Worked examples — Related rates — setting up and solving

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The scenario matrix

Every related-rates problem falls into one of these case classes. The table lists them; each example below is tagged with the cell it fills.

Cell	Case class	What's special	Example
A	Growing quantity (positive rate)	$\frac{dQ}{dt}>0$, area/volume expands	Ex 1 — balloon
B	Shrinking quantity (negative rate)	answer comes out $<0$, must read the sign	Ex 2 — melting snowball
C	Rate = 0 at a special instant	derivative vanishes even though things move	Ex 3 — ladder at the symmetric point
D	Rate → ∞ (degenerate / limiting input)	denominator → 0, speed blows up	Ex 4 — ladder near flat
E	Two variables both change → eliminate one	need Similar triangles first	Ex 5 — walking shadow
F	Angle changes (trig linking equation)	$\tan$, and why we pick it	Ex 6 — rotating searchlight
G	Real-world word problem (distance closing)	Pythagoras between two moving points	Ex 7 — two cars
H	Exam twist: asks for the value, not the rate	solve for a length/time from a given rate	Ex 8 — when is depth rising fastest

We now walk each cell. Read the Forecast and guess the sign/behaviour before reading the steps — that habit is the whole skill.

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Cell A — a growing quantity

Forecast: Volume grows, so the radius must grow → answer positive. And since a big balloon needs lots of air to gain 1 cm of radius (surface area is huge), we predict the radius rate is small and gets smaller as $r$ grows.

The figure below shows two balloon outlines — a small one (blue) and the $r=5$ one (orange). Look at how much longer the orange circle is than the blue: that circumference stands in for the surface area the incoming air must spread over. The red arrow is the single changing length $r$, and the green label reminds us air enters at a fixed $100{\text{cm}}^3/\text{s}$. The takeaway to carry into the steps: the same air pumped into a bigger skin lifts the radius less.

1. Link. For a sphere, $V=\frac{4}{3}\pi r^3$. Why this step? Volume depends only on the one changing length $r$ — a single-variable link, the simplest kind.
2. Differentiate w.r.t. $t$ (chain rule, since $r=r(t)$): $\frac{dV}{dt}=\frac{4}{3}\pi \cdot 3r^2\cdot \frac{dr}{dt}=4\pi r^2\frac{dr}{dt}.$ Why this step? We want $\frac{dr}{dt}$, so we differentiate the whole equation in time; $\frac{d}{dt}(r^3)=3r^2\frac{dr}{dt}$ is the chain rule earning its keep.
3. Solve for the unknown rate first, symbolically: $\frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}.$ Why this step? Notice $4\pi r^2$ is exactly the surface area — so radius-speed = pump-speed ÷ surface area. That confirms the forecast and matches the figure's big-skin-vs-small-skin picture: bigger surface ⇒ slower growth.
4. Substitute the instant $r=5,\frac{dV}{dt}=100$: $\frac{dr}{dt}=\frac{100}{4\pi (25)}=\frac{100}{100\pi }=\frac{1}{\pi }\approx 0.318\text{cm/s}.$ Why now? Numbers go in last, after differentiating — otherwise $r$ becomes constant and $\frac{dr}{dt}$ disappears.

Verify: Units: $\frac{{\text{cm}}^3/\text{s}}{{\text{cm}}^2}=\text{cm/s}$ ✓. Sign positive ✓ (growing). Value small ✓ as forecast.

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Cell B — a shrinking quantity

Forecast: Volume shrinks, so radius shrinks → answer negative. Watch the magnitude too: as the ball melts smaller, its skin (surface area) gets smaller, and the same cm³ removed pushes the radius in faster — so the shrink-speed should grow as $r$ drops.

The figure shows the snowball at two moments: the larger $r=4$ ball (blue) and a later, smaller ball (orange). The red arrow is the single changing length $r$, and the green label marks the melting rate $\frac{dV}{dt}=-8$. Look at how much shorter the orange circle is — a smaller skin means each cm³ removed pulls the radius in more, which is exactly the magnitude behaviour we forecast.

1. Link & differentiate — identical to Ex 1: $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\Rightarrow \frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}.$ Why this step? The geometry didn't change; only the sign of the given rate did.
2. Substitute $r=4,\frac{dV}{dt}=-8$: $\frac{dr}{dt}=\frac{-8}{4\pi (16)}=\frac{-8}{64\pi }=-\frac{1}{8\pi }\approx -0.0398\text{cm/min}.$ Why this step? A negative input yields a negative output — the formula honestly reports "shrinking."

Verify: Sign negative ✓ matches "melting." Now the clean magnitude rule: since $\frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}$, the factor $\frac{1}{r^2}$ makes $|\frac{dr}{dt}|$ larger for smaller $r$. Check at $r=8$ (a bigger ball): $\frac{dr}{dt}=\frac{-8}{4\pi (64)}=\frac{-8}{256\pi }\approx -0.00995$ — smaller magnitude than at $r=4$ ✓. So a shrinking ball speeds up its radius-loss as it vanishes, matching the forecast.

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Cell C — a rate that is exactly zero

Forecast: When the ladder stands straight up ($x=0$), the top is at the very peak. Just like a ball thrown straight up is momentarily still at the top, we predict the top's downward speed is exactly 0 at that instant.

The figure fixes the two variables in your eye: the vertical orange ladder, the red dot marking the top at height $y=10$, and the blue arrow showing the base about to slide out ($\frac{dx}{dt}=2$). Watch the red dot — at this vertical instant it has nowhere higher to go and hasn't started dropping, so its up/down speed is momentarily zero even though the base is already moving.

1. Name the variables. $x=$ distance from wall to the ladder's base; $y=$ height of the ladder's top up the wall. Both change with time. Why this step? Every symbol in the linking equation must first mean something concrete — $x$ and $y$ are the two legs of the right triangle.
2. Link (Pythagorean theorem): $x^2+y^2=10^2$. Why this step? Wall, ground, ladder form a right triangle of fixed hypotenuse (the 10 m ladder), so the two legs $x$ and $y$ are tied together.
3. Differentiate w.r.t. $t$: $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\Rightarrow \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}.$ Why this step? The right side ($100$) is constant, so its rate is $0$ — this is Implicit differentiation in disguise.
4. Substitute the instant $x=0$ (so $y=10$): $\frac{dy}{dt}=-\frac{0}{10}(2)=0\text{m/s}.$ Why this step? The factor $x$ in the numerator kills the rate. Geometrically the top is momentarily not moving vertically, exactly like the red dot in the figure.

Verify: The formula $\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$ is $0$ ⇔ $x=0$ (since $\frac{dx}{dt}\ne 0,y\ne 0$) ✓. Physically the top hovers at its highest before dropping — a genuine zero-rate instant, our Cell C.

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Cell D — a rate that blows up to infinity

Forecast: As the ladder goes flat, the top is nearly at the ground ($y\to 0$). Dividing by a tiny $y$ makes $\frac{dy}{dt}$ huge and negative → the top crashes down with speed → ∞.

This is the opposite end of the motion from Ex 3. The figure shows the orange ladder almost lying on the ground: the red top-dot sits just above the floor so $y$ is nearly zero, while the base is far out at $x=9.99$ (blue arrow still sliding). Keep that near-flat picture in mind — it is the tiny denominator $y$ that makes everything explode.

1. Reuse the rate law $\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$. Why this step? Same geometry — only the instant is degenerate now.
2. Find $y$ at $x=9.99$: $y=\sqrt{100-9.99^2}=\sqrt{100-99.8001}=\sqrt{0.1999}\approx 0.4471$ m. Why this step? We need the current $y$ to plug in; it is nearly zero.
3. Substitute: $\frac{dy}{dt}=-\frac{9.99}{0.4471}(2)\approx -44.69\text{m/s}.$ Why this step? Small denominator → enormous speed.
4. Limit as $x\to 10^{-}$: $y\to 0^{+}$, so $\frac{dy}{dt}\to -\infty$. Why this step? This is the degenerate limiting behaviour Cell D exists to expose.

Verify: At $x=9.99$, computed $\approx -44.7$ m/s ✓ (huge, negative). The denominator $y\to 0$ forces the magnitude $\to \infty$ ✓. Physically the model breaks (ladders don't reach infinite speed) — a lesson that the math is fine but the idealisation fails at the boundary.

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Cell E — two changing variables (eliminate one first)

Forecast: The shadow tip moves faster than the person (the shadow stretches ahead), so its speed is more than 1.5 m/s.

In the figure, the tall blue post and the short orange person cast a light ray (gray) that hits the ground at the red shadow tip. The two right triangles — big (post) and small (person) — share the same ray angle, which is why their sides are in proportion. Notice the two ground segments labelled $x$ (person to post) and $s$ (person to shadow tip): those are the two changing lengths we must relate before differentiating.

1. Set variables. Let $x$ = person-to-post distance, $s$ = length of shadow (person to shadow tip). Tip position from post $=x+s$. Why this step? Two lengths change ($x$ and $s$); we must relate them.
2. Link via Similar triangles. Big triangle (post, ground, ray) and small triangle (person, ground, ray) share the ray's angle: $\frac{6}{x+s}=\frac{1.8}{s}.$ Why this step? Equal-angled right triangles have proportional sides — this is the fixed geometry that ties $s$ to $x$.
3. Eliminate $s$ before differentiating: cross-multiply $6s=1.8(x+s)\Rightarrow 6s-1.8s=1.8x\Rightarrow 4.2s=1.8x\Rightarrow s=\frac{1.8}{4.2}x=\frac{3}{7}x.$ Why this step? If we kept both $x$ and $s$ and differentiated, we would need the unknown rate $\frac{ds}{dt}$ — a rate nobody gave us. Writing $s$ in terms of $x$ before differentiating removes that extra unknown entirely; this is the eliminate-first lesson from the parent note's cone example.
4. Tip distance $D=x+s=x+\frac{3}{7}x=\frac{10}{7}x$. Differentiate: $\frac{dD}{dt}=\frac{10}{7}\frac{dx}{dt}.$ Why this step? $D$ is now a plain multiple of $x$; the chain rule is trivial.
5. Substitute $\frac{dx}{dt}=1.5$: $\frac{dD}{dt}=\frac{10}{7}(1.5)=\frac{15}{7}\approx 2.14\text{m/s}.$ Why this step? Numbers last, as always.

Verify: $2.14>1.5$ ✓ (tip outruns the person, matching forecast). Note $\frac{dD}{dt}$ is independent of $x$ — the tip moves at a constant speed no matter how far out — a clean, checkable surprise ✓.

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Cell F — an angle changes (trig link)

Forecast: As $\theta$ grows toward 90°, the beam becomes nearly parallel to the wall and the spot races off — so spot-speed increases with $\theta$; at 45° it should already be a healthy number.

Let $p$ be the position of the bright spot measured along the wall from the foot of the perpendicular. Sign convention: we take $p>0$ in the direction the beam sweeps (upward along the wall in the figure), so $\frac{dp}{dt}>0$ means the spot climbs that way. Domain: this setup only makes sense for $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$; at exactly $\theta =\pm \frac{\pi }{2}$ the beam is parallel to the wall and never meets it, which is why the speed blows up as $\theta \to \frac{\pi }{2}$.

In the figure the searchlight (blue dot) sits a fixed $L=200$ m from the vertical gray wall; the orange beam strikes the wall at the red spot a height $p$ up the wall. The dashed gray line is the perpendicular (the adjacent side of length $L$), and $p$ is the opposite side. Watch what happens as the green angle $\theta$ opens toward 90°: the spot shoots up the wall.

1. Link: with $p$ the spot's position along the wall, $\tan \theta =\frac{p}{L}\Rightarrow p=L\tan \theta =200\tan \theta .$ Why this step? It isolates the moving length $p$ in terms of the moving angle $\theta$.
2. Differentiate w.r.t. $t$. Recall $\frac{d}{d\theta }\tan \theta ={\sec }^2\theta$, so by the chain rule ($\theta =\theta (t)$): $\frac{dp}{dt}=200{\sec }^2\theta \frac{d\theta }{dt}.$ Why this step? $\theta$ depends on $t$, so differentiating $\tan \theta$ carries the extra $\frac{d\theta }{dt}$ factor — the chain rule again.
3. Substitute $\theta =\frac{\pi }{4}$. Here $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$, so $\sec \frac{\pi }{4}=\frac{1}{\cos (\pi /4)}=\sqrt{2}$ and ${\sec }^2\frac{\pi }{4}=2$; with $\frac{d\theta }{dt}=0.1$: $\frac{dp}{dt}=200(2)(0.1)=40\text{m/s}.$ Why this step? Numbers last; the growing factor ${\sec }^2\theta$ is exactly why the spot races off, confirming the forecast.

Verify: Units: $\text{m}\cdot (\text{unitless})\cdot (\text{rad/s})=\text{m/s}$ ✓. Positive ✓ (spot climbs in the beam-sweep direction). Value $40$ m/s ✓. As $\theta \to {\frac{\pi }{2}}^{-}$, ${\sec }^2\theta \to \infty$, so $\frac{dp}{dt}\to \infty$ ✓ — matching the domain restriction $\theta <\frac{\pi }{2}$.

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Cell G — real-world closing distance (two moving points)

Forecast: Both cars head toward the crossing, so the gap between them is shrinking → answer negative.

The figure places the intersection at the origin, Car A (blue) directly north at $a=0.3$ and Car B (orange) directly east at $b=0.4$. The north and east roads meet at a right angle, so the two cars and the intersection form a right triangle whose hypotenuse (green) is the distance $D$ between the cars. The arrows show both cars heading toward the corner, which is why both leg-lengths are decreasing.

1. Variables. Let $a$ = A's distance north of the intersection, $b$ = B's distance east, $D$ = distance between the cars. Both decreasing: $\frac{da}{dt}=-60$, $\frac{db}{dt}=-80$ (approaching ⇒ distances shrink). Why this step? Signs must encode "getting closer" — heading toward $0$ means a negative rate.
2. Link (Pythagorean theorem, right angle at the intersection): $D^2=a^2+b^2$. Why this step? North and east directions are perpendicular, so the cars and intersection form a right triangle.
3. Differentiate w.r.t. $t$: $2D\frac{dD}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}\Rightarrow \frac{dD}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{D}.$ Why this step? Differentiate the whole relation; solve for the one rate we want.
4. Find $D$: $D=\sqrt{0.3^2+0.4^2}=\sqrt{0.09+0.16}=\sqrt{0.25}=0.5$ km. Why this step? Need the current $D$ to plug into the denominator (a 3-4-5 triangle scaled by 0.1).
5. Substitute: $\frac{dD}{dt}=\frac{(0.3)(-60)+(0.4)(-80)}{0.5}=\frac{-18-32}{0.5}=\frac{-50}{0.5}=-100\text{km/h}.$ Why this step? Numbers last; negative confirms the gap closes.

Verify: $-100$ km/h, negative ✓ (closing). Magnitude sanity: it exceeds each car's speed because both contribute — reasonable ✓. Units km/h ✓.

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Cell H — exam twist: solve for a value, not a rate

Forecast: From the parent note, at $h=4$ the rise was $\frac{1}{2\pi }\approx 0.159$ m/min. We want a faster rise ($0.5$), and the level rises faster when the surface is smaller — so we expect a shallower depth, $h<4$.

1. Rate law (from parent): differentiate $V=\frac{\pi }{12}{h}^3$: $\frac{dV}{dt}=\frac{\pi h^2}{4}\frac{dh}{dt}.$ Why this step? This is the general link between the two known/target quantities.
2. Twist — solve for $h$, not the rate. Plug the desired rate in and rearrange: $h^2=\frac{4}{\pi \frac{dh}{dt}}\frac{dV}{dt}\Rightarrow h=\sqrt{\frac{4\frac{dV}{dt}}{\pi \frac{dh}{dt}}}.$ Why this step? The unknown is now a length; ordinary algebra isolates $h$ once the rate law is written down.
3. Substitute $\frac{dV}{dt}=2,\frac{dh}{dt}=0.5$: $h=\sqrt{\frac{4(2)}{\pi (0.5)}}=\sqrt{\frac{8}{0.5\pi }}=\sqrt{\frac{16}{\pi }}\approx 2.257\text{m}.$ Why this step? Numbers last; the value comes out real and positive, as a depth must.

Verify: $h\approx 2.257<4$ ✓ (shallower, matching forecast). Plug back: $\frac{dh}{dt}=\frac{4(2)}{\pi (2.257{)}^2}=\frac{8}{\pi (5.094)}\approx 0.5$ ✓ round-trip confirmed. Units m ✓.

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Recall One-line summary of every cell

Sign of the answer tells growing vs shrinking; a factor going to $0$ in the numerator gives a zero rate; a factor going to $0$ in the denominator gives a blow-up; two moving lengths need Similar triangles to eliminate one; angles bring in $\tan$/${\sec }^2$; closing-distance problems use Pythagoras; and exam twists just re-solve the same rate law for a different unknown.

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Connections

• Related rates — setting up and solving — the parent recipe these examples exercise.
• Chain rule — supplies the $\frac{dr}{dt},\frac{d\theta }{dt}$ factors in every differentiation.
• Implicit differentiation — how we differentiate $x^2+y^2=L^2$ and $D^2=a^2+b^2$.
• Derivatives as rates of change — reading signs and magnitudes physically.
• Similar triangles — eliminates the extra variable in shadow and cone problems.
• Pythagorean theorem — the link in ladder and closing-distance cases.
• Optimization — the next place these setups reappear.