Questions on rates of change and building differential equations

Question 1

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.

Accepted Answer

dV/dt = dV/dr × dr/dt. B1 for the chain, B1 for the comment. One time rate is given, such as how fast volume is being pumped in, and the other is required. The middle factor dV/dr is the one you generate yourself, by differentiating whatever formula links the two quantities.

Question 2

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.

Accepted 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.

Question 3

A quantity y decreases at a rate proportional to the square root of y. Write down a differential equation for y, defining any constant you introduce.

Accepted Answer

The rate of decrease of y is −dy/dt, and it is proportional to √y, so −dy/dt = k√y, that is dy/dt = −k√y. B1 for the equation. Here k is a positive constant. B1 for stating k > 0. The minus sign and the positive k work together: with both in place dy/dt is negative while y is positive, which is what a decreasing quantity requires. The equation also stalls at y = 0, where the square root and the rate vanish together.

Question 4

A square metal plate is heated, and the length x cm of its side increases at a constant rate of 0.02 cm per second. Find the rate at which the area of the plate is increasing at the moment when x = 15 cm, giving your answer with its units.

Accepted Answer

The area is A = x², so dA/dx = 2x, which is 30 at x = 15. M1 for dA/dx with its value at the stated moment. The chain rule gives dA/dt = dA/dx × dx/dt = 30 × 0.02 = 0.6. A1. Centimetres squared divided by centimetres, multiplied by centimetres per second, leaves square centimetres per second, so the area is increasing at 0.6 cm² per second. B1 for the units.

Question 5

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.

Accepted 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.

Question 6

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.

Accepted 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.

Question 7

Write down a differential equation modelling a population P whose growth rate is proportional to the population itself.

Accepted Answer

dP/dt = kP, where k is a positive constant. B1 for the equation, B1 for stating k > 0. Fixing the sign of k is part of the model rather than an afterthought, since k < 0 would describe decay.

Question 8

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.

Accepted 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.

Question 9

Show that y = Ae^−3t + 5, where A is a constant, satisfies the differential equation dy/dt = 3(5 − y), and state the value that y approaches as t increases.

Accepted Answer

Differentiate the left-hand side: dy/dt = −3Ae^−3t, since the constant 5 contributes nothing and the exponent −3t brings down a factor of −3. M1 for differentiating, A1 for −3Ae^−3t. Now build the right-hand side from the same expression: 5 − y = 5 − (Ae^−3t + 5) = −Ae^−3t, so 3(5 − y) = −3Ae^−3t. M1 for forming 3(5 − y) in terms of t. The two sides agree for every t and for every value of A, so the equation is satisfied. A1. As t increases, e^−3t tends to 0, so y approaches 5. B1. The value 5 is where dy/dt = 0, so the model settles there whatever A is, coming down to it when A is positive and up to it when A is negative.

Question 10

The temperature θ °C of a component t seconds after it is switched on satisfies dθ/dt = 0.8e^−0.05t. Find the rate at which the temperature is rising when t = 10, giving your answer to 3 significant figures with its units, and describe what happens to that rate as t increases.

Accepted Answer

Substituting t = 10 gives dθ/dt = 0.8e^−0.5. M1 for the substitution. That is 0.485 to 3 significant figures. A1. The derivative is a temperature divided by a time, so its units are °C per second. B1. As t increases, −0.05t becomes more negative and e^−0.05t tends to 0, so dθ/dt tends to 0: the temperature keeps rising, because the rate stays positive, but ever more slowly. B1 for the behaviour with the reason.

Question 11

In a culture, the number N of bacteria and the mass m grams of nutrient still available are modelled by N = 400(20 − m)² for 0 ≤ m ≤ 20. The nutrient is consumed at a constant rate of 0.5 grams per hour. Find the rate at which N is increasing at the moment when m = 12, giving your answer with its units.

Accepted Answer

Differentiating the model with respect to m, and using the chain rule on the bracket, dN/dm = 2 × 400(20 − m) × (−1) = −800(20 − m). M1 for dN/dm. At m = 12 that is −800 × 8 = −6400. A1. The nutrient is being consumed, so m is falling and dm/dt = −0.5 grams per hour. B1 for the negative rate with its reason. The chain rule then gives dN/dt = dN/dm × dm/dt = (−6400) × (−0.5) = 3200. M1 for the chain. So the count is rising at 3200 bacteria per hour. A1 for the value with its units. Reading dm/dt as +0.5 would reverse the answer and describe a culture that is dying out.

Question 12

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.

Accepted 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. Leaving both r and h in the volume formula makes dV/dh meaningless, and the fixed angle is what lets one variable be eliminated.

Question 13

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.

Accepted 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.

Question 14

A ladder of length 5 m rests with its top against a vertical wall and its foot on horizontal ground. The foot slides away from the wall at a constant 0.2 m per second. Let x metres be the distance from the foot of the ladder to the wall, and y metres the height of its top above the ground. Find the rate at which the top is sliding down the wall at the moment when x = 3, giving your answer with its units. State what the model predicts about that rate as x approaches 5, and explain why the prediction should not be believed.

Accepted Answer

The wall, the ground and the ladder make a right-angled triangle, so x² + y² = 25 at every moment of the motion. M1 for the relation between x and y. Both x and y are functions of t, so differentiating that relation with respect to t gives 2x dx/dt + 2y dy/dt = 0. M1 for differentiating with respect to t with dy/dt attached to the y term. When x = 3, y = √(25 − 9) = 4. A1. Substituting, 2(3)(0.2) + 2(4) dy/dt = 0, so dy/dt = −1.2/8 = −0.15, and the top is sliding down at 0.15 m per second. A1 for the value with its units and the direction. Rearranged, dy/dt = −(x/y) × dx/dt, which has y in a denominator. As x approaches 5 the height y approaches 0, so the predicted speed of the top increases without bound. B1 for the prediction. No ladder does that: long before y reaches 0 the top leaves the wall, and the relation x² + y² = 25 has already stopped describing the situation, so the arithmetic is answering a question about a triangle rather than about a ladder. B1 for the reason.

Question 15

A searchlight stands 100 m from a long straight wall, and O is the point of the wall nearest to it. The beam turns at a constant 0.05 radians per second, and it meets the wall at a point x metres from O, where x = 100 tan θ and θ is the angle between the beam and the line from the searchlight to O. Find the rate at which the point of light moves along the wall at the moment when θ = π/6, giving your answer to 3 significant figures with its units.

Accepted Answer

Differentiating x = 100 tan θ with respect to θ gives dx/dθ = 100 sec² θ. M1 for the derivative of tan θ. At θ = π/6, cos (π/6) = √3/2, so sec² θ = 1 ÷ (3/4) = 4/3 and dx/dθ = 400/3. A1. The chain rule links the two rates: dx/dt = dx/dθ × dθ/dt = (400/3) × 0.05 = 20/3. M1 for the chain rule. That is 6.67 to 3 significant figures. A1. A distance in metres divided by a time in seconds gives metres per second. B1 for the units. The beam turns steadily but the spot does not travel steadily: sec² θ is smallest at θ = 0 and grows without bound as θ approaches π/2, so the spot is slowest opposite the searchlight and moves faster and faster along the wall as the beam swings away. The angle has to be in radians, because the derivative of tan θ assumes it.