A related-rates problem hands you the rate at which one quantity is changing and asks for the rate at which a geometrically or physically linked quantity is changing at the same instant. The only calculus involved is the chain rule: if two time-varying quantities satisfy an equation, differentiating that equation with respect to $t$ produces an equation relating their rates. This is where the chain rule and implicit differentiation stop being formal exercises and start answering questions about the world, and it rehearses the modelling habits — draw, name, relate — that optimisation will need next.

The underlying principle

Every quantity in the problem is silently a function of time. Write $x=x(t)$ , $V=V(t)$ , $\theta=\theta(t)$ and never suppress the $t$ .

Theorem (chain rule, time form). If $x$ is differentiable at $t_0$ and $f$ is differentiable at $x(t_0)$ , then $Q(t)=f\big(x(t)\big)$ is differentiable at $t_0$ and $\frac{dQ}{dt}\Big|_{t_0}=f'\big(x(t_0)\big)\cdot\frac{dx}{dt}\Big|_{t_0}.$

Implicit (constraint) form. Suppose $x$ and $y$ are differentiable on an open interval $I$ of times, the expression $F$ is differentiable at each point $\big(x(t),y(t)\big)$ — in practice $F$ is built from sums, products, quotients (with non-vanishing denominator), powers and trigonometric functions, so it is — and the constraint $F\big(x(t),y(t)\big)=c\qquad\textbf{for every } t\in I$ holds. Then the two sides are the same function of $t$ , so their derivatives agree on $I$ . Differentiating the left side with the product, quotient, power and chain rules — treating each letter as a function of $t$ — gives one linear equation in $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ . Solve it for the rate you want, which is legitimate only if the coefficient you divide by is non-zero at that instant: in $x\dfrac{dx}{dt}+y\dfrac{dy}{dt}=0$ the step $\dfrac{dy}{dt}=-\dfrac xy\dfrac{dx}{dt}$ requires $y\ne 0$ . At $y=0$ the equation collapses to $x\dfrac{dx}{dt}=0$ and says nothing whatever about $\dfrac{dy}{dt}$ ; and in the sliding-ladder model, where the foot is pulled at a constant $\dfrac{dx}{dt}>0$ and $y=\sqrt{L^2-x^2}$ , one has $\dfrac{dy}{dt}=-\dfrac{x}{\sqrt{L^2-x^2}}\dfrac{dx}{dt}\to-\infty$ as $x\to L^-$ , so no finite $\dfrac{dy}{dt}$ exists at the flat position.

Note the three hypotheses that are doing real work: differentiability of each variable, the constraint holding on an interval of times rather than at one instant, and a non-vanishing coefficient at the moment you divide.

The procedure

Draw a picture of a general instant, not the special one.
Label. Put a number on every length that never changes; put a letter on every quantity that changes. This single habit prevents most errors.
Name the variables as functions of $t$ , with units: "let $h(t)$ be the water depth in metres, $t$ in minutes."
Record the data as derivatives: what is given ( $\frac{dV}{dt}=-2$ ), what is wanted ( $\frac{dh}{dt}$ when $h=3$ ), and at which instant.
Relate. Find an equation, valid for all nearby $t$ , linking the quantities: Pythagoras, similar triangles, area/volume formulas, trigonometry, the law of cosines, or a physical law.
Reduce. If the relation contains a variable whose rate is neither known nor wanted, eliminate it (similar triangles usually do this) so that the equation involves only the quantities you care about.
Differentiate both sides with respect to $t$ .
Only now substitute the instant's numbers, solve, and state the answer with units and a sign interpretation.

Why you must substitute after differentiating

Step 5 asks for an identity in $t$ on an open interval. Only an identity can be differentiated, because $\frac{d}{dt}$ compares values at nearby times. A number obtained by evaluating a moving quantity at the single instant $t_0$ is not an identity — it is one data point — and substituting it deletes precisely the information the derivative needs.

Concretely, for a $13$ ft ladder with foot $x(t)$ and top $y(t)$ :

Wrong. At the instant $x=5$ : $\;5^2+y^2=169\Rightarrow y^2=144\Rightarrow y=12$ . Differentiate: $\dfrac{d}{dt}(12)=0$ , so $\dfrac{dy}{dt}=0$ — the ladder's top is claimed to be motionless while the foot slides away. Absurd.

Right. Differentiate the identity $x(t)^2+y(t)^2=169$ , which holds for all $t$ while the ladder touches both wall and floor: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\quad\Longrightarrow\quad \frac{dy}{dt}=-\frac{x}{y}\cdot\frac{dx}{dt}.$ Now put in $x=5,\;y=12$ .

Test for legality. A symbol may be replaced by a number before differentiating if and only if it is constant for all $t$ — the ladder's length $13$ , the lamppost's height, the fixed ratio $r/h$ of a cone. Anything that moves stays a letter until after the $\frac{d}{dt}$ .

Standard setups

Situation	Constraint (holds for all $t$ )	Differentiated in $t$
Ladder of fixed length $L$	$x^2+y^2=L^2$	$x\dfrac{dx}{dt}+y\dfrac{dy}{dt}=0$
Expanding circle	$A=\pi r^2$	$\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$
Circumference	$C=2\pi r$	$\dfrac{dC}{dt}=2\pi\dfrac{dr}{dt}$
Sphere, volume	$V=\dfrac43\pi r^3$	$\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$
Sphere, surface	$S=4\pi r^2$	$\dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}$
Cone, apex down, shape ratio $r=kh$	$V=\dfrac{\pi k^2}{3}h^3$	$\dfrac{dV}{dt}=\pi k^2h^2\dfrac{dh}{dt}$
Cylinder, fixed radius $R$	$V=\pi R^2h$	$\dfrac{dV}{dt}=\pi R^2\dfrac{dh}{dt}$
Separation at right angles	$z^2=x^2+y^2$	$z\dfrac{dz}{dt}=x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$
Separation at fixed angle $\alpha$	$z^2=x^2+y^2-2xy\cos\alpha$	$2z\dfrac{dz}{dt}=2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}-2\cos\alpha\left(y\dfrac{dx}{dt}+x\dfrac{dy}{dt}\right)$
Shadow: lamp height $H$ , walker height $p<H$ , walker at $x$ , shadow length $s$	$s(H-p)=p\,x$	$\dfrac{ds}{dt}=\dfrac{p}{H-p}\dfrac{dx}{dt}$
Angle of elevation, height $h$ , horizontal $x$	$\tan\theta=\dfrac{h}{x}$	$\sec^2\theta\dfrac{d\theta}{dt}=\dfrac{x\frac{dh}{dt}-h\frac{dx}{dt}}{x^2}$
Boyle's law (fixed temperature)	$PV=C$	$V\dfrac{dP}{dt}+P\dfrac{dV}{dt}=0$

In the shadow row, the tip of the shadow sits at distance $x+s$ from the lamp's base, so the tip moves at $\dfrac{dx}{dt}+\dfrac{ds}{dt}=\dfrac{H}{H-p}\dfrac{dx}{dt}$ — a different number from the rate the shadow lengthens.

Units and sign conventions

A rate always carries units $\dfrac{\text{unit of the quantity}}{\text{unit of time}}$ : lengths give m/s, areas m $^2$ /s, volumes m $^3$ /min, angles rad/s. Convert everything to one system before substituting; mixing cm with m is the commonest arithmetic disaster.
Fix a positive direction first. Then $\dfrac{dQ}{dt}>0$ means $Q$ is increasing in that direction.
English to symbols: "increasing at $5$ " $\Rightarrow +5$ ; "decreasing at $5$ ", "draining at $5$ ", "falling at $5$ ", "being pulled in at $5$ " $\Rightarrow -5$ . Getting this wrong flips the whole answer.
Speed is the absolute value of a rate. Report either " $\frac{dy}{dt}=-\frac56$ ft/s" or "the top slides down at $\frac56$ ft/s" — never "down at $-\frac56$ ft/s".
Angles must be in radians, because $\frac{d}{d\theta}\sin\theta=\cos\theta$ and $\frac{d}{d\theta}\tan\theta=\sec^2\theta$ are false in degrees. Convert first: $1\text{ rev}=2\pi$ rad, $1^\circ=\frac{\pi}{180}$ rad.
A distance $z=\sqrt{x^2+y^2}$ is $\ge 0$ by definition, so the sign of $\frac{dz}{dt}$ is the whole story: negative means approaching, positive means separating.

Worked example 1 — sliding ladder

A $13$ ft ladder leans against a vertical wall. The foot is pulled away from the wall at a constant $2$ ft/s. How fast is the top sliding down when the foot is $5$ ft from the wall?

Name. Let $x(t)$ be the distance from the wall to the foot and $y(t)$ the height of the top, both in feet, $t$ in seconds. Given $\dfrac{dx}{dt}=+2$ (moving away). Want $\dfrac{dy}{dt}$ when $x=5$ .

Relate. The wall is vertical and the floor horizontal, so for every $t$ with $0<x<13$ , $x(t)^2+y(t)^2=13^2=169 .$

Differentiate (chain rule on each square, the constant $169$ has derivative $0$ ): $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\quad\Longrightarrow\quad\frac{dy}{dt}=-\frac{x}{y}\cdot\frac{dx}{dt}.$

Substitute. At the instant in question $x=5$ , so $y=\sqrt{169-25}=\sqrt{144}=12$ (positive, since $y$ is a height). Hence $\frac{dy}{dt}=-\frac{5}{12}\cdot 2=-\frac{10}{12}=-\frac56\approx-0.833\ \text{ft/s}.$

Answer. The top slides down at $\dfrac56\approx0.83$ ft/s.

Sanity checks. The sign is negative, as it must be when the foot moves out. Units: $\frac{\text{ft}}{\text{ft}}\cdot\frac{\text{ft}}{\text{s}}=\frac{\text{ft}}{\text{s}}$ . And the formula predicts $\frac{dy}{dt}\to-\infty$ as $x\to13^-$ : the top accelerates without bound as the ladder flattens, which is why the idealised model stops describing a real ladder near the end.

Worked example 2 — draining cone (eliminate a variable first)

A tank is an inverted right circular cone, $6$ m deep with a top radius of $2$ m. Water drains out at $2$ m $^3$ /min. How fast is the water level falling when the water is $3$ m deep?

Name. Let $h(t)$ be the depth of water, $r(t)$ the radius of the circular surface, $V(t)$ the volume, in metres and minutes. Draining gives $\dfrac{dV}{dt}=-2$ . Want $\dfrac{dh}{dt}$ when $h=3$ .

Relate. The water forms a cone similar to the tank, so $\dfrac{r}{h}=\dfrac{2}{6}=\dfrac13$ , i.e. $r=\dfrac h3$ for every $t$ . (The ratio $\frac13$ is a genuine constant — it may be substituted before differentiating.) Then $V=\frac13\pi r^2h=\frac13\pi\left(\frac h3\right)^2h=\frac{\pi h^3}{27}.$

Reduce, then differentiate. Reducing to the single variable $h$ before differentiating avoids an unknown $\frac{dr}{dt}$ : $\frac{dV}{dt}=\frac{\pi}{27}\cdot 3h^2\frac{dh}{dt}=\frac{\pi h^2}{9}\cdot\frac{dh}{dt}.$

Substitute $h=3$ and $\dfrac{dV}{dt}=-2$ : $-2=\frac{\pi\cdot 3^2}{9}\cdot\frac{dh}{dt}=\pi\frac{dh}{dt}\quad\Longrightarrow\quad\frac{dh}{dt}=-\frac{2}{\pi}\approx-0.637\ \text{m/min}.$

Answer. The level falls at $\dfrac{2}{\pi}\approx0.64$ m/min.

Sanity check. At $h=1$ the same formula gives $\frac{dh}{dt}=-\dfrac{18}{\pi}\approx-5.7$ m/min: near the apex the cross-section is tiny, so the level must plunge for the same outflow. That qualitative behaviour is a good test of any cone answer.

Common mistakes

Wrong	Right	Why
Putting $x=5$ into $x^2+y^2=169$ and then differentiating	Differentiate, then substitute	Constants have derivative $0$ ; the constraint must hold on an interval of times
$\dfrac{d}{dt}\big(r^3\big)=3r^2$	$3r^2\dfrac{dr}{dt}$	$r$ depends on $t$ — the chain-rule factor is compulsory
$\dfrac{d}{dt}(xy)=\dfrac{dx}{dt}\dfrac{dy}{dt}$	$\dfrac{dx}{dt}\,y+x\,\dfrac{dy}{dt}$	Product rule in $t$
"Draining at $2$ " written as $\dfrac{dV}{dt}=2$	$\dfrac{dV}{dt}=-2$	Draining means the volume decreases
Using $V=\frac13\pi r^2h$ with both $r$ and $h$ varying and only one equation	Eliminate $r=kh$ first	One equation cannot pin down two unknown rates
Feeding $\dfrac{d\theta}{dt}$ in degrees per second into $\sec^2\theta\dfrac{d\theta}{dt}$	Convert to rad/s first	Trig derivative formulas are false in degrees
$\dfrac{dz}{dt}=\dfrac{dx}{dt}+\dfrac{dy}{dt}$ for a separating pair	$z\dfrac{dz}{dt}=x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$	Distance is not the sum of the legs
Reporting "the top moves down at $-0.75$ m/s"	" $\dfrac{dy}{dt}=-0.75$ m/s", i.e. down at $0.75$ m/s	The minus sign and the word "down" say the same thing once
Confusing the rate the shadow lengthens with the speed of its tip	$s'$ versus $x'+s'$	Two different questions about the same picture
Differentiating $\dfrac{H}{x+s}=\dfrac{p}{s}$ as it stands	Cross-multiply to $s(H-p)=px$ first	Clearing fractions turns a quotient rule into one product rule
Answering a question about a $13$ ft ladder "when $x=15$ "	Check the instant is geometrically possible	$x\le 13$ always; an impossible instant signals a misread
Leaving the answer as a bare number	Attach units and say increasing/decreasing	A related-rates answer is a rate, not a number

Fast self-checks

Dimensional check. Every term of the differentiated equation must carry the same units. In $z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$ each term is (length) $^2$ /time.
Sign check. Does the sign agree with the picture? A quantity that must be shrinking cannot come out positive.
Degenerate-instant check. Feed the formula an instant where the rate obviously vanishes or blows up (a cone's apex, a ladder flat on the ground, a maximal area) and see whether it behaves.
Numerical check. Parametrise the motion, e.g. $x(t)=5+2t$ , $y(t)=\sqrt{169-x(t)^2}$ , and compute $\frac{y(0.001)-y(-0.001)}{0.002}\approx-0.8333$ ; compare with your $-\frac56$ .

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