Derivative Meaning in Context: Velocity, Growth, and Marginal Change

Derivatives as “rate of change” with units and sign

A derivative connects two quantities: how an output changes when an input changes. If a function is y=f(x), then f'(x) tells the instantaneous rate of change of y with respect to x.

• Units: If y is measured in “output units” and x in “input units,” then f'(x) has units (output units)/(input units).
• Sign: f'(x)>0 means the output is increasing as the input increases; f'(x)<0 means the output is decreasing.
• Magnitude: Larger absolute value means faster change per unit of input.

Quick check: average vs instantaneous rate

Average rate of change on an interval [a,b] is (f(b)-f(a))/(b-a). Instantaneous rate at a point x=a is f'(a). Average uses two points; instantaneous describes the “right now” rate at one input value.

Motion context 1: position → velocity

Scenario

A cart moves along a straight track. Its position from a sensor is modeled by a function of time.

Function with units

Let s(t)=10- t^2, where s is position in meters (m) and t is time in seconds (s).

Compute the derivative at a point

Velocity is the derivative of position: v(t)=s'(t).

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• Differentiate: s'(t)=d/dt(10-t^2)=-2t.
• Evaluate at t=3: v(3)=-2(3)=-6.
• Units: m/s.

Interpretation: At t=3 s, velocity is −6 m/s, meaning the cart’s position is decreasing at 6 meters per second at that instant (it is moving in the negative direction).

Quick check: average vs instantaneous (with numbers)

Average velocity from t=3 to t=3.5:

• s(3)=10-9=1 m
• s(3.5)=10-12.25=-2.25 m
• Average rate: (-2.25-1)/(3.5-3)=-3.25/0.5=-6.5 m/s

This average (-6.5 m/s) is close to the instantaneous velocity v(3)=-6 m/s, but not identical because the velocity is changing over time.

Motion context 2: velocity → acceleration

Scenario

A cyclist’s velocity changes as they brake.

Function with units

Let velocity be v(t)=12-4t, where v is in meters per second (m/s) and t is in seconds (s).

Compute the derivative at a point

Acceleration is the derivative of velocity: a(t)=v'(t).

• Differentiate: v'(t)=d/dt(12-4t)=-4.
• Evaluate at t=3: a(3)=-4.
• Units: (m/s)/s = m/s^2.

Interpretation: At t=3 s, acceleration is −4 m/s^2, meaning the velocity is decreasing by 4 meters per second each second at that instant.

Sign check

Negative acceleration here indicates the cyclist is slowing down (velocity is dropping). Note: negative acceleration does not always mean “slowing down” in every situation; it means velocity is decreasing. Whether speed decreases depends on the sign of velocity too.

Non-motion context 1: population growth

Scenario

A town’s population is tracked yearly and modeled smoothly for planning.

Function with units

Let P(t)=50000+1200t-30t^2, where P is people and t is years since 2020.

Compute the derivative at a point

The derivative P'(t) is the instantaneous population growth rate.

• Differentiate: P'(t)=1200-60t.
• Evaluate at t=3: P'(3)=1200-180=1020.
• Units: people/year.

Interpretation: At t=3 years, the population growth rate is 1020 people/year, meaning the population is increasing at about 1020 people per year at that time.

Negative derivative indicates decrease

When does the model predict decline? Solve P'(t)<0:

• 1200-60t<0 ⇒ t>20.

For times beyond 20 years after 2020, the derivative is negative, indicating the population is decreasing (according to this model).

Non-motion context 2: marginal cost and marginal revenue

Scenario

A company produces q units of a product. Costs and revenue depend on production level.

Function with units

Let cost be C(q)=2000+15q+0.02q^2, where C is dollars and q is units produced.

Compute the derivative at a point (marginal cost)

Marginal cost is C'(q), the instantaneous rate at which cost changes per additional unit produced.

• Differentiate: C'(q)=15+0.04q.
• Evaluate at q=100: C'(100)=15+4=19.
• Units: dollars/unit.

Interpretation: At q=100 units, marginal cost is $19/unit, meaning producing one more unit around 100 units increases total cost by about $19.

Marginal revenue example (including negative derivative)

Suppose revenue is R(q)=40q-0.1q^2 (dollars). Then:

• R'(q)=40-0.2q (dollars/unit).
• At q=150: R'(150)=40-30=10.

Interpretation: At q=150 units, marginal revenue is $10/unit, meaning selling one more unit around 150 units increases revenue by about $10.

If production rises to q=250, then R'(250)=40-50=-10.

Interpretation: At q=250 units, marginal revenue is −$10/unit, meaning increasing sales slightly beyond 250 units would decrease total revenue by about $10 per additional unit (for example, due to price drops needed to sell more).

Quick check: average vs marginal (discrete comparison)

Average cost per unit at q=100 is C(100)/100, which is not the same as marginal cost C'(100). Average spreads total cost over all units; marginal describes the next unit’s impact near that production level.

Non-motion context 3: sensitivity (how output responds to input)

Scenario

A lab instrument’s reading depends on temperature. You want to know how sensitive the reading is to small temperature changes near a specific temperature.

Function with units

Let the reading be R(T)=5+0.8T-0.01T^2, where R is volts (V) and T is degrees Celsius (°C).

Compute the derivative at a point

Sensitivity of the reading to temperature is R'(T).

• Differentiate: R'(T)=0.8-0.02T.
• Evaluate at T=30: R'(30)=0.8-0.6=0.2.
• Units: V/°C.

Interpretation: At T=30°C, sensitivity is 0.2 V/°C, meaning increasing temperature by 1°C near 30°C increases the reading by about 0.2 V.

At T=50°C, R'(50)=0.8-1.0=-0.2 V/°C.

Interpretation: At T=50°C, sensitivity is −0.2 V/°C, meaning increasing temperature by 1°C near 50°C decreases the reading by about 0.2 V.

Common interpretation template (use this to check units and meaning)

Multi-representation tasks (practice interpreting derivatives)

1) Table → approximate derivative

A sensor records position s (m) at times t (s). Approximate velocity at t=3 using a symmetric difference quotient.

• Approximate s'(3) by (s(3.5)-s(2.5))/(3.5-2.5).
• Compute: (1.4-4.2)/1.0=-2.8 m/s.

Interpretation: At t=3 s, velocity is approximately −2.8 m/s, meaning position is decreasing at about 2.8 meters per second at that instant.

2) Graph → estimate derivative

You are given a graph of P(t) (people) versus t (years). To estimate P'(5):

• Draw (or imagine) the tangent line at t=5.
• Pick two clear points on that tangent line (not on the curve) and compute its slope ΔP/Δt.
• Attach units: people/year.

Sign check: If the tangent line slopes downward as t increases, then P'(5)<0 and the population is decreasing at that time.

3) Formula → compute derivative

Given a demand-based revenue model R(q)=60q-0.3q^2 (dollars), find marginal revenue at q=80.

• Differentiate: R'(q)=60-0.6q.
• Evaluate: R'(80)=60-48=12 dollars/unit.

Interpretation: At q=80 units, marginal revenue is $12/unit, meaning increasing sales slightly beyond 80 units increases revenue by about $12 per additional unit.