Related Rates: Step-by-Step Strategy and Classic Problems

Q: What's the link to the chain rule?

Every variable is a function of time, so (d)/(dt)x^2 = 2x(dx)/(dt) . That's the chain rule.

Yasir Hafeez September 29, 20267 min read

In a related rates problem, several quantities change over time and are linked by an equation. You know how fast some of them change and want the rate of another. The tool is implicit differentiation with respect to time $ t $.

This guide gives you a strategy that works on every problem, the intuition behind it, six fully solved classics from easy to exam-level, and the mistakes that cost the most points.

Why it works

Think of each quantity in the problem as a function of time: the ladder’s base position is $ x(t) $, its height is $ y(t) $. The equation linking them, such as $ x^2 + y^2 = 100 $, holds at every instant, not just one. Two functions that are equal for all $ t $ must also have equal derivatives, so you can differentiate both sides with respect to $ t $.

The chain rule is what turns the result into an equation about rates. Differentiating $ x^2 $ with respect to $ t $ gives $ 2x\frac{dx}{dt} $, not $ 2x $, because $ x $ itself depends on $ t $. The new equation says: at any instant, the rates are tied together in the same way the quantities are.

A good mental picture is a snapshot. The relationship holds for the whole motion, so you differentiate it while everything is still a variable. Then you freeze the moment the question asks about and plug in that snapshot’s numbers.

The 6-step strategy

Draw a picture and label every changing quantity with a variable.
Write down the given rates and the rate you want, as derivatives like $ \frac{dx}{dt} $.
Find an equation linking the variables (Pythagoras, area, volume, similar triangles).
Differentiate both sides with respect to $ t $, using the chain rule on every variable.
Substitute the values at the specific instant. Only now plug in numbers.
Solve for the unknown rate and include units.

The biggest mistake is plugging in numbers before differentiating in step 4. Values that change must stay as variables until after you differentiate.

Problem 1: the sliding ladder

A 10 m ladder leans against a wall. The bottom slides away at 3 m/s. How fast is the top sliding down when the bottom is 8 m from the wall?

Variables: $ x $ = distance of the bottom from the wall, $ y $ = height of the top.
Equation: $ x^2 + y^2 = 100 $.
Differentiate: $ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 $.
At the instant: $ x = 8 $, so $ y = 6 $, and $ \frac{dx}{dt} = 3 $.

$$\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} = -\frac{8\cdot3}{6} = -4\ \text{m/s}$$

The top is sliding down at 4 m/s (negative means $ y $ is decreasing).

Problem 2: the inflating balloon

Air is pumped into a spherical balloon at $ 100\ \text{cm}^3/\text{s} $. How fast is the radius growing when $ r = 5 $ cm?

$$V = \frac43\pi r^3 \quad\Longrightarrow\quad \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$$

$$\frac{dr}{dt} = \frac{100}{4\pi(25)} = \frac{1}{\pi} \approx 0.318\ \text{cm/s}$$

Problem 3: water in a cone

Water pours at $ 2\ \text{m}^3/\text{min} $ into an upside-down cone with height 4 m and top radius 2 m. How fast is the water level rising when the water is 2 m deep?

By similar triangles, $ \frac{r}{h} = \frac24 $, so $ r = \frac h2 $. Eliminate $ r $ before differentiating:

$$V = \frac13\pi r^2h = \frac{\pi h^3}{12} \quad\Longrightarrow\quad \frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}$$

At $ h = 2 $, the factor $ \frac{\pi h^2}{4} $ equals $ \pi $, so

$$2 = \pi\,\frac{dh}{dt} \quad\Longrightarrow\quad \frac{dh}{dt} = \frac{2}{\pi} \approx 0.64\ \text{m/min}$$

Problem 4: the moving shadow

A 6 ft person walks away from a 15 ft lamppost at 5 ft/s. How fast does the tip of the shadow move?

Let $ x $ be the person’s distance from the post and $ s $ the shadow length. Similar triangles: $ \frac{15}{x + s} = \frac{6}{s} \Rightarrow 15s = 6x + 6s \Rightarrow s = \frac23x $.

The tip is at $ x + s = \frac53x $, so its speed is $ \frac53\cdot5 = \frac{25}{3} \approx 8.33 $ ft/s, regardless of where the person is.

Problem 5: two cars and a changing distance

Car A is 6 mi north of an intersection, driving south toward it at 40 mph. Car B is 8 mi east of the intersection, driving east at 60 mph. How fast is the distance between them changing?

Variables: $ y $ = car A’s distance north, $ x $ = car B’s distance east, $ z $ = distance between the cars.
Rates: $ \frac{dy}{dt} = -40 $ (A gets closer, so $ y $ decreases) and $ \frac{dx}{dt} = 60 $.
Equation: $ z^2 = x^2 + y^2 $.
Differentiate: $ 2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt} $, then cancel the 2s.
At the instant: $ x = 8 $, $ y = 6 $, so $ z = 10 $.

$$\frac{dz}{dt} = \frac{8(60) + 6(-40)}{10} = \frac{240}{10} = 24\ \text{mph}$$

The distance is growing at 24 mph. The sign of $ \frac{dy}{dt} $ matters: forget the minus and you’d get 72 mph.

Problem 6: angle of elevation

A balloon rises straight up at 5 m/s from a point 100 m away from an observer. How fast is the angle of elevation $ \theta $ changing when the balloon is 100 m high?

Equation: $ \tan\theta = \frac{h}{100} $.
Differentiate (the derivative of tan is $ \sec^2 $): $ \sec^2\theta\,\frac{d\theta}{dt} = \frac{1}{100}\frac{dh}{dt} $.
At the instant: $ h = 100 $, so $ \tan\theta = 1 $, $ \theta = \frac\pi4 $, and $ \sec^2\theta = 2 $.

$$\frac{d\theta}{dt} = \frac{5}{100\cdot2} = 0.025\ \text{rad/s}$$

Angle rates come out in radians per unit time, because the derivative of $ \tan\theta $ is $ \sec^2\theta $ only when $ \theta $ is in radians.

Common mistakes

Plugging in too early. If you write $ x = 8 $ before differentiating, its derivative becomes 0 and the problem collapses. Substitute only after step 4.
Wrong signs. A quantity that shrinks has a negative rate. Water draining, a car approaching, and a ladder top falling all need a minus sign.
Dropping the chain rule. Every variable that changes gets its own rate: $ \frac{d}{dt}(xy) = x\frac{dy}{dt} + y\frac{dx}{dt} $ by the product rule, not $ \frac{dx}{dt}\frac{dy}{dt} $.
Keeping an extra variable. In the cone problem, $ r $ changes too. Eliminate it with similar triangles before differentiating, or you’ll have an unknown rate you can’t find.
Missing units. Rates have units of “quantity per time”: m/s, $ \text{cm}^3/\text{s} $, rad/s. Include them in the final answer.

Where it’s used

Related rates show up anywhere two changing quantities are linked by geometry or physics: radar and tracking, fluid flow into tanks, expanding ripples, and electrical circuits where resistances change. The same differentiate-then-substitute idea appears when you study motion with the position, velocity and acceleration calculator, in linear approximation, and in optimization problems, which also start by writing an equation that links the quantities.

Differentiate the relationship with the solver

Step 4 is ordinary differentiation, so the derivative calculator can check it. For the cone, $ V(h) = \frac{\pi h^3}{12} $ (type it with $ x $ as $ h $):

Step-by-step solverExact symbolic engine

Interactive Calculus Problem Solver

Derivatives, antiderivatives, definite and improper integrals, and limits. Every answer comes with the rules used, and antiderivatives are verified by differentiating them back.

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Powers x^2, roots sqrt(x), cbrt(x), absolute value |x|
Implicit multiplication works: 3x sin(2x)
sin cos tan sec csc cot, asin acos atan, sinh cosh tanh
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Enter a function and press Solve to see a full worked solution.

Practice problems

Try these before checking the answers.

Circle radius grows at 2 cm/s. $ \frac{dA}{dt} $ when $ r = 5 $?
Cube side grows at 1 cm/s. $ \frac{dV}{dt} $ when $ s = 3 $?
Ladder of Problem 1 when the bottom is 6 m out
Balloon of Problem 2 when $ r = 10 $ cm
A square’s area grows at $ 12\ \text{cm}^2/\text{s} $. How fast is its side growing when the side is 3 cm?
A 26 ft ladder slides so its top falls at 2 ft/s. How fast is the bottom moving when the top is 10 ft high?

Answers: (1) $ 20\pi\ \text{cm}^2/\text{s} $; (2) $ 27\ \text{cm}^3/\text{s} $; (3) $ \frac{dy}{dt} = -2.25\ \text{m/s} $; (4) $ \frac{1}{4\pi} \approx 0.0796\ \text{cm/s} $; (5) $ 2\ \text{cm/s} $; (6) $ \frac56 $ ft/s away from the wall (the bottom is 24 ft out).

FAQ

Why are some rates negative?

A negative rate means the quantity is decreasing, like the height of the ladder’s top.

What’s the link to the chain rule?

Every variable is a function of time, so $ \frac{d}{dt}x^2 = 2x\frac{dx}{dt} $. That’s the chain rule.

When do I plug in the numbers?

Only after differentiating. Quantities that change must stay as variables until the derivative is taken. Constants, like the ladder’s 10 m length, can go in at the start.

How do I recognize a related rates problem?

Look for two or more rates in the wording, usually phrases like “how fast,” “at what rate,” or “per second,” along with a specific instant (“when the radius is 5 cm”).

What if I have two unknown rates?

Find another equation. Usually a geometric fact such as similar triangles or a fixed length lets you eliminate one variable, as in the cone and shadow problems.

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