← Unit 4

Unit 4 Cheat Sheet

Every formula in Contextual Applications of Differentiation — and, more importantly, when to use which. This whole unit is one idea, the derivative as a rate, wearing five costumes. The pieces that look alike (speed vs. velocity, distance vs. displacement, rate vs. amount, over- vs. under-estimate) are exactly where points slip away, so each block ends with the decision that picks the right tool — and the whole-unit skill, saying it in a sentence with units, gets its own panel at the bottom.

Interpreting Derivatives in Context 4.1 · 4.3

A derivative is an instantaneous rate. Read its units and sign, and it tells a story about the real quantity.

average rate of change (secant)

instantaneous rate (the derivative)

units — always output over input

Second derivative = rate of the rate. \(f''\gt 0\) means the rate \(f'\) is rising; \(f''\lt 0\) means it's falling. That's the difference between "increasing faster" and "increasing but leveling off."

What is being asked?

"how fast"a rate → the derivative \(f'(a)\), with units output/input

"how much"an amount → the function value \(f(a)\)

"on average"the secant slope \(\tfrac{f(b)-f(a)}{b-a}\), not \(f'\)

trap

Rate \(\neq\) amount. \(V'(10)=-3\) is a rate (draining at 3/min), not "3 left."
Always attach units — a bare number earns no interpretation point.
\(f'=0\) means the rate is zero (a max/min or a pause), not \(f=0\).

Straight-Line Motion 4.2

Position \(x(t)\) on a line. Differentiate for velocity and acceleration; integrate to go back.

velocity & acceleration

speed is a magnitude

displacement (net) & distance (total)

position from velocity

Speeding up or slowing down?

same sign\(v\) and \(a\) agree \(\Rightarrow\) speeding up

opposite sign\(v\) and \(a\) disagree \(\Rightarrow\) slowing down

Direction & distance

direction\(v\gt 0\) moving right/positive; \(v\lt 0\) left/negative; \(v=0\) at rest

total distancesplit at every \(v=0\); add \(\int|v|\,dt\) piece by piece

trap

Speed \(=|v|\), always \(\ge 0\); velocity keeps its sign.
Speeding up \(\neq\) \(v\gt 0\). It's about \(v\) and \(a\) matching signs.
Distance \(\neq\) displacement whenever the particle reverses — split at the turning points.

Related Rates 4.4 · 4.5

Two quantities tied by an equation change together in time. Differentiate the relation with respect to \(t\).

Equation: write a relation among the quantities (geometry, formula).
Differentiate both sides with respect to \(t\) — the chain rule brings out each \(\tfrac{d}{dt}\).
Substitute the instant's values (numbers go in now, not before).
Solve for the unknown rate.

circle area · sphere volume

Pythagorean (ladders, distances)

Two changing dimensions?

reduce firstuse a constraint (similar triangles, e.g. \(r=\tfrac h2\)) to write \(V\) in one variable before differentiating

trap

Differentiate before you substitute. Plugging numbers in early turns a variable into a constant and kills its rate.
Every variable gets a \(\tfrac{d}{dt}\) — that's the chain rule; a missing rate is the usual error.
Signs carry meaning: a negative rate means the quantity is shrinking.

Linear Approximation 4.6

Near a point, a curve looks like its tangent line. Use that line to estimate nearby values.

the tangent line (linearization)

differential — estimated change

Over- or underestimate?

\(f''\gt 0\)concave up → tangent below → underestimate

\(f''\lt 0\)concave down → tangent above → overestimate

trap

Concavity decides over/under — not whether \(f\) is increasing.
It's only local: accurate when \(x\) is close to \(a\); the error grows as you move away.
Get the slope \(f'(a)\) right — a wrong slope wrecks every estimate.

L’Hôpital’s Rule 4.7

For an indeterminate quotient, the limit is the ratio of the rates.

the rule (differentiate top & bottom separately)

Check the form FIRST

\(\tfrac00\) or \(\tfrac{\pm\infty}{\pm\infty}\)apply the rule; re-check the form and repeat if needed

\(0\cdot\infty\)rewrite as a fraction \(\tfrac{f}{1/g}\), then apply

\(\infty-\infty\)combine into one fraction first

\(1^{\infty},\,0^0,\,\infty^0\)take \(\ln\), find that limit, then exponentiate

trap

Never apply to a determinate form. \(\tfrac{0+1}{0+2}=\tfrac12\) already — using the rule gives a wrong \(1\).
Not the quotient rule — differentiate \(f\) and \(g\) on their own.
Re-check after each pass; keep going only while it stays indeterminate.

Read the signs, name the behavior the phrase decoder

The direction comes from \(f'\); the "at a(n) __ rate" comes from \(f''\). Two signs, four stories — this is the language AP wants back in your sentence.

\(f'\gt 0,\ f''\gt 0\)

Increasing at an increasing rate

rising, and speeding up — concave up

\(f'\gt 0,\ f''\lt 0\)

Increasing at a decreasing rate

rising, but leveling off — concave down

\(f'\lt 0,\ f''\lt 0\)

Decreasing at an increasing rate

falling, and dropping ever faster — concave down

\(f'\lt 0,\ f''\gt 0\)

Decreasing at a decreasing rate

falling, but slowing its fall — concave up

Say it in a sentence the whole-unit skill

Most lost points in this unit aren't algebra — they're interpretations left vague. Every "interpret" or "justify" answer needs the same four ingredients.

1 · the value

the number you found

e.g. \(H'(5)\approx 0.55\).

2 · the units

output per input

meters per hour — never omit them.

3 · the quantity

what is changing

"the water depth," not "the function."

4 · direction / time

increasing/decreasing, at when

"is increasing … at \(t=5\) hours."

put together

"At \(t=5\) hours, the water depth is increasing at about \(0.55\) meters per hour."

value + units + quantity + direction/time = full credit.

Justification checklist

motion sign workstate the sign of \(v\) (and \(a\)) explicitly — "since \(v(2.5)\lt 0\) and \(a(2.5)\gt 0\), … slowing down"

over/undername the concavity — "an overestimate because \(f\) is concave down"

extremumcite \(f'=0\) and the sign change (or \(f''\)) — don't just assert "it's a max"

The reliable move: after any calculation, ask "what does this number mean, in what units, for which real thing, and which way is it going?" Answer all four and the justification points are automatic.

Unit 4 is one idea in five costumes: the derivative is a rate of change. Reading it in context is naming that rate; motion is that rate on a line; related rates is two rates chained through an equation; linear approximation trusts that rate to predict nearby values; and L’Hôpital resolves \(\tfrac00\) by comparing two rates. Master "how fast, which way, in what units" and the whole unit is one question asked five ways.

Test yourself on the exam set →

Rates in Context Straight-Line Motion Related Rates Linear Approximation L’Hôpital’s Rule