Calculus I › Applications of derivatives › full formula sheet
Related rates
Everything moves — relate the rates, then solve for the one you want.
The golden rule: differentiate first, substitute after. Every quantity that changes is secretly a function of time t.
Before this lesson: Chain rule · Implicit differentiation
Where it comes from
The problem: a 13-ft ladder slides down a wall. The bottom moves out at 2 ft/s — how fast is the top sliding down when the bottom is 5 ft from the wall? Let x = bottom’s distance, y = top’s height: x² + y² = 169. The naive move — plug in x = 5 first — murders the problem:
The intuition: x and y aren’t numbers — they’re functions of time, x(t) and y(t), and the equation x(t)² + y(t)² = 169 holds at every instant. Differentiating the relationship (not the frozen snapshot) is what links the rates. The minus sign says the top slides down at 5/6 ft/s. ✓
Before reading on: in x² + y² = 169, both x and y change with time. If you plug in x = 5 before differentiating, which term do you silently kill — and why is that wrong?
Derivation
There’s no new theorem — related rates is the chain rule, with time as the inside function.
Why the order is sacred: differentiating turns variables into rates (x → dx/dt). Substituting turns variables into constants (x → 5), and constants have no rates. Substitute first and there’s nothing left to differentiate — the rates die before they’re born.
Before reading on: the ladder’s bottom slides out at 2 ft/s. Is the top sliding down faster or slower than 2 ft/s when x = 5? Guess, then check the sign.
How to use it
The procedure, every time:
- Draw and label. Sketch the situation; name every changing quantity (x, y, r, θ…); list what’s given (rates and values) and what’s wanted.
- Write the relating equation — the geometry that ties the variables: Pythagoras (ladders), similar triangles (shadows), area/volume formulas (expanding shapes). No numbers yet.
- Differentiate both sides w.r.t. t. Every time-dependent variable earns its d/dt factor via the chain rule.
- Substitute the known values — only now. (You may need the relating equation itself to find a missing value, like y = 12 above.)
- Solve, with units and a sign check. Negative = decreasing in the story’s terms; if the sign contradicts the story, recheck.
When to reach for it
Any “how fast is ___ changing when ___” where quantities are geometrically linked. The relating equation is the whole battle — once it’s written, the rest is chain rule and algebra.
Classic relating equations
Ladders: x² + y² = L². Shadows: similar triangles (lamp height)/(total distance) = (person height)/(shadow). Spheres: V = (4/3)πr³. Cones: V = (1/3)πr²h (plus a similar-triangle link between r and h).
Worked examples
Four rates, easiest first. In each one, read every step — the why of each move is the lesson.
Example 1 — the ladder: 13-ft, bottom sliding out at 2 ft/s, x = 5 ft
- Relate. x² + y² = 169. (Why? Right triangle: wall, ground, ladder.)
- Differentiate w.r.t. t. 2x·dx/dt + 2y·dy/dt = 0.
- Substitute. x = 5 ⇒ y = √(169−25) = 12; dx/dt = 2. So 2(5)(2) + 2(12)·dy/dt = 0.
- Solve. 20 + 24·dy/dt = 0 ⇒ dy/dt = −20/24 = −5/6 ft/s. The top slides down at 5/6 ft/s. ✓ (Sign check: bottom out → top down — negative correct.)
Your turn: A 10-ft ladder: bottom slides out at 3 ft/s. How fast is the top sliding when x = 6 ft?
Answer: dy/dt = −9/4 ft/s.
x² + y² = 100; x = 6 gives y = √(100 − 36) = 8. 2x dx/dt + 2y dy/dt = 0: 2(6)(3) + 2(8)dy/dt = 0, so dy/dt = −36/16 = −9/4. Negative — the top slides down ✓.
Example 2 — expanding circle: dr/dt = 3 cm/s, find dA/dt at r = 10 cm
- Relate. A = πr².
- Differentiate w.r.t. t. dA/dt = 2πr·dr/dt. (Why the dr/dt? r is a function of t — chain rule.)
- Substitute. r = 10, dr/dt = 3: dA/dt = 2π(10)(3) = 60π ≈ 188.5 cm²/s.
Your turn: A circle expands with dr/dt = 2 cm/s. Find dA/dt when r = 5 cm.
Answer: 20π ≈ 62.8 cm²/s.
A = πr², dA/dt = 2πr dr/dt = 2π(5)(2) = 20π ✓.
Example 3 — the shadow: 6-ft person walks from a 15-ft lamp at 4 ft/s; how fast does the shadow’s tip move?
- Relate. Let x = person’s distance, s = shadow length; tip at x+s. Similar triangles: 15/(x+s) = 6/s. Cross-multiply: 15s = 6x + 6s ⇒ 9s = 6x, i.e. s = (2/3)x. (Why? The lamp’s ray grazes the person’s head — two similar right triangles.)
- Differentiate w.r.t. t. 9·ds/dt = 6·dx/dt.
- Substitute. dx/dt = 4: ds/dt = (6/9)(4) = 8/3 ft/s.
- The tip. Tip position T = x + s, so dT/dt = dx/dt + ds/dt = 4 + 8/3 = 20/3 ≈ 6.67 ft/s. (The tip outruns the walker — the shadow stretches as he goes.)
Your turn: A 5-ft person walks away from a 12-ft lamp at 3 ft/s. How fast does the shadow’s tip move?
Answer: 36/7 ≈ 5.14 ft/s.
Similar triangles: 12/(x+s) = 5/s gives 7s = 5x, so ds/dt = (5/7)(3) = 15/7. Tip T = x + s: dT/dt = 3 + 15/7 = 36/7 ✓.
Example 4 — inflating sphere: dV/dt = 100 cm³/s, find dr/dt at r = 5 cm
- Relate. V = (4/3)πr³.
- Differentiate w.r.t. t. dV/dt = 4πr²·dr/dt.
- Substitute. 100 = 4π(25)·dr/dt = 100π·dr/dt.
- Solve. dr/dt = 100/(100π) = 1/π ≈ 0.318 cm/s. ✓ (Sanity: volume pours in fast but the radius creeps — plausible for a big balloon.)
Your turn: A sphere inflates with dV/dt = 50 cm³/s. Find dr/dt when r = 2 cm.
Answer: 25/(8π) ≈ 0.995 cm/s.
dV/dt = 4πr² dr/dt: 50 = 4π(4)dr/dt = 16π dr/dt, so dr/dt = 50/(16π) = 25/(8π) ✓.
Memorization tips
- The golden rule: differentiate first, substitute after. Say it before every problem.
- Every variable gets a d/dt: d/dt[x²] = 2x·dx/dt — the rate factor is the chain rule’s signature. No bare 2x survives.
- Draw first: the relating equation comes from the picture (Pythagoras, similar triangles, volume). No picture, no equation, no solution.
- Sign = direction: negative rate means decreasing in the story’s language. Check it every time.
- Units ride along: ft/s, cm²/s — if your units are wrong, the setup is wrong.
- Related rates = chain rule: there’s no new calculus here, only new bookkeeping. If you can chain-rule, you can related-rate.
Final challenge
Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and related rates are yours.
← Back to the Calculus I formula sheet
How to learn a formula here
- Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
- Work the examples with the answers covered, then uncover one step at a time and compare.
- Finish with the final challenge — five mixed questions including the classic traps.
- Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.
Frequently asked questions
What is the related rates method?
Write the equation relating the changing quantities, differentiate both sides with respect to t (chain rule on every time-dependent variable), then substitute the known values, and solve for the unknown rate.
Why can’t I plug in the numbers before differentiating?
Plugging in freezes variables into constants, and the derivative of a constant is 0 — the rates vanish. In the ladder problem, substituting x = 5 first gives 2y·dy/dt = 0, i.e. dy/dt = 0: dead wrong. Differentiate first, substitute after.
Where does the chain rule appear in related rates?
Every variable is secretly a function of t: d/dt[x²] = 2x·dx/dt. The dx/dt factor is the chain rule’s “derivative of the inside” — x’s inside is time itself.
What is the standard related rates procedure?
1) Draw and label; list what’s changing. 2) Write the relating equation (Pythagoras, similar triangles, volume formulas). 3) Differentiate w.r.t. t. 4) Substitute known values. 5) Solve, with units and a sign sanity check.
What do signs mean in related rates answers?
Sign is direction: dy/dt = −5/6 ft/s means the ladder top slides down at 5/6 ft/s. A positive rate would mean rising. Always sanity-check the sign against the story.
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