Difficulty: Introductory-Intermediate | Prerequisites: basic differentiation rules (power rule), function evaluation
This topic sits at the intersection of differentiation technique and real-world modelling. You have already learnt what a derivative is in the abstract; here, you apply it. The problems below use polynomial functions to model weight gain in infants and population growth in cities, then ask you to differentiate, evaluate, and interpret. If the power rule is not yet automatic for you, review that first. This is the kind of material that shows up on exams as "applied rate of change" or "interpret the derivative" questions.
A derivative tells you the rate at which something is changing at a specific moment. In these problems, you take polynomial functions that model weight (in pounds) or population (in people) over time, differentiate them using the power rule, and then plug in a value of t to find how fast the quantity is changing at that instant. The key skill is moving from "what is the value?" to "how fast is the value changing?"
Derivative (f'(t) or df/dt)
The instantaneous rate of change of a function with respect to its variable. In simple terms, it tells you how quickly the output is moving at one precise moment, not over a stretch.
Rate of change
How much a quantity changes per unit of the independent variable (usually time). Think of it as the speedometer reading for whatever quantity you are tracking.
Power rule
The differentiation rule stating that d/dt [t^n] = n * t^(n-1). In simple terms, bring the exponent down as a multiplier and reduce the exponent by one. This is the workhorse rule for polynomial derivatives.
Polynomial function
A function made up of terms of the form a * t^n, added together. Examples: W(t) = 7.15 + 1.81t - 0.0015t² + 0.000709t³ and P(t) = 400,000 + 6000t². Think of it as a sum of power terms, each with its own coefficient.
Growth rate (dP/dt)
The derivative of a population (or similar) function with respect to time. It tells you how many people (or units) are being added per year (or per unit time) at a given moment.
Constant term
A term with no variable attached (e.g., 7.15 or 400,000). Its derivative is always zero because a fixed number does not change.
The median weight of a boy aged 0 to 36 months is modelled by:
W(t) = 7.15 + 1.81t - 0.0015t^2 + 0.000709t^3where t is in months and W is in pounds.
Differentiate each term separately using the power rule.
d/dt [7.15] = 0 (constant drops out)
d/dt [1.81t] = 1.81 (coefficient stays, t¹ becomes t⁰ = 1)
d/dt [-0.0015t²] = -0.0015 × 2 × t = -0.003t
d/dt [0.000709t³] = 0.000709 × 3 × t² = 0.002127t²
Putting it together:
W'(t) = 1.81 - 0.003t + 0.002127t^2This derivative tells you the rate at which the boy's weight is changing (in pounds per month) at any given month t.
Plug t = 12 into the original function:
W(12) = 7.15 + 1.81(12) - 0.0015(12)² + 0.000709(12)³
W(12) = 7.15 + 21.72 - 0.216 + 1.224
W(12) ≈ 28.88 pounds
This is the estimated median weight at 12 months (one year).
Plug t = 12 into the derivative:
W'(12) = 1.81 - 0.003(12) + 0.002127(12)²
W'(12) = 1.81 - 0.036 + 0.306
W'(12) ≈ 2.08 pounds per month
Interpretation: at 12 months of age, the boy's weight is increasing at approximately 2.08 pounds per month.
A city's population after t years is modelled by:
P(t) = 400{,}000 + 6{,}000\,t^2The initial population (at t = 0) is 400,000.
Differentiate each term:
d/dt [400,000] = 0 (constant)
d/dt [6,000t²] = 6,000 × 2t = 12,000t
Result:
\frac{dP}{dt} = 12{,}000\,tNotice that the growth rate itself depends on t. The population grows faster the further out you go, which makes sense for a quadratic model.
P(15) = 400,000 + 6,000(15)²
P(15) = 400,000 + 6,000(225)
P(15) = 400,000 + 1,350,000
P(15) = 1,750,000
The population is projected to be 1,750,000 after 15 years.
For reference, at t = 10: P(10) = 400,000 + 6,000(100) = 1,000,000.
P'(t) = 12,000t
P'(15) = 12,000 × 15 = 180,000 people per year
At 15 years, the population is growing at a rate of 180,000 people per year.
The growth rate of 180,000 at t = 15 means that at exactly that moment, the population is increasing at a pace of 180,000 people per year. It does not mean 180,000 people were added in year 15 specifically; it is the instantaneous rate, not an average over a year. This distinction between instantaneous and average rate of change is a central concept.
Power rule (general form)
\frac{d}{dt}[t^n] = n \cdot t^{n-1}Derivative of a constant
\frac{d}{dt}[c] = 0Weight model and derivative
W(t) = 7.15 + 1.81t - 0.0015t^2 + 0.000709t^3W'(t) = 1.81 - 0.003t + 0.002127t^2Population model and derivative
P(t) = 400{,}000 + 6{,}000\,t^2P'(t) = 12{,}000\,tPaediatricians use weight-vs-age curves (growth charts) built from functions like W(t) to track whether an infant is growing at a healthy rate. The derivative W'(t) is what tells the doctor whether growth is accelerating, steady, or slowing, which matters more clinically than the weight itself at a single visit.
Urban planners and demographers model city populations with functions like P(t) to forecast infrastructure needs: roads, schools, water supply. The growth rate dP/dt drives budget decisions because a city growing at 180,000 people per year needs a very different investment plan from one growing at 20,000.
Students often confuse f(t) with f'(t). Plugging into the original function gives you the value at time t. Plugging into the derivative gives you the rate of change at time t. These answer different questions.
Students sometimes forget that the derivative of a constant is zero, and either leave the constant in the derivative or try to apply the power rule to it.
A common arithmetic slip: when applying the power rule to a term like 0.000709t³, students multiply the coefficient by the exponent but forget to reduce the exponent by one, writing 0.002127t³ instead of 0.002127t².
Students often interpret the derivative at a point as a total change ("the baby gained 2.08 pounds in month 12") rather than as an instantaneous rate ("at month 12, weight is increasing at 2.08 pounds per month"). These are close but conceptually distinct.
⚠️ "Find the rate of change at t = __" is one of the most common exam question formats. It means: differentiate, then substitute.
⚠️ "Interpret your answer" questions require units and context. For W'(12) ≈ 2.08, a full answer is: "At 12 months, the baby's weight is increasing at approximately 2.08 pounds per month." Missing the units or the context costs marks.
⚠️ You may be asked to distinguish between average rate of change (slope of a secant line) and instantaneous rate of change (slope of the tangent line, i.e. the derivative). Know both definitions.
⚠️ Polynomial differentiation with decimals is common on Brief Calculus exams. Arithmetic errors on coefficients like 0.000709 × 3 are easy to make under time pressure. Write out intermediate steps.
True or false: The derivative of a constant term is 1. (False. It is 0.)
Fill in the blank: To differentiate 5t⁴, multiply 5 by ___ and reduce the exponent to ___. (4; 3. Result: 20t³.)
True or false: W'(12) = 2.08 means the baby weighs 2.08 pounds at 12 months. (False. It means the weight is increasing at a rate of 2.08 pounds per month at t = 12.)
Fill in the blank: If P(t) = 400,000 + 6,000t², then dP/dt = ___. (12,000t.)
True or false: A growth rate of 180,000 people per year at t = 15 means exactly 180,000 people were added during year 15. (False. It is the instantaneous rate at t = 15, not the total added over that year.)
Q: Given W(t) = 7.15 + 1.81t - 0.0015t² + 0.000709t³, find W'(t).
A: W'(t) = 1.81 - 0.003t + 0.002127t²
Q: What is the median weight of a boy at 12 months, according to this model?
A: W(12) = 7.15 + 21.72 - 0.216 + 1.224 ≈ 28.88 pounds.
Q: At what rate is the boy's weight changing at 12 months? Include units and interpretation.
A: W'(12) ≈ 2.08 pounds per month. At 12 months of age, the median weight is increasing at approximately 2.08 pounds per month.
Q: If P(t) = 400,000 + 6,000t², find the growth rate dP/dt.
A: dP/dt = 12,000t.
Q: What is the population after 15 years?
A: P(15) = 400,000 + 6,000(225) = 1,750,000.
Q: Find and interpret the growth rate at t = 15.
A: P'(15) = 12,000(15) = 180,000 people per year. At year 15, the population is growing at an instantaneous rate of 180,000 people per year.
Q: Why does the growth rate increase as t increases in the population model?
A: Because P'(t) = 12,000t is a linear function of t. As t gets larger, so does the derivative. The quadratic term in the original function means the population accelerates over time.
This material connects directly to limits, since the derivative is defined as a limit of the difference quotient. If you are shaky on limits, revisit that chapter, as it is the theoretical foundation for everything here.
It also connects forward to optimisation (finding maxima and minima), where you set the derivative equal to zero and solve. The same power rule mechanics apply, just with a different question at the end.
In integral calculus, you will reverse this process: given a rate of change, find the original quantity. That is the antiderivative, and understanding differentiation well makes integration far more intuitive.
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