Related Rates Problems: Sliding Ladders, Filling Cones and Falling Demand

In 30 seconds

A related rates problem links two or more quantities that all change with time, and uses the rate you know to find the rate you want.
The recipe: draw and label, write an equation linking the quantities, differentiate both sides with respect to $ t $, then substitute the numbers for that instant.
Substitute last. Values that are true only at one moment must go in after differentiating, never before.
The sign of the answer carries meaning: a negative $ \dfrac{dy}{dt} $ means $ y $ is falling.
Exam point: related rates appear in AP Calculus Unit 4, Cambridge 9709 Paper 1 (“connected rates of change”) and IB AA HL.

The case: a slipping ladder on a building site

A decorator leans a 6.5 m ladder against a vertical wall on level ground. The rubber feet are worn, and the base starts to slide away from the wall at a steady 0.3 m/s. The top slides down the wall as a result. The site supervisor wants to know: how fast is the top of the ladder moving down the wall at the moment the base is 2.5 m from the wall? And is that speed steady, or does it change as the ladder slips further?

The numbers are illustrative, but the situation is the classic one. We know how fast one thing is changing (the base) and want how fast a connected thing is changing (the top). Neither speed is given as a formula in time, so we cannot just differentiate a function of $ t $. What we do have is a fixed geometric link between the two positions.

The concept: rates that are tied together

Call the distance from the wall to the base $ x $ and the height of the top $ y $. Both change as time passes, so both are really functions of time, $ x(t) $ and $ y(t) $. The ladder’s length does not change, and Pythagoras ties the two together at every instant:

\[ x^2 + y^2 = 6.5^2 \]

If an equation links $ x $ and $ y $ at every moment, then the way they change must be linked too. Push the base out a little and the top must drop a little. Differentiating the equation with respect to time turns the link between the quantities into a link between their rates. That is the whole idea, and it is just the chain rule applied to a function of time^[5].

Key termRelated rates: rates of change, usually with respect to time, of quantities that are connected by an equation. Differentiating the connecting equation with respect to $ t $ gives an equation between the rates, which lets one unknown rate be found from the others.

The formula

The engine of every related rates problem is the chain rule written for time^[1]:

\[ \frac{d}{dt}\big[f(y)\big] = f'(y)\,\frac{dy}{dt} \]

Here $ y $ is any quantity that depends on time $ t $, $ f $ is a function of it (such as $ y^2 $ or $ \tfrac{1}{3}\pi y^3 $), and $ \dfrac{dy}{dt} $ is its rate of change. In words: differentiate as usual, then multiply by the rate of whatever was inside. So $ \dfrac{d}{dt}(x^2) = 2x\dfrac{dx}{dt} $, not just $ 2x $. The Cambridge syllabus often phrases the same idea as $ \dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt} $^[3].

Applied to the ladder, differentiating $ x^2 + y^2 = 42.25 $ term by term gives

\[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \]

The right-hand side is 0 because the length 6.5 m is a constant.

Chapter 11 of Mathematics Made Simple covers implicit differentiation and related rates, one idea at a time.Get the textbook · $12

Why this method, and when it is not needed

The method fits because the case gives us a relationship rather than a timetable. We do not know $ y $ as a formula in $ t $, but we do not need it. Differentiating the relationship handles any motion of the base, steady or not, and we only ever plug in the values at the instant we care about.

It is not always necessary. If a question already gives a quantity as an explicit function of time, such as $ y = 6-0.2t^2 $, just differentiate it directly. And if one quantity is a simple function of the other, as in a cone where the radius is always a fixed fraction of the depth, it is usually cleaner to use that fact to remove a variable before differentiating, so the equation has only one changing quantity. The second worked example shows this.

Setting it up

Use the same five steps every time:

Draw and label. Give letters to everything that changes ($ x $, $ y $) and write numbers only for what stays fixed (6.5 m).
Write down the rates. Known: $ \dfrac{dx}{dt} = +0.3 $ m/s (positive, because $ x $ is growing). Wanted: $ \dfrac{dy}{dt} $ when $ x = 2.5 $.
Link the quantities with an equation that holds at every moment: $ x^2 + y^2 = 42.25 $.
Differentiate both sides with respect to $ t $, using the chain rule on every changing term.
Now substitute the values for that instant, including any you must work out first, such as $ y $.

Solving it

Worked example

The 6.5 m ladder’s base slides away from the wall at 0.3 m/s. Find the rate at which the top slides down when the base is (a) 2.5 m and (b) 6 m from the wall.

Step 1Link: $ x^2 + y^2 = 42.25 $. Differentiate with respect to $ t $: $ 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0 $.

Step 2Make the wanted rate the subject: $ \dfrac{dy}{dt} = -\dfrac{x}{y}\,\dfrac{dx}{dt} $.

Step 3(a) Find $ y $ at this instant: $ y = \sqrt{42.25-2.5^2} $ $ = \sqrt{42.25-6.25} $ $ = \sqrt{36} = 6 $ m.

Step 4Substitute: $ \dfrac{dy}{dt} = -\dfrac{2.5}{6} \times 0.3 = -0.125 $ m/s.

Step 5(b) Now $ x = 6 $: $ y = \sqrt{42.25-36} = \sqrt{6.25} = 2.5 $ m.

Step 6$ \dfrac{dy}{dt} = -\dfrac{6}{2.5} \times 0.3 = -0.72 $ m/s.

When the base is 2.5 m out, the top is sliding down at 0.125 m/s (12.5 cm per second). By the time the base is 6 m out, the top is dropping at 0.72 m/s, almost six times as fast, even though the base has not sped up at all.

How fast the top of the 6.5 m ladder slides down m/s

speed of top = 0.3x / √(42.25 − x²)

Figure 1. The foot always moves at 0.3 m/s, but the top’s speed depends on where the ladder is. It is slower than the foot until the ladder reaches 45° (x = y), then races ahead as the top nears the floor.

Worked example

A water tank is an inverted cone 5 m deep with a top radius of 2 m. Water is pumped in at 0.4 m³ per minute. How fast is the water level rising when the water is 3 m deep?

Step 1Let the water have depth $ h $ and surface radius $ r $. Known: $ \dfrac{dV}{dt} = 0.4 $. Wanted: $ \dfrac{dh}{dt} $ when $ h = 3 $.

Step 2Similar triangles fix the shape: $ \dfrac{r}{h} = \dfrac{2}{5} $, so $ r = \dfrac{2h}{5} $ at every depth.

Step 3Remove $ r $ from the volume formula: $ V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi \left(\dfrac{4h^2}{25}\right) h = \dfrac{4\pi h^3}{75} $.

Step 4Differentiate with respect to $ t $: $ \dfrac{dV}{dt} = \dfrac{12\pi h^2}{75}\,\dfrac{dh}{dt} = \dfrac{4\pi h^2}{25}\,\dfrac{dh}{dt} $.

Step 5Substitute $ h = 3 $: $ 0.4 = \dfrac{4\pi (9)}{25}\,\dfrac{dh}{dt} = \dfrac{36\pi}{25}\,\dfrac{dh}{dt} $.

Step 6$ \dfrac{dh}{dt} = \dfrac{0.4 \times 25}{36\pi} = \dfrac{10}{36\pi} \approx 0.0884 $ m per minute.

At 3 m deep the level is rising at about 0.088 m, or 8.8 cm, per minute. A useful check: $ \dfrac{4\pi h^2}{25} $ is $ \pi r^2 $, the area of the water surface. So the formula says inflow = surface area × rate of rise, which makes physical sense.

Worked example

(Exam style, economics.) A shop finds that weekly demand for a gadget is $ q = 5000-2p^2 $, where $ p $ is the price in dollars. The price is being raised by $0.50 a week. At the moment when $ p = 30 $, find the rate of change of (a) the quantity demanded and (b) weekly revenue $ R = pq $.

Step 1Known: $ \dfrac{dp}{dt} = 0.5 $. At $ p = 30 $: $ q = 5000-2(900) = 3200 $.

Step 2(a) Differentiate the demand equation with respect to $ t $: $ \dfrac{dq}{dt} = -4p\,\dfrac{dp}{dt} $.

Step 3$ \dfrac{dq}{dt} = -4(30)(0.5) = -60 $ units per week.

Step 4(b) Revenue is a product, so use the product rule: $ \dfrac{dR}{dt} = q\,\dfrac{dp}{dt} + p\,\dfrac{dq}{dt} $.

Step 5$ \dfrac{dR}{dt} = 3200(0.5) + 30(-60) $ $ = 1600-1800 = -200 $ dollars per week.

Sales are falling by 60 units a week, and revenue is falling by $200 a week even though the price is rising. The extra 50 cents on each unit still sold (+$1600) is outweighed by the lost sales (−$1800). In economics terms, demand is price-elastic here: the elasticity is $ \dfrac{dq}{dp}\cdot\dfrac{p}{q} = \dfrac{-120 \times 30}{3200} = -1.125 $.

The Practice Questions book has related rates drills with full working: ladders, cones, shadows and more.Get 1,565 questions · $9

What the answers tell us

Signs and units. Every answer came out with a sign and a unit. The ladder’s $ -0.125 $ m/s is negative because $ y $ is shrinking, so the top is moving down. If you are asked for the speed, quote 0.125 m/s; if asked for the rate of change of the height, keep the minus. The tank’s answer is in metres per minute because the inflow was in cubic metres per minute.

Is the ladder answer reasonable? At $ x = 2.5 $ the ladder is steep, nearly upright. Sliding the base out a little barely lowers the top, so the top should move more slowly than the base, and 0.125 is less than 0.3. Near the floor the ladder is almost flat, and a small slide of the base makes the top plunge. The ratio $ x/y $ controls everything: the two ends move at the same speed exactly when $ x = y $, at about 4.60 m, when the ladder makes 45° with the ground (Figure 1).

Sensitivity. The formula $ \dfrac{dy}{dt} = -\dfrac{x}{y}\dfrac{dx}{dt} $ grows without limit as $ y \to 0 $. The real ladder does not reach infinite speed: the model assumes the top stays against the wall, which stops being true just before it hits the ground. For the tank, $ \dfrac{dh}{dt} = \dfrac{2.5}{\pi h^2} $, so doubling the depth cuts the rate of rise to a quarter (Figure 2). That is why a cone-shaped glass seems to fill quickly at first, then slowly.

Rate the water level rises in the conical tank m per minute

dh/dt = 2.5 / (πh²)

Figure 2. With a steady inflow of 0.4 m³ per minute, the level shoots up while the cone is narrow and crawls near the top, where the water surface is widest.

The full chapter on implicit differentiation, inverse functions and related rates, with plenty of worked examples, is in the Mathematics Made Simple textbook.

Common mistakeSubstituting too early. If you put $ x = 2.5 $ into $ x^2 + y^2 = 42.25 $ before differentiating, $ x $ becomes a constant and its rate disappears, giving $ \dfrac{dy}{dt} = 0 $. Only the fixed length (6.5 m) can go in at the start. Values that are true just at one instant go in after differentiating.

Common mistakeForgetting the chain rule factor. $ \dfrac{d}{dt}(y^2) $ is $ 2y\dfrac{dy}{dt} $, not $ 2y $. Every term that contains a changing quantity must pick up that quantity’s rate.

Exam tipRelated rates are AP Calculus topics 4.4 and 4.5 in the AB and BC course^[2], where free-response questions often give a rate in context and ask for another “at the instant when”. In Cambridge 9709 Paper 1 they appear as connected rates of change^[3], and in IB Mathematics: analysis and approaches they sit in HL topic 5.14^[4]. Write the list of known and wanted rates with units before any calculus; it earns method marks and stops sign errors.

Get both Mathematics Made Simple books for $16 and practise related rates until they feel routine.Both books · $16

Practice questions

Try each question on paper before opening the answer. For hundreds more, all with full working, see the Mathematics Made Simple practice questions book.

EasyQ1. Water is poured into a cylindrical glass of radius 4 cm at 20 cm³ per second. How fast is the water level rising? Show worked answer

$ V = \pi r^2 h = 16\pi h $.
Differentiate: $ \dfrac{dV}{dt} = 16\pi\dfrac{dh}{dt} $.
$ \dfrac{dh}{dt} = \dfrac{20}{16\pi} \approx 0.398 $ cm/s, at every depth (the glass has straight sides).

EasyQ2. $ y = x^3 + 2x $, and $ x $ is increasing at 4 units per second. Find $ \dfrac{dy}{dt} $ when $ x = 1 $. Show worked answer

$ \dfrac{dy}{dt} = (3x^2 + 2)\dfrac{dx}{dt} $.
At $ x = 1 $: $ (3 + 2) \times 4 = 20 $ units per second.

MediumQ3. A 1.8 m tall person walks away from a 6 m street lamp at 1.5 m/s. How fast is the length of their shadow increasing? Show worked answer

Let $ x $ be the distance from the lamp and $ s $ the shadow length. Similar triangles: $ \dfrac{s}{1.8} = \dfrac{x + s}{6} $.
So $ 6s = 1.8x + 1.8s $, giving $ 4.2s = 1.8x $ and $ s = \dfrac{3x}{7} $.
$ \dfrac{ds}{dt} = \dfrac{3}{7} \times 1.5 \approx 0.643 $ m/s, whatever the distance.

MediumQ4. Two cyclists leave a crossroads at the same time, one riding north at 15 km/h and one east at 20 km/h. How fast is the distance between them increasing when the first is 3 km and the second 4 km from the crossroads? Show worked answer

Let north distance $ a $, east distance $ b $, gap $ D $: $ D^2 = a^2 + b^2 $.
Differentiate: $ 2D\dfrac{dD}{dt} = 2a\dfrac{da}{dt} + 2b\dfrac{db}{dt} $.
At the instant, $ D = \sqrt{9 + 16} = 5 $ km.
$ \dfrac{dD}{dt} = \dfrac{3(15) + 4(20)}{5} = \dfrac{125}{5} = 25 $ km/h.

ExamQ5. A funnel is an inverted cone of height 12 cm and top radius 4 cm. (a) Show that when the water is $ h $ cm deep its volume is $ V = \dfrac{\pi h^3}{27} $. [2] (b) Water drips out of the funnel at 3 cm³ per second. Find the rate of change of the depth when $ h = 9 $, to 3 significant figures. [3] (c) Find the rate of change of the area of the water surface at that instant. [3] Show worked answer

(a) Similar triangles: $ \dfrac{r}{h} = \dfrac{4}{12} $, so $ r = \dfrac{h}{3} $ [1]. $ V = \tfrac{1}{3}\pi \left(\dfrac{h}{3}\right)^2 h = \dfrac{\pi h^3}{27} $ [1].
(b) $ \dfrac{dV}{dt} = \dfrac{\pi h^2}{9}\dfrac{dh}{dt} $ [1], with $ \dfrac{dV}{dt} = -3 $ [1].
At $ h = 9 $: $ -3 = 9\pi\dfrac{dh}{dt} $, so $ \dfrac{dh}{dt} = -\dfrac{1}{3\pi} \approx -0.106 $ cm/s [1].
(c) $ A = \pi r^2 = \dfrac{\pi h^2}{9} $ [1]. $ \dfrac{dA}{dt} = \dfrac{2\pi h}{9}\dfrac{dh}{dt} $ [1] $ = 2\pi \times \left(-\dfrac{1}{3\pi}\right) = -\dfrac{2}{3} \approx -0.667 $ cm²/s [1].

Frequently asked questions

What is the first step in a related rates problem?

Draw a diagram and give letters to every quantity that changes. Then write down, with units and signs, the rate you are given and the rate you want. Most errors come from skipping this and mixing up which quantity is which.

Why can I not substitute the numbers straight away?

Because a value like $ x = 2.5 $ is true for only one instant. If you put it in before differentiating, you treat $ x $ as a constant and lose its rate. Fixed quantities, such as a ladder’s length or a cone’s shape, can go in at any time.

Why is my answer negative?

A negative rate means the quantity is decreasing. That is often exactly right: the top of a slipping ladder goes down and a leaking funnel empties. Keep the sign when asked for a rate of change, and drop it only when the question asks for a speed.

Do related rates need implicit differentiation?

Yes, in effect. Differentiating $ x^2 + y^2 = 42.25 $ with respect to $ t $ is implicit differentiation with time as the variable. If you are comfortable finding $ \dfrac{dy}{dx} $ for a circle, related rates use the same skill.

References

Strang, G. and Herman, E. (2016) Calculus Volume 1, section 4.1 “Related Rates”. Houston, TX: OpenStax. https://openstax.org/details/books/calculus-volume-1
College Board (2020) AP Calculus AB and AP Calculus BC Course and Exam Description, Unit 4 “Contextual Applications of Differentiation”. New York: College Board.
Cambridge University Press & Assessment. Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. cambridgeinternational.org
International Baccalaureate Organization (2019) Mathematics: analysis and approaches guide (first assessment 2021). Geneva: IBO.
Stewart, J. (2016) Calculus: Early Transcendentals, 8th edn, section 3.9 “Related Rates”. Boston, MA: Cengage Learning.

Related Rates Problems: Sliding Ladders, Filling Cones and Falling Demand

The case: a slipping ladder on a building site

The concept: rates that are tied together

The formula

Why this method, and when it is not needed

Setting it up

Solving it

What the answers tell us

Practice questions

Frequently asked questions

What is the first step in a related rates problem?

Why can I not substitute the numbers straight away?

Why is my answer negative?

Do related rates need implicit differentiation?

References

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