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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.

Year 12-13EDEXCEL 9MA0 7.4, 7.6

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.
  • Verify proposed solutions of a differential equation by differentiating.

WHAT YOU PROBABLY THINK

If two quantities are linked, their rates of change are equal.

Connected rates

When quantities are linked by a formula, their rates are linked by the chain rule: dV/dt = dV/dr × dr/dt for a volume depending on a radius. Linked, but rarely equal, and the opening lie dies at the first balloon. The volume can grow steadily while the radius growth collapses, because the connecting factor dV/dr itself grows.

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 formula says why: the connecting factor 4πr2 is the balloon's surface area, and more surface shares the volume out thinner.

A balloon inflated at a steady 100 cubic centimetres per second: the radius grows fast when the balloon is small and ever more slowly as it swells, with the growth rate at radius 5 equal to 1 over pir = 5dr/dt = 1/π when r = 5dr/dt = 100/(4πr²)same puff, less growth
FIG. 1The radius growth rate against radius for the 100 cm³/s balloon: fast when small, crawling when large, exactly 1/π at r = 5.

YOUR TURN

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.

The minus signs carried the physics: shrinking volume, shrinking side. Quote the rate's size with its direction in words if the question asks for the rate of decrease.

Sentences into equations

A differential equation states how fast something changes, and building one is translation work. “Rate of change of P” becomes dP/dt; “proportional to” brings in k; “decreasing” plants a minus sign. The exponentials lesson met the flagship case, rate proportional to the quantity itself, as dP/dt = kP.

Translating a sentence into a differential equation: the rate of change of P becomes d P by d t, is proportional to becomes equals k times, and decreasing plants a minus sign“the rate of change of P”dP/dt“proportional to P”= kP“decreasing”minus signassemble:dP/dt = −kPk > 0
FIG. 2The translation table: rate words to derivative, proportionality to k, decrease to a minus sign, assembling into 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; “inversely proportional to r2” is k/r2.

So −dr/dt = k/r2, that is dr/dt = −k/r2, with k > 0.

Stating k > 0 alongside the minus sign is what keeps the model honest; the two signs between them say the mint shrinks, fastest near the end.

TRY IT UNSEEN

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. The two directions between them are the whole of differential equations at this level.

THE EXAM BIT

  • Write the connecting formula, differentiate it, then chain: the three lines earn three marks in order.
  • Track signs in words and symbols together; a rate of decrease is a negative derivative.
  • Evaluate connected rates only after the general chain is built; numbers go in last.
  • In translation questions define every symbol, include k with its sign convention, and state units.
  • “Show that P satisfies…” means differentiate and compare, never 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.

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.
  • Verify proposed solutions of a differential equation by differentiating.

No animated video for this topic yet; these notes stand alone.