Derivatives in Real-World Rates: Modeling, Interpretation, and Communication

What it means to use derivatives in real-world rates

In applications, a derivative is a rate of change with units. If a quantity y = f(x) depends on an input x, then f'(a) describes how fast y is changing at x = a. The most important skill is not just computing f'(a), but communicating it clearly: state the variables and units, compute the derivative value, and interpret it in a complete sentence tied to the scenario.

In this chapter, each scenario follows a consistent workflow:

• Define variables with units (what is input, what is output?).
• Write the derivative with units (output units per input unit).
• Compute f'(a) at a meaningful input a.
• Interpret f'(a) as a sentence about “per 1 unit increase in input, output changes by about … near a.”
• Approximate with a reasonable interval (choose a small but realistic change in input) and explain limitations.

Communication checklist (use every time)

• Label the function: “Let f(v) be braking distance (m) at speed v (m/s).”
• Label the derivative: “f'(v) has units m per (m/s), i.e., seconds.”
• Evaluate: “f'(20) = 8 (m per (m/s)).”
• Interpret: “Near v=20 m/s, increasing speed by 1 m/s increases braking distance by about 8 m.”
• State scope: “This is local; it may not hold far from 20 m/s or under different road conditions.”

Scenario 1: Braking distance vs speed

Model and variables

Suppose a simplified model for braking distance on dry pavement is

d(v) = 0.08v^2

where:

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• v = speed in m/s
• d(v) = braking distance in meters (m)

This model captures the common idea that stopping distance grows roughly like the square of speed.

Derivative and units

d'(v) measures how braking distance changes with speed. Units:

• d is in m
• v is in m/s
• So d'(v) is in m / (m/s) = s (seconds). This unit can feel surprising, but it is correct: it is “meters of extra distance per (meter/second) of extra speed.”

Compute d'(a) and interpret

Differentiate: d'(v) = 0.16v.

Evaluate at a = 20 m/s (about 72 km/h):

d'(20) = 0.16(20) = 3.2 m per (m/s).

Interpretation sentence: Near v = 20 m/s, increasing speed by 1 m/s increases braking distance by about 3.2 meters.

Task: choose an interval for approximation

You want to estimate the change in braking distance when speed increases from 20 m/s to 21 m/s.

• Derivative-based estimate (local linear): Δd ≈ d'(20)·Δv = 3.2·1 = 3.2 m.
• Check against the model (optional): d(21)-d(20)=0.08(441-400)=0.08(41)=3.28 m, close to 3.2 m.

Explain interval choice: Using Δv = 1 m/s is small relative to 20 m/s (a 5% change), so local linearity is plausible.

Limitations to state: The model may fail if road conditions change (wet/ice), brakes fade, tires differ, or if speeds are far outside the range used to fit the model. The derivative statement is local: it describes behavior near 20 m/s, not necessarily at 35 m/s.

Task: extrapolation risk

Use the derivative at 20 m/s to estimate the change from 20 to 30 m/s: Δv = 10 m/s gives Δd ≈ 3.2·10 = 32 m. Explain why this is risky: the slope d'(v)=0.16v increases with v, so using the slope at 20 underestimates the increase over a wide interval; also the physical model may not hold over such a large change.

Scenario 2: Profit vs number of items (marginal profit)

Model and variables

Let profit depend on the number of items produced and sold. Suppose

P(q) = -0.02q^2 + 30q - 500

where:

• q = number of items (items)
• P(q) = profit in dollars ($)

This model includes diminishing returns (the negative quadratic term) and a fixed cost (the -500).

Derivative and units

P'(q) is the rate of change of profit with respect to quantity. Units:

• P is in $
• q is in items
• So P'(q) is in $ per item

This is often called marginal profit: the approximate additional profit from producing/selling one more item near q.

Compute P'(a) and interpret

Differentiate: P'(q) = -0.04q + 30.

Evaluate at a = 400 items:

P'(400) = -0.04(400) + 30 = -16 + 30 = 14 $/item.

Interpretation sentence: Near q = 400 items, producing and selling one additional item increases profit by about $14 per item.

Task: reasonable interval and discrete nature

Because q counts items, changes happen in whole numbers. A reasonable “small interval” is often Δq = 1 item or Δq = 5 items.

• Using Δq = 1: ΔP ≈ P'(400)·1 = 14 dollars.
• Using Δq = 5: ΔP ≈ 14·5 = 70 dollars.

Limitations to state: The derivative treats q as continuous, but real production is discrete. Also, the model may ignore capacity constraints, bulk discounts, overtime pay, or demand limits; marginal profit can change abruptly if costs change at certain thresholds.

Task: interpret negative marginal profit

Find when marginal profit becomes negative: solve P'(q) < 0 → -0.04q + 30 < 0 → q > 750. Explain in context: beyond about 750 items, producing more items decreases profit (each extra item loses money) according to this model.

Scenario 3: Concentration vs time (reaction/decay rate)

Model and variables

Let C(t) be the concentration of a substance in a tank during a process. Suppose

C(t) = 12e^{-0.3t}

where:

• t = time in minutes (min)
• C(t) = concentration in mg/L

Derivative and units

C'(t) measures how concentration changes over time. Units:

• C is mg/L
• t is min
• So C'(t) is (mg/L)/min

A negative value means concentration is decreasing.

Compute C'(a) and interpret

Differentiate: C'(t) = 12(-0.3)e^{-0.3t} = -3.6e^{-0.3t}.

Evaluate at a = 5 min:

C'(5) = -3.6e^{-1.5} ≈ -3.6(0.2231) ≈ -0.803 (mg/L)/min.

Interpretation sentence: At t = 5 minutes, the concentration is decreasing at about 0.803 mg/L per minute.

Task: choose an interval for approximation and justify

To estimate concentration after 30 seconds (0.5 min) from t=5, use Δt = 0.5 min:

ΔC ≈ C'(5)·Δt ≈ (-0.803)(0.5) ≈ -0.402 mg/L.

So C(5.5) ≈ C(5) - 0.402 mg/L.

Explain interval choice: 0.5 min is short compared with the time scale of change, so the rate at 5 min is a reasonable approximation over that small window.

Limitations to state: If the process changes (stirring stops, inflow changes, temperature shifts), the exponential model may no longer apply. Also, using the derivative at 5 min to predict far ahead (e.g., to 20 min) is unreliable because the rate itself changes with time.

Task: relative rate interpretation

Compute the relative rate C'(t)/C(t):

\frac{C'(t)}{C(t)} = \frac{-3.6e^{-0.3t}}{12e^{-0.3t}} = -0.3 per minute.

Interpretation: at any time, concentration is decreasing at about 30% per minute (according to this model). State that this is model-based and assumes exponential decay behavior.

Scenario 4: Temperature vs depth (geothermal gradient)

Model and variables

Suppose temperature increases with depth below the surface. A simple linear model might be

T(z) = 15 + 0.025z

where:

• z = depth in meters (m)
• T(z) = temperature in °C

Derivative and units

T'(z) measures how temperature changes with depth. Units:

• T is °C
• z is m
• So T'(z) is °C/m

Compute T'(a) and interpret

Differentiate: T'(z) = 0.025 °C/m.

Evaluate at a = 800 m: T'(800) = 0.025 °C/m.

Interpretation sentence: Near a depth of 800 m, temperature increases by about 0.025 °C for each additional meter of depth.

Task: choose an interval and discuss model validity

Estimate the temperature change from 800 m to 820 m (Δz = 20 m):

ΔT ≈ T'(800)·20 = 0.025·20 = 0.5 °C.

Explain interval choice: 20 m is small relative to 800 m, and for many sites the gradient is roughly steady over short ranges.

Limitations to state: The gradient can vary by rock type, groundwater flow, and local geology. A linear model may be reasonable over a limited depth range but can fail over large depths; extrapolating to several kilometers without site data can be misleading.

Choosing an input interval for derivative-based approximations

When you use f'(a) to approximate a change, you are using a local linear model:

f(a+Δx) ≈ f(a) + f'(a)Δx

Picking Δx is a modeling decision. Use these guidelines:

• Small relative change: choose Δx so that |Δx| is small compared with |a| (when a is not near 0).
• Meaningful in context: 1 item, 0.5 min, 1 m/s, 10 m depth—use increments that actually occur.
• Stay within the model’s valid range: do not use a derivative from a model fit on 10–25 m/s to predict at 50 m/s.
• Watch for changing conditions: if the situation can change abruptly (pricing tiers, phase changes, friction changes), local linearity may not hold even for small intervals.

Capstone: mixed problems (rules, tangent-line approximation, interpretation)

For each problem: (1) define variables with units, (2) compute f'(x), (3) compute f'(a), (4) write a complete interpretation sentence, (5) use f'(a) to approximate a nearby change using a justified interval, and (6) state at least one limitation.

Problem 1: Braking distance with reaction distance included

A driver’s total stopping distance is modeled as

S(v) = 0.08v^2 + 1.2v

where v is in m/s and S is in meters. Evaluate and interpret S'(25). Then estimate the change in stopping distance from 25 m/s to 26 m/s, and explain why using the same derivative to estimate from 25 m/s to 35 m/s is less reliable.

Problem 2: Revenue, cost, and profit

Revenue is R(q) = 50q - 0.01q^2 dollars and cost is K(q) = 12q + 800 dollars, where q is items. Let P(q)=R(q)-K(q).

• Compute P'(q) and evaluate P'(600).
• Interpret P'(600) in a sentence with units.
• Choose Δq to estimate the profit change from 600 items to 605 items, and justify your choice.

Problem 3: Concentration with a changing input (composition)

A sensor outputs voltage V based on concentration C by V(C)=0.5\sqrt{C} (volts), and concentration changes with time by C(t)=16e^{-0.2t} (mg/L), with t in minutes.

• Find V(t)=V(C(t)) and compute V'(t).
• Evaluate V'(10) and interpret it as “volts per minute” at 10 minutes.
• Use V'(10) to approximate the voltage change over the next 0.2 minutes; explain why 2 minutes would be a less appropriate interval.

Problem 4: Temperature gradient that is not constant

In a region, temperature is modeled by

T(z)=14+0.03z-0.000002z^2

with z in meters and T in °C.

• Compute T'(z) and evaluate T'(1000).
• Interpret T'(1000) in a complete sentence.
• Estimate T(1020) using a tangent-line approximation at 1000 m, and state one limitation of using this approximation.

Problem 5: Clear labeling and final statement practice

A company models the time to assemble a product as t(n)=40+\frac{200}{n} minutes, where n is the number of workers on a task (treat n as continuous for modeling).

• Compute t'(n) and evaluate t'(5).
• Write an interpretation sentence that includes: the point (n=5), the direction of change, and the units.
• Choose a reasonable interval (e.g., from 5 to 6 workers) to estimate the time change; explain why the model may be questionable for very small n or very large n.

Problem 6: One-page report style (communication emphasis)

Pick one of the four scenarios (braking, profit, concentration, temperature) and write a short “technical note” that includes:

• Function definition with units
• Derivative with units
• A computed value f'(a) at a realistic a
• A one-sentence interpretation for a non-specialist audience
• An approximation over a justified small interval
• Two limitations (one about local validity near a, one about extrapolation or changing conditions)