Maths › Differentiation › Rates of change and building differential equations
Rates of change and building differential equations
Real problems hand over rates, not formulae: the balloon fills at a known rate, the question asks how fast the radius grows. The chain rule connects linked rates, and translating rate sentences into differential equations is the skill that later feeds the integration unit its problems to solve.
Builds on Implicit and parametric differentiation and Exponential functions and e.
IN THIS TOPIC
- Connect rates through the chain rule and evaluate them at an instant.
- Translate rate sentences into differential equations, signs and constants included.
COMMON MISCONCEPTION
If two quantities are linked, their rates of change are equal.
Connected rates
Quantities linked by a formula have rates linked by the chain rule, so dV/dt = dV/dr × dr/dt for a volume that depends on a radius. Linked, though, is a long way from equal. Volume can grow at a perfectly steady rate while the radius growth collapses, because the connecting factor dV/dr is itself growing.
WORKED EXAMPLE
The balloon, timed
A spherical balloon is inflated at a steady 100 cm3 per second. Find the rate at which its radius grows when r = 5 cm.
The link: V = (4/3)πr3, so dV/dr = 4πr2.
The chain: dV/dt = dV/dr × dr/dt, so 100 = 4πr2 × dr/dt.
At r = 5: dr/dt = 100/(100π) = 1/π ≈ 0.32 cm per second.
The same puff of air stretches a big balloon less, and the algebra says why. That connecting factor 4πr2 is the balloon's surface area, and more surface shares the volume out thinner.
GUIDED PRACTICE
A melting cube
An ice cube of side x melts so its volume decreases at a steady 2 cm3 per minute. Find the rate at which x decreases when x = 10 cm, before opening the working.
Show the working
The link: V = x3, so dV/dx = 3x2.
The chain, with the melt written as dV/dt = −2: dx/dt = (dV/dt)/(dV/dx) = −2/(3x2).
At x = 10: dx/dt = −2/300 = −1/150 cm per minute.
Minus signs carried the physics here, shrinking volume and shrinking side. If the question asks for the rate of decrease, quote the size and say in words which way it is going.
Sentences into equations
A differential equation states how fast something changes, and building one is translation work, phrase by phrase. “Rate of change of P” becomes dP/dt. “Proportional to” brings in a constant k. “Decreasing” plants a minus sign. The exponentials lesson already met the flagship case, rate proportional to the quantity itself, as dP/dt = kP.
WORKED EXAMPLE
The mint, modelled
A spherical mint dissolves so that its radius decreases at a rate inversely proportional to the square of the radius. Write a differential equation for r.
“Rate of decrease of r” is −dr/dt, and “inversely proportional to r2” is k/r2.
So −dr/dt = k/r2, that is dr/dt = −k/r2, with k > 0.
State k > 0 alongside the minus sign. Those two together are what make the model describe decay, and they say the mint shrinks, fastest near the end.
INDEPENDENT PRACTICE
Verify, do not yet solve
Show that P = 500e0.2t satisfies the differential equation dP/dt = 0.2P.
Show the working
Differentiate: dP/dt = 500 × 0.2e0.2t = 0.2 × 500e0.2t.
The right-hand side is 0.2P, so the equation is satisfied. ∎
Verifying is differentiation. Solving is integration, and the integration unit finishes this story.
ASSESSMENT FOCUS
- Write the connecting formula, differentiate it, then chain. Three lines, three marks, in that order.
- Track signs in words and symbols together. A rate of decrease is a negative derivative.
- “Show that P satisfies…” means differentiate and compare. It never means solve.
CHECK YOURSELF
The area of a circular oil slick grows at a steady 12 m2 per hour. Find the rate at which the radius grows when r = 6 m.
Show a hint
A = πr² links them; chain and evaluate.
Show the answer
dA/dr = 2πr, and dA/dt = dA/dr × dr/dt.
So 12 = 2π × 6 × dr/dt, giving dr/dt = 12/(12π) = 1/π ≈ 0.32 m per hour.
The growing circumference 2πr is the sharing-out factor, the slick's edge playing the balloon's surface.
Linked quantities chain their rates: build the link, differentiate it, multiply.
Rate sentences translate word by word: derivative, k, and a sign that tells the truth.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the rates of change and building differential equations questions page.
CHECK YOUR PROGRESS
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- Connect rates through the chain rule and evaluate them at an instant.
- Translate rate sentences into differential equations, signs and constants included.
Open the full revision checklist to see every objective in the course in one place.