems on paper to prepare for exams, which will be on paper.
    • The instructor encourages asking questions if unsure about how many decimals to use or whether decimals or exact forms are preferred.

  • Exponential growth models and the Rule of 70

    • Exponential growth has the general form: f(t)=c ektf(t) = c \, e^{k t} or, equivalently, f(t)=p<em>0 ektf(t) = p<em>0 \, e^{k t} where $k$ is the continuous growth rate and $p0$ is the initial amount.
    • If compounding is annual (discrete), the form is f(t)=p0 (1+r)tf(t) = p_0 \, (1 + r)^{t} where $r$ is the annual interest rate (as a decimal).
    • Relationship between the continuous rate $k$ and the discrete rate $r$: k=ln⁡(1+r)k = \ln(1 + r) and conversely 1+r=ek.1 + r = e^{k}.
    • Doubling time (Rule of 70): for continuous growth with rate $k$, the time to double is
      T2=ln⁡2k≈0.693k.T_{2} = \frac{\ln 2}{k} \approx \frac{0.693}{k}.
    • For a given annual percentage rate $p$ (as a percent), the approximate doubling time is
      T2≈70p.T_{2} \approx \frac{70}{p}.
    • Intuition: when the rate is small, $\ln(1+r) \approx r$, so the continuous and discrete models are close to each other.
    • The natural log of $2$ is about $0.693$, which underpins the Rule of 70 approximation.
    • Historical note: interest and compound growth have been around for millennia; the rule of 70 is a practical approximation, not a fundamental law.

  • Derivations and interpretations of exponential growth

    • From f(t)=p<em>0  ektf(t) = p<em>0 \; e^{k t}, taking natural logs yields a linear relationship: ln⁡f(t)=ln⁡p</em>0+kt ,\ln f(t) = \ln p</em>0 + k t\,, i.e., a line with slope $k$ in a plot of $\ln f$ vs. $t$.
    • For discrete compounding, f(t)=p<em>0(1+r)tf(t) = p<em>0 (1+r)^t leads to ln⁡f(t)=ln⁡p</em>0+t  ln⁡(1+r)\ln f(t) = \ln p</em>0 + t \; \ln(1+r), again linear in $t$ with slope $\ln(1+r)$.
    • Practically, to find $k$ or $p_0$, use two data points and eliminate unknowns by dividing or subtracting equations; example below.
    • When solving problems numerically, be mindful that early approximations (e.g., using decimals) can propagate errors; sometimes it’s better to retain exact expressions until the end.

  • Worked example: two data points, project to target value

    • Given: after 2 years, value is 5000=p<em>0e2k5000 = p<em>0 e^{2k} and after 5 years, value is 7000=p</em>0e5k7000 = p</em>0 e^{5k}.
    • Step 1 (eliminate $p_0$): divide the two equations to get
      e3k=70005000=75.e^{3k} = \frac{7000}{5000} = \frac{7}{5}.
      Therefore, k=13  ln⁡(75).k = \frac{1}{3} \; \ln\left(\frac{7}{5}\right).
    • Step 2 (solve for $p0$): from $5000 = p0 e^{2k}$, we get
      p0=5000e2k.p_0 = \frac{5000}{e^{2k}}.
    • Step 3 (find $t$ such that $f(t) = 10000$): solve
      10000=p<em>0ekt⟹ekt=10000p</em>0.10000 = p<em>0 e^{k t} \Longrightarrow e^{k t} = \frac{10000}{p</em>0}.
      Hence, t=1k  ln⁡(10000p0).t = \frac{1}{k} \; \ln\left(\frac{10000}{p_0}\right).
    • Step 4 (numerical approximations):
    • $k = \frac{1}{3} \ln\left(\frac{7}{5}\right) \approx 0.112$ (roughly).
    • $p_0 = 5000 / e^{2k} \approx 5000 / e^{0.224} \approx 5000 / 1.251 \approx 3997$.
    • $t \approx \frac{1}{0.112} \ln\left(\frac{10000}{3997}\right) \approx \frac{1}{0.112} \ln(2.502) \approx \frac{0.916}{0.112} \approx 8.2$ years.
    • Important practical note: using decimals early can accumulate errors; when grading, exact fractions or more decimals may be preferable to prevent rounding errors.
    • Instructor stance on grading: for a test, students typically receive credit for correct process even if some arithmetic is off; small arithmetic mistakes are tolerated unless they undermine the problem’s difficulty.
    • Key takeaway: to solve exponential problems, keep track of the two unknowns ($p_0$, $k$) with two data points, then solve for the desired time when the target value is reached.

  • Precision and forms: decimals vs exact forms

    • You can present answers as decimals or exact expressions; both are often acceptable, but be mindful of error tolerances in the grading scheme.
    • If an exact expression (e.g., a rational log value) is available, using it can reduce cumulative numerical error.
    • For interpreting rates: avoid rounding to too few decimals when the rate is itself small or when precision matters (e.g., a few tenths or hundredths of a percent).
    • When discussing growth rates, continuous rate $k$ and annual rate $r$ can be close when $r$ is small; the difference grows with larger $r$.

  • Logarithms, plots, and the base-change idea

    • Exponential form to log form:
    • If f(t)=c  ektf(t) = c \; e^{k t}, then ln⁡f(t)=ln⁡c+kt\ln f(t) = \ln c + k t, which is linear in $t$. The slope of this line is $k$.
    • On a log plot (log of $y$ vs $x$ with base $e$), you see a straight line for an exponential model with slope $k$.
    • Desmos caveat: plotting a log plot with base 10 on some tools introduces a constant factor because log⁡10y=ln⁡yln⁡10.\log_{10} y = \frac{\ln y}{\ln 10}.
    • Change-of-base formula: for any $b>0$, log⁡by=ln⁡yln⁡b.\log_b y = \frac{\ln y}{\ln b}. This means slopes on different log bases are proportional by a constant factor, not a fundamental difference in the linearity.
    • Power functions: For f(x)=c  xpf(x) = c \; x^{p}, taking logs gives
      ln⁡f(x)=ln⁡c+p  ln⁡x\ln f(x) = \ln c + p \; \ln x, which is linear in $\ln x$ (a log-log plot).
    • On a log-log plot, a power function appears as a straight line with slope $p$; this is a central way to identify power-law behavior in data.
    • Real-world note: power laws appear in natural phenomena (e.g., metabolism roughly scales with mass to the 0.75 power).

  • Power functions: definition, behavior, and domain considerations

    • General form: f(x)=C xpf(x) = C \, x^{p} where $C$ is a constant and $p$ is the power.
    • Important distinction: exponential vs power.
    • Exponential: base is constant, exponent is variable: f(x)=C  axf(x) = C \; a^{x}.
    • Power: base is variable, exponent is constant: f(x)=C  xpf(x) = C \; x^{p}.
    • What is a power function? Includes integer and non-integer powers, negative powers, and fractional powers such as x−1=frac1xx^{-1} = frac{1}{x} and x1/2=xx^{1/2} = \sqrt{x}.
    • Domain considerations:
    • For integer powers, the domain is all real numbers (e.g., x2,x3x^{2}, x^{3}).
    • For fractional or irrational powers, the natural domain is usually x≥0x \ge 0 unless specified otherwise.
    • For negative powers, the function is defined for all nonzero $x$ (e.g., x−1=1/xx^{-1} = 1/x) and typically has a vertical asymptote at $x = 0$.
    • Graphical shapes depending on $p$:
    • $p = 1$: straight line $y = C x$;
    • $p > 1$: grows faster than linear; becomes steeper as $x$ increases;
    • $0 < p < 1$: concave and increasing, but slower than linear;
    • $p < 0$: vertical asymptote at $x = 0$ and decay as $x \to \infty$; can be symmetric in certain cases (e.g., $x^{-1}$ is odd, not symmetric about the y-axis).
    • Visual intuition: frugal to think of power curves as “how fast $y$ grows as a power of $x$” with the steepness controlled by $p$.
    • If you include a constant $C$ in front, it stretches the graph vertically; changing the sign of $C$ reflects across the x-axis.
    • Practical caveat: when using fractional powers or negative powers, pick the domain carefully to avoid undefined values and to reflect the intended interpretation of the model.

  • Connections to problem-solving and visualization

    • The instructor emphasizes keeping a mental image of the curve when solving problems, especially for exponential and power functions.
    • When solving, you should be able to interpolate between known points and reason about growth rates from the slope of the log-transformed relationship.
    • The value of a problem often lies in recognizing the right transformed form (linear in a log scale) rather than blindly applying algebra.

  • Real-world relevance and broader implications

    • Exponential growth models underpin population dynamics, compound interest, and many natural processes; even small rates can lead to large effects over time.
    • The use of log plots and log-log plots is common in data analysis to linearize nonlinear relationships and to estimate growth exponents.
    • Understanding the difference between continuous compounding (exponential) and discrete compounding (power-like) informs finance, biology, and physics.
    • The idea of approximations (e.g., (\ln(1+r) \approx r) for small $r$) invites further study into error analysis, limits, and when approximations break down.

  • Practical exam-oriented tips and expectations

    • Practice solving problems on paper to mimic test conditions and to develop step-by-step reasoning.
    • For exact versus decimal answers, weigh the precision requirements of the task; use exact forms where appropriate to avoid cumulative rounding errors.
    • When presenting solutions, show the key steps (e.g., taking logs, isolating variables) rather than only the final number; partial credit is common if the method is correct but an arithmetic error occurs.
    • Be mindful of domain restrictions when dealing with power functions, especially with fractional or negative powers.

  • Quick reference formulas (summary)

    • Exponential with continuous growth: f(t)=c  ektf(t) = c \; e^{k t}, or f(t)=p0 ektf(t) = p_0 \, e^{k t}.
    • Discrete (annual) growth: f(t)=p0  (1+r)tf(t) = p_0 \; (1 + r)^t with k=ln⁡(1+r)k = \ln(1 + r).
    • Linear log form: ln⁡f(t)=ln⁡c+kt .\ln f(t) = \ln c + k t\,.
    • Doubling time (continuous): T<em>2=ln⁡2kT<em>{2} = \frac{\ln 2}{k}; note ln⁡2≈0.693,\ln 2 \approx 0.693, so typically T</em>2≈0.693kT</em>{2} \approx \frac{0.693}{k}.
    • Rule of 70 (heuristic for annual rate $p$): T2≈70p.T_{2} \approx \frac{70}{p}.
    • Exponential-to-log relation: for $f(t) = c e^{k t}$, on a plot of $\ln f(t)$ vs. $t$, the slope is $k$.
    • Power functions: f(x)=C  xpf(x) = C \; x^{p}; on a log-log plot, ln⁡f(x)=ln⁡C+p  ln⁡x\ln f(x) = \ln C + p \; \ln x; slope on that plot is $p$.
    • Change of base: log⁡by=ln⁡yln⁡b.\log_b y = \frac{\ln y}{\ln b}.
    • Domain caveats for powers: for integer $p$, domain is all real; for fractional or irrational $p$, domain typically $x \ge 0$ unless specified; negative $p$ yields vertical asymptote at $x=0$.