Notes for: Quizzes, Rates of Change, Derivatives, and Limits

Course Logistics and Quiz Structure

• No class on Monday; next meeting is in recitation, which is where the first standard quiz happens.
• On Tuesday, students take the first standard quiz.
• Quizzes are intended to be low-stakes: they assess whether core material is being learned and practiced, not whether you’re “good at math” or whether you’ll pass the course.
• A quiz indicates what you’ve understood from the past few days and what needs more practice.
• Each standard quiz is typically short, designed to take about
  ext{10--15 minutes}
  maximum.
• Quizzes usually contain about three questions related to a particular context. However, depending on the tentative calendar, there can be weeks with more than one quiz if multiple standards are covered in that week.
• You may have a few attempts to pass a quiz, so if you don’t pass on the first try, you’ll have additional tries.
• Quizzes are intended to align with homework and WebAssign problems: some WebAssign problems are designed to practice the standard, while others resemble quiz-type questions.
• You can review WebAssign lecture notes and additional practice problems at the end of each section to practice that standard.
• Examples for this time were provided in emails/announcements.
• Calculator policy: you may use a calculator on the quiz; buying a calculator is not required. LAs have an extra calculator available for anyone who doesn’t have one. If you bring your own, you may need to share it with others; you can use it to figure out what you need and then pass it on. The instructor designs exams so you generally do not need a calculator for them; quizzes may allow one.
• There are no exams in this course; the assessment structure revolves around quizzes and other questions.
• Recitation on Tuesday is a quiz session; students should bring the note/paper from today to recitation.
• If you have questions about logistics, you can ask the instructor in class.

Core Concepts Highlighted in Today’s Session

• Slope of the secant line represents the average rate of change of a function over a given interval.
• The average rate of change between two inputs a and b is rac{f(b) - f(a)}{b - a}.
• The average rate of change tells you how much the output changes per one unit of input over that interval.
• As intervals shrink (e.g., from [0,10] to [9,10]), the average rate of change can approximate the instantaneous rate of change at a point.
• The instantaneous rate of change at a point is the limit of the average rate of change as the interval around the point shrinks to zero. This leads to the concept of the derivative.
• The derivative at a point gives the slope of the tangent line at that point: f'(x0) = ext{Slope of the tangent line at } x0.
• The derivative is defined as a limit:
  f'(x) = oxed{
  abla ext{or}
  abla rac{f(x+h) - f(x)}{h} ext{ as } h o 0}
• The tangent line slope (instantaneous rate) is the limit of secant slopes as the second point approaches the first: the tangent line is the limiting position of secant lines.
• Limits (Standard 2) connect the behavior of a function to its limits and vice versa; the limit describes where the function is headed as x grows large or approaches a point.
• The mean value theorem and the intermediate value theorem are foundational theorems that will be covered later and underpin the ideas of approximating derivatives via limits.

Worked BAC (Blood Alcohol Concentration) Example: Average Rate of Change

• Context: We studied a problem about BAC and computed the average rate of change between time points to understand how BAC changes per minute.
• Let C(t) denote BAC, with t in minutes. The average rate of change over an interval [a, b] is:
  ext{ARC}_{[a,b]} = rac{C(b) - C(a)}{b - a}.
• Example discussion points:
  • The instructor reminded students that the average rate of change tells you how much the BAC is changing per minute on average over that interval.
  • Between 0 and 10 minutes, ARC gives a broad view of how BAC changes in the first ten minutes.
  • Between 9 and 10 minutes, ARC is a better, more localized estimate of the instantaneous rate of change at t = 10 than ARC over [0,10].
  • This leads toward the concept of the derivative as an instantaneous rate of change at a point as the interval shrinks.
• Practical interpretation of units:
  • Outputs (BAC) are typically measured in ext{mg/mL} (often written as per mL) and inputs are time in minutes.
  • The units of ARC are rac{ ext{mg/mL}}{ ext{min}} = ext{mg} ig/ ( ext{mL} imes ext{min}).
  • In the discussed calculation, the ARC between 9 and 10 minutes was approximately 0.008 ext{ mg/(mL·min)}.
• Calculation example (outline rather than exact numbers):
  • To compute ARC{[9,10]}, evaluate the BAC at t = 9 and t = 10: C(9), C(10). Then compute ext{ARC}{[9,10]} = rac{C(10) - C(9)}{10 - 9}.
  • The instructor walked through evaluating C(10) and C(9) by either plugging into a given model (e.g., a formula involving exponentials) or reading values from a graph. An example given was:
    C(10) = 0.0225 imes 10 imes e^{-0.467 imes 10},
    with a note that e is the base of natural logarithms and can be entered via a calculator.
  • After computing the outputs, the ARC is computed and interpreted as the BAC changing by that amount per minute on the interval [9,10].
• A crucial observation:
  • Using smaller intervals generally yields better approximations to the instantaneous rate of change at a specific time (e.g., t = 10). This motivates the move from secant lines to tangent lines and to derivatives.

Concrete Problem Walk-Through: A Context-Free Function (WebAssign Style)

• The instructor introduced a context-free problem to illustrate secants and tangents.
• The curve given: a point (3, -3) lies on the curve f(x) =
  f(x) = rac{3}{2 - x}.
• Example points:
  • For x = 2.1, the output is f(2.1) = -30 (since 3/(2 - 2.1) = 3/(-0.1) = -30).
  • The average rate of change between blue point (2.1, -30) and yellow point (3, -3) is:
    ext{ARC}_{[2.1, 3]} = rac{f(3) - f(2.1)}{3 - 2.1} = rac{-3 - (-30)}{0.9} = rac{27}{0.9} = 30.
  • The green point is another point on the curve; the instructor states the ARC between the green point and the yellow point is 6, illustrating how steeper vs less steep secants relate to larger vs smaller ARC values.
• Tangent (instantaneous rate) approximation near x = 3:
  • Slopes of secant lines approaching x = 3 were given as approximations: about 3.3 repeating and about 2.72 (to illustrate narrowing around the tangent slope).
  • The slope of the tangent line at x = 3 is approximately 3 (more precisely, it lies between those secant slopes).
• Constructing the tangent line:
  • Use the known point (3, -3) and slope m ≈ 3 to write:
  • Point-slope form: y - (-3) = 3igl(x - 3igr).
  • Slope-intercept form: rearranging gives y = 3x + b. Using (3, -3): -3 = 3(3) + b \,\Rightarrow\ b = -12. Thus:
    y = 3x - 12.
• Conceptual takeaway:
  • The derivative at x = 3 is the limit of the slopes of secant lines as the second point approaches x = 3; the tangent line slope is the instantaneous rate of change at that point.
  • This example motivates the idea of limits as the foundation for derivatives.

Standard 2: Limits and Graphs

• The instructor introduced a matching activity to connect limit statements with graphs.
• Example description (one of the descriptions):
  • "The values of f(x) can be made as close to -1 as x goes to infinity" meaning the function’s outputs approach -1 as x → ∞.
• Graphs were labeled 1, 2, 3, 4, etc. The task was to match each description to the graph that exhibits that limit behavior.
• Observations for matching:
  • The graph that shows the function’s output approaching -1 as x increases without bound corresponds to that description.
  • Other graphs with outputs tending to +1, or tending to -∞, do not describe the given limit behavior.
• The goal of this activity is to deepen intuition about limits and to connect verbal descriptions to visual behavior on graphs.
• The instructor notes that a given description can describe more than one graph and a single graph can fit more than one description; this requires careful interpretation of the limit statements.

Connections, Theorems, and Study Notes for Exam Preparation

• Key definitions:

  • Average rate of change over [a, b]: ext{ARC}_{[a,b]} = rac{f(b) - f(a)}{b - a}.
  • Instantaneous rate of change at x: f'(x) = oxed{rac{d}{dx}f(x)} = ext{the limit of the average rate of change as } h o 0: f'(x) =

  \lim_{h \to 0} rac{f(x+h) - f(x)}{h}.

  • Tangent line at x0 has slope f'(x0) and passes through (x0, f(x0)). Its equation can be written as y = f'(x0)(x - x0) + f(x_0) or in slope-intercept form once simplified.

• Conceptual connections:

  • The process of using shorter intervals to approximate the instantaneous rate of change is a concrete application of the limit concept.
  • Secant lines provide approximations to the tangent line; as the second point approaches the first, the secant line approaches the tangent line.
  • The derivative is a local property of the function, describing its behavior in an infinitesimally small neighborhood around a point.

• Practical exam tips:

  • Always identify what is changing (the output) and what is being held constant (the input) to determine the correct units for the rate of change.
  • When interpreting a rate of change in context, translate the units back into meaningful quantities (e.g., BAC per minute).
  • Distinguish between the function value (an output) and the rate of change (a ratio of changes in output to changes in input).
  • For limit-based questions, be clear about what x is approaching (∞, a finite point, or a specific value) and what that implies for the outputs.

• Important equations to memorize:

  • ext{ARC}_{[a,b]} = rac{f(b) - f(a)}{b - a}
  • f'(x) = oxed{rac{d}{dx}f(x)} = ext{limit of arc as } h o 0: f'(x) = \lim_{h o 0} rac{f(x+h) - f(x)}{h}
  • For a line through a given point with slope m: y = mx + b;\ b = y0 - m x0 ext{ (use a known point } (x0,y0) ext{)}.

• Example recap: a function f(x) and point (3, -3) on it:

  • If a secant slope near x = 3 is approximated by various nearby points, the sequence of slopes converges to the tangent slope at x = 3.
  • The tangent line found through (3, -3) with slope m ≈ 3 is y = 3x - 12.

Quick Practice Prompts for Self-Study

• Compute the average rate of change for a provided function over a specified interval and interpret the units.
• For a given function, identify a point where the secant slope is steeper versus flatter and explain what that says about the function’s rate of change over that interval.
• Given a graph, determine which description corresponds to a limit as x → ∞ and justify your choice with a short explanation.
• Practice deriving the tangent line equation from a point and a nearby slope approximation, then compare with the exact derivative if known.
• Distinguish between a function’s value at a point and its instantaneous rate of change at that point in context-rich problems (e.g., BAC, population growth, chemical concentration).

Notation and Quick References (LaTeX)

• Average rate of change: ext{ARC}_{[a,b]} = rac{f(b) - f(a)}{b - a}
• Instantaneous rate of change (derivative): f'(x) = oxed{rac{d}{dx}f(x)} = \lim_{h \to 0} rac{f(x+h) - f(x)}{h}
• Tangent line at x = x0: y = f'(x0)(x - x0) + f(x0) or equivalently y = m x + b,\ b = f(x0) - f'(x0)x_0.