# Chapter 8: Implicit Differentiation and Related Rates

> Not every curve lets you solve for y. Implicit differentiation unlocks circles, trig curves and more — and the same move, differentiated with respect to time, powers every related-rates problem.

**Canonical:** https://calculus.kitzo.work/chapter/implicit-differentiation  
**Level:** Core  
**Estimated time:** 50 minutes  
**Exercises:** 8 (auto-marked; answers stay server-side)

## Topics
- Implicit differentiation
- Second derivatives implicitly
- Related rates setup
- Ladder problems
- Cone & sphere problems
- Real-world related rates

**Why this matters**
Most interesting curves cannot be written as y = f(x), and most real machines change several quantities at once. Implicit differentiation is the single manoeuvre — treat y as a function of x, and every y-term grows a y′ factor — behind both of these workhorses of calculus.

## 1 · The technique: differentiate y-terms as chain rules

When y is defined implicitly by an equation, differentiate both sides with respect to x. Every time you meet y, the chain rule attaches a dy/dx: d/dx(y²) = 2y·y′, d/dx(y³) = 3y²·y′, and d/dx(sin y) = cos y·y′. Then collect y′-terms on one side and solve. Nothing else is new — just patience.

**Worked example 8.1 — Three equations, one manoeuvre**
**(a) Circle x² + y² = 25 at (3, 4):** 2x + 2y·y′ = 0, so y′ = −x/y = −3/4.
**(b) Rectangular hyperbola xy = 1:** product rule → y + x·y′ = 0, so y′ = −y/x; at (2, ½) the slope is −¼.
**(c) Trig implicit sin y = x + y²:** cos y·y′ = 1 + 2y·y′ → y′(cos y − 2y) = 1 → y′ = 1/(cos y − 2y).

## 2 · Finding second derivatives implicitly

Differentiate twice, and the key is substitution, not algebra heroics: after the first pass you own an expression for y′ — feed it back in wherever y′ appears in the second. For y² = 4x: 2y·y′ = 4 gives y′ = 2/y; differentiating again y″ = −2y′/y² = −4/y³, so at (1, 2), y″ = −½ (concave down, as a sideways-opening parabola should be on its upper branch). On the circle, the same loop yields y″ = −r²/y³.

## 3 · Related rates: the five-step setup

- **Draw** the situation and **label** every changing quantity with a variable (never substitute changing numbers yet).

- **Write** one equation linking the variables — a rate is never in it, a geometry or volume formula always is.

- **Check** which quantities are constants (ladder length, pole height) and keep them as constants.

- **Differentiate** every term with respect to time t: x becomes dx/dt, y² becomes 2y·dy/dt.

- **Only now** substitute the instant's known numbers — position, then rates — and solve.

## 4 · The four archetypes

**Worked example 8.2 — One method, four classics**
**(a) Expanding circle:** A = πr² → dA/dt = 2πr·dr/dt. If r grows at 3 cm/s, at r = 10 the area grows at 60π ≈ 188.5 cm²/s.
**(b) Ladder sliding:** x² + y² = 100 (10 ft ladder). If dx/dt = 4 ft/s, then dy/dt = −(x/y)·dx/dt; at x = 6, y = 8, so dy/dt = −3 ft/s — the top descends at 3 ft/s.
**(c) Walking person, lamp 6 m, person 2 m, walking at 1.5 m/s:** similar triangles give 6s = 2(x + s) where s is shadow length, so s = x/2 and ds/dt = 0.75 m/s. The tip of the shadow moves at dx/dt + ds/dt = 2.25 m/s.
**(d) Cone being filled:** r = h/2 always (fixed cone), V = πh³/12 → dV/dt = (πh²/4)·dh/dt; at h = 6 with dh/dt = 2 cm/s, dV/dt = 18π ≈ 56.5 cm³/s.

## 5 · Handling time derivatives

Every variable in a related-rates problem is secretly a function of t: writing dy/dx where the problem means dy/dt is the single most common mechanical slip. Differentiate each piece as a t-derivative — constants survive, variables become a rate — then replace measured values. If the cone in 8.2(d) poured at a known *volume* rate instead, you would substitute dV/dt and solve for dh/dt.

## 6 · When the equation will not solve for y

Tough related-rates setups hide behind non-solvable equations — think two boats with the distance between them growing by the law of cosines, or air filling a sphere. The escape is always the same: differentiating implicitly needs no explicit formula, so set up the geometric relation honestly and let the chain rule do the rest.

**Watch out**
Substituting known numbers before differentiating silently deletes the rate information — 6 m and 2 m are constants, but x and s are not. Differentiate the symbolic equation first, and only then feed in the snapshot values: position at that instant, not last Tuesday's.

**Chapter formula sheet**
d/dx(yⁿ) = n yⁿ⁻¹·y′,  d/dx(sin y) = cos y·y′
x² + y² = r²  ⇒  y′ = −x/y,  y″ = −r²/y³
related rates: dy/dt = (dy/dx)·(dx/dt)
A = πr² ⇒ dA/dt = 2πr·dr/dt ;  V = (1/3)πr²h
second derivative: differentiate twice, substitute y′

## Common mistakes

- Forgetting the y′ factor on y-terms — d/dx(y²) is 2y·y′, never 2y.

- Substituting x and y numbers before differentiating, which kills every rate.

- Treating dy/dx as if it were dy/dt (or vice versa) — name the variable before differentiating.

- Using the ladder's constant length as if x or y were constant — only 100 = x² + y² is permanent.

- Confusing the shadow's length rate with the shadow tip's speed, which differ by the walking rate.

- Declaring y′ = 0 at a maximum without checking where the curve genuinely flattens.

## Practice exercises

1. On the circle x² + y² = 25, the slope of the tangent at (3, 4) is:

- A. −3/4
- B. 3/4
- C. −4/3
- D. 4/3

2. For the curve xy = 12, express dy/dx in terms of x and y.

3. A circle's radius grows at 2 cm/s. At the instant r = 5 cm, how fast is the area growing (cm²/s)? Round to two decimals.

4. A 10 ft ladder leans on a wall with its base sliding away at 3 ft/s. When the base is 6 ft from the wall, find dy/dt (ft/s), where y is the ladder's height on the wall.

5. For xy = 1, compute the second derivative d²y/dx², expressed in terms of x and y.

6. A 6 m lamppost lights a 2 m pedestrian who walks directly away at 1.5 m/s. How fast (m/s) does the tip of the shadow move?

7. Water fills a cone whose radius always equals half its height (r = h/2) at dh/dt = 2 cm/s. When h = 6 cm, how fast is the volume growing (cm³/s)? Round to two decimals.

8. In a related-rates problem, when should you substitute the known values of the variables?

- A. Before differentiating, to simplify the algebra
- B. After differentiating the symbolic equation
- C. Never — keep everything symbolic throughout
- D. Only the rate values, never the position values

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