Contextual Applications of Differentiation
~3 weeks · October · 5–10%Unit 5 asked what derivatives say about graphs. This unit asks what they say about the world. The calculus is mostly the same; what's being taught is modeling — turning a situation into an equation you can differentiate. That translation is the actual skill, and it's the one that transfers.
L'Hôpital's rule also lands here in most courses, because it needs derivatives. Full treatment below.
Motion along a line
Position s(t) → velocity v = s′ → acceleration a = v′ = s″.
| Question | Answer |
|---|---|
| Speed | |v| — a magnitude, never negative |
| Moving right / left | v > 0 / v < 0 |
| Changes direction | where v changes sign, not merely where v = 0 |
| At rest | v = 0 |
| Speeding up | v and a have the same sign |
| Slowing down | v and a have opposite signs |
Speed is |v|, so speeding up means |v| is growing. If the object is moving left (v < 0) and acceleration is also negative, it's being pushed further left — moving faster in the negative direction. |v| grows. So negative acceleration can mean speeding up.
The clean statement: acceleration in the same direction as motion adds speed; acceleration opposing motion removes it. Same-sign, speeding up. Opposite-sign, slowing down. This is asked on nearly every AP exam and missed constantly.
A particle has s(t) = t³ − 6t² + 9t for t ≥ 0. Describe its motion.
| v = 3t² − 12t + 9 = 3(t−1)(t−3) | factor for the sign chart |
| v = 0 at t = 1, 3 | candidate direction changes |
| t=0: v=+9 · t=2: 3(1)(−1)=−3 · t=4: 3(3)(1)=+9 | sign chart: + − + |
| moves right, then left after t=1, then right after t=3 | both are genuine sign changes, so both are direction reversals |
| a = 6t − 12 = 6(t−2) | negative before t=2, positive after |
| on (1,2): v < 0, a < 0 → same sign | speeding up, despite moving backward and decelerating in the everyday sense |
| on (2,3): v < 0, a > 0 → opposite signs → slowing down |
Note the interval (1,2): moving left and speeding up. Everyday language calls negative acceleration "slowing down," and here it's the opposite. Trust the sign rule, not the vocabulary.
Displacement vs. total distance needs integrals and belongs in Unit 6. For the record: displacement over [a,b] is ∫v dt, and total distance travelled is ∫|v| dt. They differ whenever the object reverses. Above, the particle ends up somewhere modest but has covered more ground than that, because it doubled back between t=1 and t=3.
Related rates
Two or more quantities linked by an equation, all changing in time. Differentiate the equation with respect to t; the chain rule attaches a rate to every variable.
In implicit differentiation, y is secretly a function of x. In related rates, everything is secretly a function of t. Same rule, different hidden variable. The curriculum presents them as separate topics; telling her they're one thing is worth doing.
The procedure
- Sketch it. Label anything that varies with a letter; label fixed quantities with numbers.
- Write the relation among the variables — geometry, or a physical law.
- Eliminate any variable whose rate you don't know, usually via similar triangles. Do this before differentiating.
- Differentiate with respect to t.
- Now substitute the instantaneous values.
- Solve. Check the sign and the units.
Triangles with the same angles have proportional sides. If a small triangle sits inside a larger one sharing an apex and a direction, then small-height/small-base = big-height/big-base.
Example: a cone of total height 10 with top radius 5 has r/h = 5/10 at every depth, so r = h/2 always. That one substitution is what makes the cone problem solvable.
Water fills a cone (top radius 5 m, height 10 m) at 3 m³/min. How fast is the depth rising when h = 4?
| V = (1/3)πr²h | volume of a cone — two varying quantities, only one known rate |
| r = h/2 | similar triangles; eliminate r before differentiating |
| V = (1/3)π(h/2)²h = πh³/12 | now one variable only |
| dV/dt = (πh²/4)·dh/dt | differentiate in t; chain rule puts dh/dt on the h³ |
| 3 = (π·16/4)·dh/dt | now substitute dV/dt = 3 and h = 4 |
| dh/dt = 3/(4π) ≈ 0.239 m/min | units: m³/min ÷ m² = m/min ✓ |
If you plug h = 4 into the volume formula first, you've turned a variable into a constant and its derivative into zero. The answer comes out wrong and looks perfectly clean — there's no error message. Numbers go in only after the differentiation is done. This single mistake accounts for most related-rates failures.
A 13 ft ladder leans on a wall. The base slides away at 2 ft/s. How fast is the top falling when the base is 5 ft from the wall?
| x² + y² = 169 | Pythagorean; 13 is fixed, so it's a number, while x and y vary |
| 2x·(dx/dt) + 2y·(dy/dt) = 0 | differentiate in t; the constant 169 gives 0 |
| at x = 5: y = √(169−25) = 12 | find the other side at this instant |
| 2(5)(2) + 2(12)(dy/dt) = 0 | substitute now, after differentiating |
| dy/dt = −20/24 = −5/6 ft/s | negative because the top is descending |
The sign carries meaning and should not be discarded. Also note the equation 2x·ẋ + 2y·ẏ = 0 says the two rates always trade off — which is why the top accelerates dramatically as the base nears the wall's far reach.
Standard setups
| Situation | Relation to differentiate |
|---|---|
| Ladder on a wall | x² + y² = L², L constant |
| Streetlight shadow | Similar triangles |
| Two vehicles, right angles | z² = x² + y² |
| Angle of elevation | tan θ = y/x → sec²θ·(dθ/dt) = … |
| Inflating sphere | V = (4/3)πr³ → dV/dt = 4πr²·(dr/dt) |
| Cone or trough filling | Similar triangles first, then the volume formula |
| Expanding ripple | A = πr² → dA/dt = 2πr·(dr/dt) |
Sign convention: decreasing quantities get negative rates. Declare it at setup rather than patching the sign at the end.
Textbook related-rates problems are contrived because the honest ones need multivariable calculus. One that isn't: cumulative arrival and departure curves at a transit platform.
Riders arrive at rate λ(t); the cumulative count A(t) is its integral. Departures D(t) are a second curve. The vertical gap A − D is the queue length; the horizontal gap is an individual rider's wait; the area between the curves is total passenger-delay. The related rate — how fast is the queue growing? — is just A′(t) − D′(t).
Everything in that picture is a calculus object with a physical meaning she can point at, which the ladder is not.
Differentiating V = (4/3)πr³ gives dV/dt = 4πr²·(dr/dt). Check the units before the arithmetic: cm² × cm/s = cm³/s, a volume per unit time ✓.
If your answer came out in cm²/s you differentiated the surface area formula by mistake — a substitution slip, not an arithmetic one, and the units are the only thing that catches it.
Linear approximation
This is the tangent line at a, used as a stand-in for f near a. It's Unit 2's second definition of the derivative deployed as a tool.
Estimate √4.1.
| f(x) = √x, choose a = 4 | pick the nearest point where you know the answer exactly |
| f(4) = 2, f′(x) = 1/(2√x), f′(4) = 1/4 | |
| L(x) = 2 + ¼(x − 4) | |
| L(4.1) = 2 + ¼(0.1) = 2.025 | |
| √4.1 ≈ 2.025 (true value 2.02485…) | accurate to four decimals |
Is it an over- or underestimate? f″ = −1/(4x3/2) < 0, so f is concave down, so the tangent lies above the curve — this is an overestimate. And 2.025 > 2.02485 ✓. This over/under question is asked routinely and answered entirely by the sign of f″.
Newton’s method is not tested on the AP Calculus exams — College Board’s own sample BC syllabus lists it, alongside Simpson’s rule, trigonometric substitution, and volume by cylindrical shells, as material beyond the Course Description. It appears here anyway because it is three lines long, it is the clearest possible demonstration that linear approximation is a tool and not just a fact, and it is the direct ancestor of Euler’s method in Unit 7 and of essentially all numerical optimization.
Read it for the idea; don’t spend drill time on it.
The tangent to y = √x at x = 4 estimates √4.1 ≈ 2.025. Since √x is concave down, the tangent line sits above the curve — so the estimate has to come out slightly high. It does: the true value is 2.02485.
Knowing which side of the curve your tangent line is on turns linear approximation from a guess into a bound.
Newton's method
You want a root of f. You have a guess xn. Replace f by its tangent line there — the linear approximation again — and solve that for zero instead, since solving a line is easy.
The tangent is y = f(xn) + f′(xn)(x − xn). Set y = 0 and solve for x: x = xn − f(xn)/f′(xn). That's the formula, derived in one line. Then repeat from the new point.
Convergence is quadratic — the number of correct digits roughly doubles each step. That's why it's still the workhorse root-finder three and a half centuries later.
Solve x² − 2 = 0 starting from x₀ = 1.
| f(x) = x² − 2, f′(x) = 2x | |
| x₁ = 1 − (1−2)/2 = 1 + 0.5 = 1.5 | |
| x₂ = 1.5 − (2.25−2)/3 = 1.5 − 0.08333 = 1.41667 | 3 correct digits |
| x₃ = 1.41667 − (0.006945)/2.83334 = 1.414216 | 6 correct digits |
| √2 = 1.4142136… — digits doubling each step |
L'Hôpital's rule — the proper treatment
You saw the statement and several worked examples in the limits unit. Everything there still stands. What follows is the part that needed derivatives: why it works, and the ways it fails.
Near a, each function is well approximated by its tangent line. If f(a) = g(a) = 0, both tangent lines pass through zero there, so near a:
The ratio is f′(a)(x−a) / g′(a)(x−a), and the (x−a) cancels, leaving f′(a)/g′(a).
L'Hôpital is cancelling the common factor again — the same move as factoring in Unit 1, executed with tangent lines instead of algebra. That's exactly why it requires the 0/0 condition: without it the tangent lines don't both pass through zero and there's no shared factor to cancel.
(The honest proof uses the Cauchy Mean Value Theorem, a two-function version of MVT, rather than this hand-wave. The intuition is right and the machinery is Unit 5's.)
The three failure modes
limx→0 (x + 2)/(x + 1) is just 2/1 = 2. Apply L'Hôpital anyway and you get 1/1 = 1. Wrong, with no warning. Always check the form first.
| limx→∞ x/√(x²+1) is ∞/∞, so the rule is legal | |
| → 1/[x/√(x²+1)] = √(x²+1)/x | which is the reciprocal of the original |
| applying again returns the original | infinite loop |
| Algebra instead: divide by x → 1/√(1+1/x²) → 1 |
Legal is not the same as useful. If two applications haven't simplified anything, stop and look for algebra.
limx→∞ (x + sin x)/x is 1 — divide by x and the sin x/x term dies. But L'Hôpital gives (1 + cos x)/1, which oscillates forever and has no limit. The rule says "if the limit of f′/g′ exists, then it equals the original." When it doesn't exist, the rule tells you nothing — it does not tell you the original fails to exist.
Find limx→0 (ex − 1 − x)/x².
| substitute: (1 − 1 − 0)/0 = 0/0 ✓ | check the form before anything else |
| → (ex − 1)/(2x) | differentiate top and bottom separately |
| still 0/0 at x = 0 | re-check the form each time |
| → ex/2 | apply again |
| = 1/2 | now substitutable |
The answer ½ is the coefficient of x² in the Taylor series for ex. That's not a coincidence — Unit 10 will show that L'Hôpital and Taylor series are two views of the same fact.
Next stop — multivariable calculus. With several inputs, "slope" is meaningless but "best linear approximation" still works: the derivative becomes a gradient vector pointing in the direction of steepest increase, and then a Jacobian matrix when the output is also multidimensional.
Then — optimization and machine learning. Gradient descent is literally "compute the gradient, step downhill, repeat" — Newton's method's cousin, run in ten million dimensions. Backpropagation is the chain rule applied through a composed function with millions of layers. If she is at all interested in AI, this is the honest answer to "what is calculus for": the entire training procedure of a neural network is Unit 3's chain rule plus Unit 4's linear approximation, at scale.
The actual proof of L'Hôpital, via the Cauchy Mean Value Theorem — a two-function version of the MVT from Unit 5.
there is a c with [f(b) − f(a)]·g′(c) = [g(b) − g(a)]·f′(c)
Setting g(x) = x recovers the ordinary MVT, so it's a genuine generalization. Now suppose f(a) = g(a) = 0. Cauchy MVT gives, for each x near a, some c between a and x with
As x → a, c is squeezed to a as well, so the left side approaches whatever f′/g′ approaches. That's L'Hôpital, properly. The tangent-line argument in the main text is the honest intuition; this is the machinery that makes it airtight — and it explains why the theorem needs the 0/0 hypothesis so specifically.
Error propagation as relative error — the differentials material done the way a physicist would.
If y = f(x) and x carries a small error dx, then dy ≈ f′(x)dx. But the useful quantity is usually the relative error dy/y:
So relative error is governed by the derivative of the logarithm — which is exactly why logarithmic differentiation exists as a technique.
Worked: a sphere's radius is measured as 10 cm ± 1%. What's the relative error in the volume?
| V = (4/3)πr³, so ln V = ln(4π/3) + 3 ln r | take logs first |
| dV/V = 3·(dr/r) | differentiate |
| = 3 × 1% = 3% | a cubed quantity triples the relative error |
The general rule falls straight out: raising to a power multiplies relative error by that power; multiplying quantities adds their relative errors. Two facts that cover most of experimental science, both consequences of ln turning products into sums.
Unit 4 — Contextual Applications
Motion
- s → v = s′ → a = v′. Speed = |v|.
- Speeding up ⇔ v and a same sign. Slowing down ⇔ opposite signs.
- Direction change requires v to change sign, not merely to vanish.
Related rates
- sketch → relation → eliminate unknown-rate variables → differentiate in t → then substitute → check sign and units
- Decreasing quantities get negative rates; declare it at setup.
Approximation
- L(x) = f(a) + f′(a)(x − a)
- Concave up (f″>0) ⇒ tangent below curve ⇒ underestimate; concave down ⇒ overestimate
- Differentials: dy = f′(x)dx
L’Hôpital
- Only for 0/0 or ∞/∞. Verify the form every time, including on repeat applications.
- Differentiate numerator and denominator separately.
- 0·∞ → make a fraction · ∞−∞ → common denominator · 1∞, 00, ∞0 → take ln, then exponentiate at the end
- Fails by: wrong form · endless cycling · f′/g′ having no limit
Practise Unit 4 → Fresh problems from this unit, with worked solutions.