Related Rates & Optimization
How fast is the ladder sliding? What box holds the most? Related rates link changing quantities through implicit differentiation; optimization turns a constraint into a max-or-min hunt with the derivative tests.
1 Related rates: geometry first, calculus second
Two quantities change with time, and geometry ties them together. The golden rule: write the relating equation first, then differentiate — never the other way around.
Step 1 is pure geometry (Pythagoras, volume formulas). Step 2 uses implicit differentiation: every x becomes dx/dt, every y becomes dy/dt, via the chain rule.
A 10-ft ladder slides: the foot moves from the wall at dx/dt = 2 ft/s. How fast is the top sliding down when the foot is 6 ft out?
- Relate: x2 + y2 = 102.
- Differentiate w.r.t. t: 2x·dx/dt + 2y·dy/dt = 0.
- Substitute: at x = 6, y = √(100 − 36) = 8. So 2(6)(2) + 2(8)(dy/dt) = 0.
- Answer: dy/dt = −24/16 = −1.5 ft/s. Negative = the height is decreasing, exactly as the picture suggests.
2 Optimization: one variable, then hunt
Maximize or minimize something subject to a constraint. The move: use the constraint to eliminate a variable, so the thing you optimize depends on one variable.
Critical points are only candidates. Compare values at critical points and endpoints (closed interval), or use the first/second derivative test.
Cut equal squares of side x from each corner of a 12×12 sheet and fold up an open box. What x maximizes the volume?
- Primary: V = x(12 − 2x)2, for 0 < x < 6.
- Differentiate: V′ = (12 − 2x)2 − 4x(12 − 2x) = (12 − 2x)(12 − 6x).
- Critical points: x = 6 (endpoint, V = 0) or x = 2.
- Verify: V(2) = 2(8)2 = 128, while V = 0 at both endpoints. So x = 2 in gives the maximum, V = 128 in³.
Pattern worth memorizing: for an s×s sheet the optimum is always x = s/6.
Water pours into a conical tank with r = h/2. The height rises at 2 m/min; find dV/dt when h = 6 m.
- Relate: V = (1/3)πr2h = (1/3)π(h/2)2h = πh3/12.
- Differentiate: dV/dt = (πh2/4)(dh/dt).
- Substitute and answer: dV/dt = π(36)(2)/4 = 18π m³/min.
3 Common mistakes
Three traps, each with the wrong version and the fix.
Wrong: jumping to dV/dt = 4πr2 without deciding what V equals. Right: geometry first: write V = (4/3)πr3, then differentiate with respect to t.
Wrong: “V′ = 0 at x = 2, so x = 2 is the maximum.” Right: critical points are candidates — compare V at critical points and endpoints (or run a derivative test).
Wrong: reporting dy/dt = +1.5 ft/s for the falling ladder top. Right: decreasing means negative: dy/dt = −1.5 ft/s. Let the sign tell the story.
Key vocabulary
- Related rates — rates of change linked by a geometric equation, found via implicit differentiation in t.
- Implicit differentiation — differentiating an equation term by term, attaching dx/dt (chain rule) to each x.
- Primary equation — the quantity to optimize (area, volume, cost).
- Constraint — the fixed condition (perimeter, material) used to eliminate a variable.
- Critical point — where the derivative is zero or undefined; a candidate for max/min, never the verdict.
4 Quick checks
Cover the answers, decide, then reveal.