4.3 Rates of Change in Applied Contexts other than Motion

First, identify the population model they give (a formula, table, or graph). Then choose the right rate concept: - Average rate of change over [a,b]: (P(b) − P(a)) / (b − a). Units: people per unit time. Use this for phrases like “on average over 5 years.” - Instantaneous rate of change at t (the population growth rate): P′(t). If they give P(t) explicitly, differentiate and evaluate. Interpret with units (e.g., people/year). This is CHA-3.C: use the derivative to interpret rate of change. Common models and their derivatives (use when given): - Exponential: P(t)=P0 e^{kt} ⇒ P′(t)=k P(t) (growth proportional to population). - Logistic: P(t)=L/(1+Ae^{−kt}) ⇒ P′(t)=k P(t) (1 − P(t)/L) (growth slows near carrying capacity L). On the AP: show work (write derivative or difference quotient), include units, and state the interpretation (increasing/decreasing, how fast). For examples and targeted practice, see the Topic 4.3 study guide (https://library.fiveable.me/ap-calculus/unit-4/rates-change-applied-contexts-other-than-motion/study-guide/ZaLyHm3IcLaORcqYsxxB), the Unit 4 overview (https://library.fiveable.me/ap-calculus/unit-4), and lots of practice problems (https://library.fiveable.me/practice/ap-calculus).

There isn’t one single “magic” formula—use implicit differentiation/chain rule on the relationship tying the quantities together. In practice: - If the quantities satisfy an equation F(x,y,…) = 0, differentiate w.r.t. time t: ∂F/∂x · dx/dt + ∂F/∂y · dy/dt + … = 0. - If z = f(x,y), use the multivariable chain rule: dz/dt = f_x(x,y)·dx/dt + f_y(x,y)·dy/dt. So pick the relation that models the context (volume, concentration, cost, revenue, population, etc.), differentiate every term with respect to t, then solve for the unknown rate (like dy/dt). Always keep units and interpret the result (CED CHA-3.C: interpret rates of change). This method covers non-motion related AP problems (marginal cost/revenue, concentration, production rates). For worked examples and extra practice, see the Topic 4.3 study guide (https://library.fiveable.me/ap-calculus/unit-4/rates-change-applied-contexts-other-than-motion/study-guide/ZaLyHm3IcLaORcqYsxxB) and more practice problems (https://library.fiveable.me/practice/ap-calculus).

Use regular division (a difference quotient) when you want an average rate over an interval: (change in output)/(change in input)—e.g., ΔQ/Δt gives the average production rate over that time. Units are important: “vehicles per hour” from total vehicles divided by hours. Use the derivative when you need an instantaneous or sensitivity rate—the limit of those difference quotients. f′(t) gives the instantaneous rate (how fast something is changing right now), which is what AP wants you to interpret in applied contexts (CHA-3.C, instantaneous rate of change, marginal cost/revenue, etc.). Use implicit differentiation and the chain rule for related-rates problems where multiple quantities change together. If you only have table data, approximate f′ with symmetric or forward differences (e.g., [f(t+h)−f(t−h)]/(2h)). On the exam, show units and interpret the meaning (e.g., “decreasing by 216 vehicles/hour per hour at t=5”). For more practice and examples, see the Topic 4.3 study guide (https://library.fiveable.me/ap-calculus/unit-4/rates-change-applied-contexts-other-than-motion/study-guide/ZaLyHm3IcLaORcqYsxxB) and the practice problems page (https://library.fiveable.me/practice/ap-calculus).

Short answer: usually yes—when the problem asks for an instantaneous rate of change or you need sensitivity (marginal cost, growth rate, concentration change, etc.), you take the derivative. But not always—sometimes you only need an average rate (use a difference quotient or an integral over an interval), or you use the derivative implicitly (related rates/implicit differentiation), or you combine derivative and algebra (local linearization) to answer the context question. How to decide quickly: - If the prompt says “instantaneous rate,” “rate at time t,” or asks for marginal something → take d/dt (or appropriate variable). - If it asks “average rate over [a,b]” → use (f(b)−f(a))/(b−a) or (1/(b−a))∫ f. - If quantities are linked (radius and height, concentration and volume) → write relation and differentiate implicitly (related rates). - Always interpret units and speak the meaning (CED CHA-3.C: interpret rates in context). For more examples and AP-style practice, see the Topic 4.3 study guide (https://library.fiveable.me/ap-calculus/unit-4/rates-change-applied-contexts-other-than-motion/study-guide/ZaLyHm3IcLaORcqYsxxB) and Unit 4 resources (https://library.fiveable.me/ap-calculus/unit-4). For lots of practice problems, use (https://library.fiveable.me/practice/ap-calculus).

Think of these problems as rate/accumulation pairs—the derivative gives the flow rate, and integrating gives total volume. If r(t) is the rate water flows in (units: e.g., liters/min), then - instantaneous rate: dV/dt = r(t) (CHA-3.C.1). - total volume from t=a to t=b: V(b) = V(a) + ∫_a^b r(t) dt. Always track units (liters/min × minutes = liters). If the tank’s shape matters (cone, cylinder, etc.), write V in terms of the changing geometric variable(s) (for a cone V = (1/3)πr^2h), then differentiate implicitly for related rates: - Example: V = (1/3)πr^2h → dV/dt = (1/3)π(2r dr/dt · h + r^2 dh/dt). Use chain rule/implicit differentiation to relate dr/dt, dh/dt, and dV/dt. On the AP: you’ll be asked to interpret dV/dt (instantaneous) vs average rate and to set up/solve integrals or related-rates equations (Topic 4.3 & related 4.4–4.5). For worked examples and practice, check the Topic 4.3 study guide (https://library.fiveable.me/ap-calculus/unit-4/rates-change-applied-contexts-other-than-motion/study-guide/ZaLyHm3IcLaORcqYsxxB), the Unit 4 overview (https://library.fiveable.me/ap-calculus/unit-4), and plenty of practice problems (https://library.fiveable.me/practice/ap-calculus).

Average rate of change = change in the quantity over an interval: (f(b) − f(a))/(b − a). It tells you the overall slope between two times—e.g., average cars/hour crossing a bridge from 9–11 AM. Units come from the context (vehicles/hour, dollars/day). Instantaneous rate of change = the derivative f′(t) at a single time. It’s the limit of average rates over shorter intervals and gives the exact rate “right now”—e.g., how fast traffic flow is increasing at 10:00 AM (vehicles/hour per hour) or marginal cost (dollars per additional unit). On the AP, you must interpret both kinds in context (CED CHA-3.C): compute average rates from data or integrals and use derivatives for instantaneous/marginal rates. Always state units and meaning (what is changing, per what). For more examples and AP-style practice on Topic 4.3, see the Fiveable topic study guide (https://library.fiveable.me/ap-calculus/unit-4/rates-change-applied-contexts-other-than-motion/study-guide/ZaLyHm3IcLaORcqYsxxB) and hundreds of practice problems (https://library.fiveable.me/practice/ap-calculus).

Steps you can follow every time (works for volume, surface area, etc.): 1. Draw a picture and label all variables (r, h, V, A...). 2. Write an equation that relates those variables (use geometry: V = πr^2h for a cylinder, V = (4/3)πr^3 for a sphere, SA = 4πr^2, etc.). This is your constraint. 3. Differentiate both sides with respect to time t (implicit differentiation). Use the chain rule: when you differentiate r(t) you get dr/dt, etc. Example: cylinder V = πr^2h gives dV/dt = 2πr h (dr/dt) + πr^2 (dh/dt). 4. Substitute the known values (current r, h, dV/dt, dr/dt, …) and solve for the unknown rate. Keep units (e.g., cm^3/s → cm/s). 5. Check plausibility (signs: positive = increasing, negative = decreasing). Quick examples: - Sphere: V = (4/3)πr^3 → dV/dt = 4πr^2 dr/dt so dr/dt = dV/dt ÷ (4πr^2). - Cylinder (given dr/dt and dh/dt): use the formula above to find dV/dt. On the AP exam this is Topic 4.3: you’ll use implicit differentiation and the chain rule (CED CHA-3.C.1). For extra practice and worked examples, see the Topic 4.3 study guide (https://library.fiveable.me/ap-calculus/unit-4/rates-change-applied-contexts-other-than-motion/study-guide/ZaLyHm3IcLaORcqYsxxB), the Unit 4 overview (https://library.fiveable.me/ap-calculus/unit-4), and many practice problems (https://library.fiveable.me/practice/ap-calculus).

Pick the quantity the problem treats as the independent variable—that’s what you differentiate with respect to. Usually: - If the context is changing over time (traffic, flow, population, funnel), differentiate with respect to t (time). Units will be e.g., vehicles/hour per hour or ounces/minute per minute. - If a function is given as C(q) (cost vs. quantity), take d/dq—marginal cost = dC/dq. - If the relationship is written implicitly (x and y tied by an equation) and the rates are “how fast x and y change,” pick the shared independent variable (often time) and do implicit differentiation with respect to that variable, using the chain rule (dy/dt appears when differentiating y). Always write units and interpret: the derivative’s units tell you what the rate measures (AP CED CHA-3.C: interpret rates in context). For practice on non-motion rates and related rates, see the Topic 4.3 study guide (https://library.fiveable.me/ap-calculus/unit-4/rates-change-applied-contexts-other-than-motion/study-guide/ZaLyHm3IcLaORcqYsxxB) and hit the practice question bank (https://library.fiveable.me/practice/ap-calculus) to get lots of examples.

Use implicit differentiation whenever the quantity you need (often y or another dependent variable) is tied to the independent variable through an equation that isn’t solved for that dependent variable, or when multiple quantities (x and y, radius and height, concentration and volume, etc.) all change with time. In related-rates/rate-of-change problems you usually treat every variable as a function of time and differentiate both sides with respect to t, applying the chain rule (so dy/dt appears when differentiating any y-term). Quick checklist: - Is the relationship given implicitly (e.g., x^2 + y^2 = 25) or hard to solve for one variable? Use implicit differentiation. - Replace each d/dt of a variable with its derivative (dx/dt, dy/dt). - Solve algebraically for the desired rate. Example: x^2 + y^2 = 25, dx/dt known → 2x·dx/dt + 2y·dy/dt = 0 so dy/dt = −(x/y)·dx/dt. This appears in the CED keywords (implicit differentiation, chain rule, related rates) and shows up on AP problems about interpreting rates (Topic 4.3) and related-rates tasks in Topics 4.4–4.5. For a quick study guide, see the Topic 4.3 study guide (https://library.fiveable.me/ap-calculus/unit-4/rates-change-applied-contexts-other-than-motion/study-guide/ZaLyHm3IcLaORcqYsxxB) and more unit review/practice at (https://library.fiveable.me/ap-calculus/unit-4) and (https://library.fiveable.me/practice/ap-calculus).

They’re asking for a rate of change—how fast one quantity changes with respect to another. On the AP CED (CHA-3.C) that usually means using a derivative. - Instantaneous rate: f′(a) = the derivative at a point. It tells you the exact speed of change at that moment (with units). Example: if C(t) is cost in dollars and t is hours, C′(3)=5 means cost is increasing at $5 per hour at t=3. - Average rate: [f(b)−f(a)]/(b−a). It’s the overall change per unit over an interval. - Sign matters: positive → increasing; negative → decreasing. Magnitude tells how fast. - In applied problems you must include units and interpret in context (e.g., “population is decreasing by 200 people per year at t=10”). - For related rates or implicit contexts use chain rule/implicit differentiation to convert between variables. See the Topic 4.3 study guide for examples and practice (https://library.fiveable.me/ap-calculus/unit-4/rates-change-applied-contexts-other-than-motion/study-guide/ZaLyHm3IcLaORcqYsxxB). For broader unit review and lots of practice problems go to (https://library.fiveable.me/ap-calculus/unit-4) and (https://library.fiveable.me/practice/ap-calculus).

Think of temperature/heat questions as plain rate problems: the derivative dT/dt is the instantaneous rate of temperature change (units: °C/min, °F/hr, etc.). Start by writing a model or relationship (energy in/out, Newton’s law, geometry). Common tools from the CED you’ll use: chain rule, implicit differentiation, related rates, and separable ODEs. Quick recipe: - Identify the quantity T(t) and units. Ask: what is dT/dt asking for (instantaneous vs average)? - Write the equation. Example: Newton’s Law of Cooling: dT/dt = −k(T − Ts) (k>0, Ts ambient). Solve by separation to get T(t) = Ts + (T0 − Ts)e−kt. - If geometry matters (heat through a changing area), use chain rule/related rates: dQ/dt = (dQ/dT)(dT/dt) or relate radius/area with dr/dt and implicit differentiation. - Interpret answer with units and context (CED CHA-3.C). For AP review and examples on Topic 4.3, see the Fiveable study guide (https://library.fiveable.me/ap-calculus/unit-4/rates-change-applied-contexts-other-than-motion/study-guide/ZaLyHm3IcLaORcqYsxxB) and practice problems (https://library.fiveable.me/practice/ap-calculus).

You find dV/dt because radius r is changing in time, but V (volume) is the quantity that’s related algebraically to r. On the AP (CHA-3.C) you use implicit differentiation/chain rule to connect rates: differentiate the relation V = f(r, …) with respect to t to get an equation that includes both dV/dt and dr/dt. Then you solve that equation for the rate you want (dr/dt). Example: V = πr^2 h (h may be constant or also depend on t). Differentiate: dV/dt = 2πr h · dr/dt + πr^2 · dh/dt. If the problem gives dV/dt (or dh/dt), you can algebraically solve for dr/dt. If it gives other info instead, you might first compute dV/dt from that info and then find dr/dt. Always keep units (e.g., cm^3/s and cm/s) and state you used implicit differentiation and the chain rule. For more practice with related rates see the Topic 4.3 study guide (https://library.fiveable.me/ap-calculus/unit-4/rates-change-applied-contexts-other-than-motion/study-guide/ZaLyHm3IcLaORcqYsxxB) or Unit 4 review (https://library.fiveable.me/ap-calculus/unit-4).

Think of marginal cost (MC) and marginal revenue (MR) as instantaneous rates of change: MC(q) = C′(q) and MR(q) = R′(q), where C(q) and R(q) are cost and revenue as functions of quantity q. Process: 1. Identify C(q) and/or R(q) from the problem (units matter: dollars per unit, etc.). 2. Differentiate to get MC(q) and MR(q) (use chain rule/implicit differentiation if needed). 3. Interpret values: MC(50)=3 means the cost is increasing at $3 per additional unit when q=50 (dollars per unit produced). 4. Use equalities or comparisons: profit is maximized where MR(q)=MC(q) (check second-derivative or sign chart to confirm a max). If asked marginal profit, P′(q)=R′(q)−C′(q)=MR−MC. 5. For approximate changes, use linearization: ΔC ≈ MC(q)·Δq. On the AP exam, show derivative work, units, and interpretation (CED CHA-3.C). For extra practice and examples see the Topic 4.3 study guide (https://library.fiveable.me/ap-calculus/unit-4/rates-change-applied-contexts-other-than-motion/study-guide/ZaLyHm3IcLaORcqYsxxB) and more practice problems (https://library.fiveable.me/practice/ap-calculus).

Short checklist to stop getting mixed up: 1. Pick variables and units. Write what each function means (e.g., P(t) = population, C(q) = cost). Units tell you if a given number is a rate (vehicles/hour, $/unit, people/year). 2. Translate phrases into derivatives: - “rate of change of P with respect to t” → P′(t). - “how fast P is changing at t = 5” → instantaneous rate P′(5). - “average rate over [a,b]” → (P(b)−P(a))/(b−a). - “marginal cost” → C′(q) (approx. change in cost per one more unit). 3. Identify what’s GIVEN vs WHAT TO FIND: - Given: any explicit formula (P(t)=..., C(q)=...), a numeric derivative (dV/dt = 3), or a verbal rate (“increasing at 2 units/min”). - Find: if the question asks “find d?/dt”, “how fast”, “when is the rate”, you need to differentiate (use implicit diff and chain rule for related rates). 4. For related-rates: write relation between variables (geometry/physics), differentiate implicitly w.r.t time, then plug given rates/values to solve for the unknown derivative. 5. Units & interpretation: always attach units to your derivative answer and write a one-sentence interpretation (CED CHA-3.C—derivative = instantaneous rate in context). If you want more worked examples and practice, check the Topic 4.3 study guide (https://library.fiveable.me/ap-calculus/unit-4/rates-change-applied-contexts-other-than-motion/study-guide/ZaLyHm3IcLaORcqYsxxB) and try problems on Fiveable’s practice page (https://library.fiveable.me/practice/ap-calculus).

A negative rate of change (derivative) just means the quantity is decreasing, and you always read it with units. For example: if P(t) is population and P′(t) = -200 people/year at t = 5, the population is falling at 200 people per year at that instant. If R(x) is revenue and R′(x) = -$3 per unit, producing one more unit *decreases* revenue by $3 (marginal revenue is negative). Distinguish instantaneous vs. average: P′(t) is the instantaneous speed of change; (P(b)-P(a))/(b-a) is the average rate over [a,b]. On AP free-response you’ll be asked to interpret derivatives in context (CED Objective CHA-3.C), so always state the sign, magnitude, units, and precise meaning (instant or average). Practice translating answers on the Topic 4.3 study guide (https://library.fiveable.me/ap-calculus/unit-4/rates-change-applied-contexts-other-than-motion/study-guide/ZaLyHm3IcLaORcqYsxxB) and try related problems at (https://library.fiveable.me/practice/ap-calculus).