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Rates of change and building differential equations questions

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.

8 original questions · 29 marks · the rates of change and building differential equations notes · Differentiation

Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.

2 marks · warm-up
A quantity V depends on r, and r depends on time t. Write down the chain rule linking dV/dt, dV/dr and dr/dt, and state which of the three a typical question hands you directly.
Worked answer
dV/dt = dV/dr × dr/dt. B1 for the chain, B1 for the comment. Questions usually give one time rate, such as how fast volume is being pumped in, and ask for the other. The middle factor dV/dr is the one you generate yourself, by differentiating whatever formula links the two quantities.
3 marks · warm-up
A circular oil slick spreads so that its area increases at a constant rate of 10 m² per hour. Find the rate at which the radius is increasing when the radius is 4 m, giving your answer to 3 significant figures.
Worked answer
A = πr², so dA/dr = 2πr. The chain rule gives dA/dt = dA/dr × dr/dt, that is 10 = 2π × 4 × dr/dt, so dr/dt = 10/(8π) = 5/(4π) = 0.398 m per hour. M1 for dA/dr, M1 for the chain, A1 for the value. The same amount of area arrives every hour but it has to spread around an ever longer rim, so the radius grows more and more slowly.
4 marks
A spherical balloon is inflated so that its volume increases at a constant rate of 120 cm³ per second. Find the rate at which the radius is increasing when the radius is 4 cm, giving your answer to 3 significant figures.
Worked answer
V = (4/3)πr³, so dV/dr = 4πr², which is 64π at r = 4. Then 120 = 64π × dr/dt, so dr/dt = 120/(64π) = 15/(8π) = 0.597 cm per second. M1 for dV/dr, A1 for 64π, M1 for the chain, A1 for 0.597. Differentiate the linking formula with respect to r, not t. The chain supplies the time part, and trying to differentiate V with respect to t directly leaves an r that nothing can be done with.
4 marks
A cube expands so that its volume increases at a constant rate of 27 cm³ per second. Find the rate at which the edge length is increasing at the moment when the edge is 3 cm, and the rate at which the total surface area is increasing at that same moment.
Worked answer
With edge x, V = x³ gives dV/dx = 3x² = 27 at x = 3, so 27 = 27 × dx/dt and dx/dt = 1 cm per second. For the surface, S = 6x² gives dS/dx = 12x = 36 at x = 3, so dS/dt = 36 × 1 = 36 cm² per second. M1 A1 for the edge rate, M1 A1 for the surface rate. One edge rate feeds both answers, because every rate in the cube has to travel through dx/dt.
2 marks · write
Write down a differential equation modelling a population P whose growth rate is proportional to the population itself.
Worked answer
dP/dt = kP, where k is a positive constant. B1 for the equation, B1 for stating k > 0. The sentence names the rate, names what it is proportional to, and the constant absorbs the proportionality. Fixing the sign of k is part of the model rather than an afterthought, since k < 0 would describe decay.
4 marks
A hot drink cools so that the rate of decrease of its temperature T °C is proportional to the amount by which T exceeds the room temperature of 20 °C. Write down a differential equation for T, defining any constant you introduce.
Worked answer
The rate of decrease of T is −dT/dt and the excess over the room is T − 20, so −dT/dt = k(T − 20), that is dT/dt = −k(T − 20), where k is a positive constant. M1 for identifying T − 20, M1 for the minus sign, A1 for the equation, B1 for defining k. The minus sign and the positive k work together to keep the drink cooling while it is hotter than the room, and the model stalls correctly at T = 20 because the bracket vanishes there.
6 marks · stretch
Water is poured into an inverted right circular cone of semi-vertical angle 30° at a constant rate of 8 cm³ s⁻¹. Find the rate at which the depth of the water is increasing when the depth is 6 cm, giving your answer to 3 significant figures.
Worked answer
Let the depth be h and the surface radius r. The semi-vertical angle gives r = h tan 30° = h/√3. The volume of water is V = (1/3)πr²h = (1/3)π(h²/3)h = πh³/9. Differentiating, dV/dh = πh²/3, which is 12π at h = 6. The chain rule gives 8 = 12π × dh/dt, so dh/dt = 8/(12π) = 2/(3π) = 0.212 cm s⁻¹. M1 for relating r and h, A1 for V in terms of h alone, M1 for differentiating, A1 for 12π, M1 for the chain, A1 for 0.212. Everything hangs on the first step. Leaving both r and h in the volume formula makes dV/dh meaningless, and the fixed angle is what lets one variable be eliminated.
4 marks · stretch
Water flows into a tank at a constant rate of 30 litres per minute and leaks out at a rate proportional to the volume V litres currently in the tank. Write down a differential equation for V, and given that the volume in the tank eventually settles at 600 litres, find the constant of proportionality.
Worked answer
The net rate of change is the inflow minus the outflow, so dV/dt = 30 − kV with k a positive constant. The volume settles when it stops changing, so dV/dt = 0 at V = 600, giving 30 = 600k and k = 0.05 per minute. M1 for the structure inflow minus outflow, A1 for the equation, M1 for setting dV/dt = 0, A1 for k = 0.05. The settled value is a steady state rather than a limit you have to solve the equation to find, and spotting that saves all the work of separating variables.

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