Mathematics guide, Unit 4, step 15 of 33Previous step Next step

Step 15: Rates of Change, Motion and Related Rates

By the EconTutorials team. 12-minute read with 6 practice questions.

SyllabusAP 4.1–4.5 · 9709 P1 1.7 · IB SL 5.9, HL 5.14

In 30 seconds

A derivative is a rate of change with units: units of $ y $ per unit of $ x $. In context, say when, what, which direction and how fast, as in “at $ t = 5 $ minutes the temperature is falling at 9.7 °C per minute”.
For a particle moving on a line with position $ s(t) $: velocity $ v = s'(t) $, acceleration $ a = v'(t) = s”(t) $, and speed $ = |v| $. It changes direction where $ v $ changes sign.
The particle is speeding up when $ v $ and $ a $ have the same sign, and slowing down when their signs differ. A negative acceleration on its own does not mean slowing down.
Exam point: in a related-rates problem, write the equation linking the quantities, differentiate it with respect to time, and only then substitute the values at the instant you care about.

Video lesson coming soonA short animated lesson for this step will appear here.

A derivative in context

So far we have mostly asked for a derivative and stopped. In applied questions the derivative is only half the answer; the other half is saying what it means. If $ y $ is measured in one unit and $ x $ in another, then $ \dfrac{dy}{dx} $ is measured in units of $ y $ per unit of $ x $, and $ \dfrac{d^2y}{dx^2} $ in units of $ y $ per unit of $ x $ per unit of $ x $.

Key termInstantaneous rate of change: the value of the derivative at one instant, such as $ W'(10) $. It tells you how fast, and in which direction, a quantity is changing at that moment, in units of the output per unit of the input.

An AP interpretation sentence has four parts: the instant (“at $ t = 5 $ minutes”), the quantity (“the temperature of the cake”), the direction (“is decreasing”) and the rate with units (“at 9.7 °C per minute”). Writing “the derivative is $ -9.7 $” earns nothing.

Worked example

A cake is taken out of the oven. Its temperature after $ t $ minutes is $ T(t) = 20 + 160e^{-0.1t} $ °C. (a) Find $ T'(5) $ and interpret it. (b) Find $ T”(5) $ and say what it tells you. (c) Use $ T'(5) $ to estimate $ T(5.5) $.

Step 1(a) $ T'(t) = -16e^{-0.1t} $, so $ T'(5) = -16e^{-0.5} \approx -9.70 $. At $ t = 5 $ minutes the temperature of the cake is decreasing at about 9.70 °C per minute.

Step 2(b) $ T”(t) = 1.6e^{-0.1t} $, so $ T”(5) \approx 0.970 $ °C per minute per minute. It is positive, so the rate of change is increasing: the cake is still cooling, but more slowly as time goes on.

Step 3(c) $ T(5) = 20 + 160e^{-0.5} \approx 117.04 $. Over half a minute the rate stays close to $ -9.70 $, so $ T(5.5) \approx 117.04 + 0.5 \times (-9.70) \approx 112.19 $ °C.

The true value is $ T(5.5) \approx 112.31 $ °C. The estimate is a little low because the graph bends upwards (Figure 1); Step 16 makes that idea precise.

A cake cooling: T = 20 + 160e^(−0.1t) °C

T = 20 + 160e^(−0.1t)tangent at t = 5: gradient ≈ −9.70 °C per minute

Figure 1. The gradient of the temperature curve is the rate of cooling. At t = 5 the tangent falls about 9.7 °C per minute; later the curve flattens, because the rate of cooling shrinks as the cake nears room temperature.

An economics example: marginal cost. If $ C(Q) $ dollars is the cost of making $ Q $ items, $ C'(Q) $ is measured in dollars per item, and economists call it marginal cost. For $ C(Q) = 2000 + 15Q + 0.02Q^2 $, $ C'(Q) = 15 + 0.04Q $, so $ C'(100) = 19 $. At an output of 100 items, cost is rising at $19 per item: the 101st item costs about $19. The exact extra cost, $ C(101)-C(100) $, is $19.02, so the derivative is an excellent stand-in for “the cost of one more”. The same reading works for marginal revenue (dollars per item), population growth (people per year) or a leaking tank (litres per minute).

Motion along a line

The most examined rate of all is motion. A particle moves along a straight line, and $ s(t) $ is its position, measured from a fixed origin, at time $ t $. Then its velocity is the rate of change of position, and its acceleration is the rate of change of velocity.

\[ v(t) = s'(t) \qquad a(t) = v'(t) = s”(t) \qquad \text{speed} = |v(t)| \]

Condition	Meaning
$ v > 0 $	Moving in the positive direction (right, or up)
$ v < 0 $	Moving in the negative direction
$ v = 0 $	At rest for an instant; a change of direction if $ v $ changes sign
$ v $ and $ a $ the same sign	Speeding up
$ v $ and $ a $ opposite signs	Slowing down

Why compare signs? Speed is the size of $ v $. If $ v = -5 $ and $ a < 0 $, the velocity goes $ -5, -6, -7 $, so the speed goes 5, 6, 7: it is rising, even though the acceleration is negative. Acceleration pulls $ v $ away from zero when it points the same way as $ v $, and towards zero when it points the other way.

Worked example

A particle moves along a line with position $ s(t) = t^3-9t^2 + 24t $ metres for $ 0 \leq t \leq 5 $ seconds. (a) When is it at rest? (b) When is it speeding up? (c) Find the total distance it travels.

Step 1$ v = 3t^2-18t + 24 = 3(t-2)(t-4) $ and $ a = 6t-18 $.

Step 2(a) At rest when $ v = 0 $: $ t = 2 $ and $ t = 4 $ seconds.

Step 3(b) $ v > 0 $ on $ [0, 2) $ and $ (4, 5] $; $ v < 0 $ on $ (2, 4) $. $ a < 0 $ for $ t < 3 $ and $ a > 0 $ for $ t > 3 $. Same signs on $ (2, 3) $ and $ (4, 5) $: speeding up there, slowing down on $ (0, 2) $ and $ (3, 4) $.

Step 4(c) It changes direction at $ t = 2 $ and $ t = 4 $, so split the journey there: $ s(0) = 0 $, $ s(2) = 20 $, $ s(4) = 16 $, $ s(5) = 20 $.

Step 5Distance $ = |20-0| + |16-20| + |20-16| = 20 + 4 + 4 = 28 $ m.

The displacement, $ s(5)-s(0) = 20 $ m, is not the distance. Figures 2 and 3 show the whole story.

Position of the particle: s = t³ − 9t² + 24t m

s(t)

Figure 2. The particle moves right to s = 20, back left to s = 16, then right again to s = 20. Its displacement over 0 ≤ t ≤ 5 is 20 m, but it travels 20 + 4 + 4 = 28 m.

Velocity and acceleration of the particle

v = 3t² − 18t + 24 (m/s)a = 6t − 18 (m/s²)

Figure 3. Gold bands: v and a have the same sign, so the speed is increasing. On 2 < t < 3 both are negative; on 4 < t < 5 both are positive. Elsewhere they have opposite signs and the particle is slowing down.

Common mistakeWriting “$ a < 0 $, so the particle is slowing down”. On $ 2 < t < 3 $ above, $ a < 0 $ but the particle speeds up, because $ v < 0 $ too. Always quote both signs: “$ v(2.5) < 0 $ and $ a(2.5) < 0 $, so the speed is increasing.”

Exam tipParticle motion along a line is AP topic 4.2 and a regular pattern in past AP free-response questions, often with $ v(t) $ given instead of $ s(t) $. IB SL 5.9 covers the same kinematics with displacement $ s $, velocity $ v $ and acceleration $ a $. In every syllabus, “distance travelled” means you must find the turning points first.

Related rates: the method

In a related-rates problem (Cambridge calls them connected rates of change) two or more quantities change with time, and an equation links them. You know how fast one changes and want the rate of another. The idea is the one from Step 14: differentiate the linking equation, this time with respect to $ t $. Every variable is a function of time, so each one picks up its own rate by the chain rule, just as $ y $ picked up $ \dfrac{dy}{dx} $.

Key termRelated rates: rates of change with respect to time of quantities linked by an equation. Differentiating the equation with respect to $ t $ gives an equation linking the rates, for instance $ A = \pi r^2 \Rightarrow \dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt} $.

A reliable routine:

Stage	What to do
1. Draw	Sketch the situation. Letters for quantities that change, numbers for those that stay fixed.
2. Name the rates	Write the given rate and the wanted rate as derivatives with respect to $ t $, with units.
3. Link	Find an equation connecting the variables: a volume or area formula, Pythagoras, similar triangles.
4. Differentiate	Differentiate with respect to $ t $, using the chain rule on every variable.
5. Substitute	Only now put in the values at the instant you care about, and solve for the wanted rate.

Worked example

A spherical snowball melts so that its volume decreases at 10 cm³ per minute. How fast is its radius changing when the radius is 5 cm? How fast is its surface area changing then?

Step 1Given $ \dfrac{dV}{dt} = -10 $ cm³/min (negative, because $ V $ is decreasing). Want $ \dfrac{dr}{dt} $ when $ r = 5 $.

Step 2Link: $ V = \tfrac{4}{3}\pi r^3 $. Differentiate with respect to $ t $: $ \dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt} $.

Step 3At $ r = 5 $: $ -10 = 100\pi\dfrac{dr}{dt} $, so $ \dfrac{dr}{dt} = -\dfrac{1}{10\pi} \approx -0.0318 $ cm per minute.

Step 4Surface area $ S = 4\pi r^2 $, so $ \dfrac{dS}{dt} = 8\pi r\dfrac{dr}{dt} = 40\pi \times \left(-\dfrac{1}{10\pi}\right) = -4 $ cm² per minute.

When the radius is 5 cm, it is shrinking at about 0.032 cm per minute and the surface area is shrinking at 4 cm² per minute.

Worked example

A plane flies horizontally at a height of 6 km, at 500 km/h, directly away from a radar station on the ground. How fast is its distance from the station increasing when it is 8 km away horizontally?

Step 1Let $ x $ be the horizontal distance and $ z $ the direct distance. Given $ \dfrac{dx}{dt} = 500 $ km/h; want $ \dfrac{dz}{dt} $ when $ x = 8 $. The height, 6 km, is fixed.

Step 2Pythagoras: $ z^2 = x^2 + 36 $. Differentiate: $ 2z\dfrac{dz}{dt} = 2x\dfrac{dx}{dt} $.

Step 3When $ x = 8 $, $ z = \sqrt{64 + 36} = 10 $. So $ 10\dfrac{dz}{dt} = 8 \times 500 $, giving $ \dfrac{dz}{dt} = 400 $ km/h.

The distance grows more slowly than the plane flies, because part of its motion is “sideways” to the line of sight. Far away, $ \dfrac{dz}{dt} $ approaches 500 km/h.

Related rates with similar shapes, and in economics

When a container is filling, the linking equation often has one variable too many. Use similar triangles to remove it before differentiating.

Worked example

A water trough is 4 m long. Its cross-section is an isosceles triangle, vertex down, 1 m wide at the top and 0.5 m deep. Water flows in at 0.2 m³ per minute. How fast is the water level rising when the water is 0.25 m deep?

Step 1Let the depth be $ h $ and the width of the water surface $ w $. Given $ \dfrac{dV}{dt} = 0.2 $; want $ \dfrac{dh}{dt} $ when $ h = 0.25 $.

Step 2Similar triangles: $ \dfrac{w}{h} = \dfrac{1}{0.5} $, so $ w = 2h $.

Step 3Volume = cross-section area × length $ = \tfrac{1}{2}wh \times 4 = \tfrac{1}{2}(2h)(h)(4) = 4h^2 $.

Step 4Differentiate: $ \dfrac{dV}{dt} = 8h\dfrac{dh}{dt} $. At $ h = 0.25 $: $ 0.2 = 2\dfrac{dh}{dt} $, so $ \dfrac{dh}{dt} = 0.1 $ m per minute.

As the trough fills, the surface gets wider, so the same inflow raises the level more slowly.

Common mistakeSubstituting the instant’s values too early. If you put $ h = 0.25 $ into $ V = 4h^2 $ first, $ V $ becomes the constant 0.25 and its derivative is 0, which is nonsense. Substitute values for changing quantities only after differentiating. Fixed quantities (the 4 m length, the 6 km height) can go in at any time.

An economics example: is revenue rising or falling? Demand for a product is $ Q = 1200-30P $, and the firm is raising its price by $0.40 a week. Differentiating the demand equation with respect to time, $ \dfrac{dQ}{dt} = -30\dfrac{dP}{dt} = -12 $ units per week. Revenue is $ R = PQ $, so the product rule gives

\[ \frac{dR}{dt} = P\frac{dQ}{dt} + Q\frac{dP}{dt}. \]

When $ P = 25 $, $ Q = 450 $, and $ \dfrac{dR}{dt} = 25(-12) + 450(0.4) = -300 + 180 = -120 $ dollars per week. Revenue is falling: the $180 gained from charging more on the units still sold is outweighed by the $300 lost on the units no longer sold. That fits the elasticity there, $ \varepsilon = -30 \times \dfrac{25}{450} \approx -1.67 $. Demand is elastic, so a price rise cuts revenue.

Exam tipRelated rates are AP topics 4.4 and 4.5, Cambridge 9709 Paper 1 (topic 1.7, “connected rates of change”, usually through $ \dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt} $) and IB HL 5.14. Give units with every rate, and use the sign: a negative $ \dfrac{dV}{dt} $ for a draining tank or a melting snowball is part of the answer.

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. The edge of a cube is increasing at 0.5 cm per second. How fast is its volume increasing when the edge is 4 cm? Show worked answer

$ V = x^3 $, so $ \dfrac{dV}{dt} = 3x^2\dfrac{dx}{dt} $.
At $ x = 4 $: $ 3 \times 16 \times 0.5 = 24 $.
The volume is increasing at 24 cm³ per second.

EasyQ2. A particle has position $ s(t) = t^2-6t + 5 $ metres at time $ t $ seconds. Find its velocity and acceleration at $ t = 4 $, and say whether it is speeding up or slowing down then. Show worked answer

$ v = 2t-6 $, so $ v(4) = 2 $ m/s.
$ a = 2 $ m/s² for all $ t $.
$ v(4) > 0 $ and $ a(4) > 0 $: same sign, so the particle is speeding up.

MediumQ3. An oil slick stays circular while its area grows at 50 m² per minute. How fast is the radius increasing when the radius is 10 m? Show worked answer

$ A = \pi r^2 $, so $ \dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt} $.
At $ r = 10 $: $ 50 = 20\pi\dfrac{dr}{dt} $.
$ \dfrac{dr}{dt} = \dfrac{5}{2\pi} \approx 0.796 $ m per minute.

MediumQ4. A particle has position $ s(t) = 2t^3-15t^2 + 36t $ metres for $ 0 \leq t \leq 4 $ seconds. Find the total distance it travels. Show worked answer

$ v = 6t^2-30t + 36 = 6(t-2)(t-3) $, so it changes direction at $ t = 2 $ and $ t = 3 $.
$ s(0) = 0 $, $ s(2) = 28 $, $ s(3) = 27 $, $ s(4) = 32 $.
Distance $ = 28 + 1 + 5 = 34 $ m. (The displacement is only 32 m.)

MediumQ5. A firm’s daily cost is $ C = 500 + 8Q + 0.01Q^2 $ dollars when it produces $ Q $ units a day, and its daily output $ Q $ is rising by 20 units per week. How fast is daily cost rising, per week, when $ Q = 300 $? Show worked answer

$ \dfrac{dC}{dt} = \dfrac{dC}{dQ} \times \dfrac{dQ}{dt} = (8 + 0.02Q) \times 20 $.
At $ Q = 300 $: marginal cost $ = 8 + 6 = 14 $ dollars per unit.
$ \dfrac{dC}{dt} = 14 \times 20 = 280 $.
Daily cost is rising at $280 per week.

ExamQ6. A hot-air balloon rises vertically from the ground at a constant 3 m/s. An observer stands on level ground 40 m from the launch point. At time $ t $ seconds the balloon is $ h $ m high, its distance from the observer is $ z $ m, and its angle of elevation from the observer is $ \theta $ radians. (a) Find $ \dfrac{dz}{dt} $ when $ h = 30 $. [3] (b) Find $ \dfrac{d\theta}{dt} $ when $ h = 30 $, with units. [4] (c) Is $ \theta $ increasing faster when $ h = 30 $ or at the moment of launch? Justify your answer. [2] Show worked answer

(a) $ z^2 = h^2 + 1600 $, so $ 2z\dfrac{dz}{dt} = 2h\dfrac{dh}{dt} $. [M1]
At $ h = 30 $: $ z = 50 $. [B1]
$ 50\dfrac{dz}{dt} = 30 \times 3 $, so $ \dfrac{dz}{dt} = 1.8 $ m/s. [A1]
(b) $ \tan\theta = \dfrac{h}{40} $, so $ \sec^2\theta\,\dfrac{d\theta}{dt} = \dfrac{1}{40}\dfrac{dh}{dt} $. [M1 A1]
At $ h = 30 $: $ \tan\theta = \tfrac{3}{4} $, so $ \sec^2\theta = 1 + \tfrac{9}{16} = \tfrac{25}{16} $. [M1]
$ \dfrac{d\theta}{dt} = \dfrac{3}{40} \times \dfrac{16}{25} = 0.048 $ radians per second. [A1]
(c) At launch $ h = 0 $, $ \sec^2\theta = 1 $, so $ \dfrac{d\theta}{dt} = \dfrac{3}{40} = 0.075 $ rad/s. [M1]
$ 0.075 > 0.048 $, so $ \theta $ increases faster at launch. [A1]
Total: 9 marks.

Check yourself

1. At some instant a particle has $ v = -3 $ m/s and $ a = -2 $ m/s². What is happening to its speed?

$ v $ and $ a $ have the same sign, so the velocity is moving away from zero and the speed $ |v| $ is increasing.

2. A circle’s radius grows at 1 cm per second. How fast is its area growing when the radius is 3 cm?

$ \dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt} = 2\pi(3)(1) = 6\pi $. The value $ 9\pi $ is the area itself, not its rate of change.

3. Position $ s $ is in metres and time $ t $ in seconds. What are the units of $ \dfrac{d^2s}{dt^2} $?

$ \dfrac{ds}{dt} $ is metres per second; differentiating again with respect to time gives metres per second per second, m/s².

Your score: 0 out of 3

The short version

A derivative is a rate in units of $ y $ per unit of $ x $. Interpret it with the instant, the quantity, the direction and the rate with units.
Motion: $ v = s’ $, $ a = v’ = s” $, speed $ = |v| $. The particle changes direction where $ v $ changes sign.
Speeding up when $ v $ and $ a $ share a sign; slowing down when they differ. Total distance means splitting the journey at every change of direction.
Related rates: draw, name the rates, link the variables, differentiate with respect to $ t $, and only then substitute.
Economics: marginal cost $ C'(Q) $ approximates the cost of one more unit, and $ \dfrac{dR}{dt} = P\dfrac{dQ}{dt} + Q\dfrac{dP}{dt} $ splits a revenue change into its quantity and price effects.
The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.

Step 15: Rates of Change, Motion and Related Rates

A derivative in context

Motion along a line

Related rates: the method

Related rates with similar shapes, and in economics

Practise: 6 questions

Check yourself

The short version

Further reading on EconTutorials

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