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Related rates, critical points and optimisation with a constraint, linear approximation, the mean value theorem, and reading a curve's direction and bend from its two derivatives.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you can differentiate a relation with respect to time to get one rate from another, find and classify critical points, solve an optimisation problem by using its constraint to reach a single variable, and use a tangent line as an estimate whose direction of error you know. You can also state what the mean value theorem promises, and read a curve's direction and its bend off the two derivatives without confusing them.
You can differentiate anything this course has offered so far, including composites and relations. This section is about what a derivative is for: finding how fast one thing changes when another does, locating the best possible value of something, and estimating what a function will do next.
Related rates: two quantities tied by a relation, both changing with time.
Critical point: where $f' = 0$ or $f'$ does not exist.
Local extremum: a highest or lowest value in a neighbourhood.
Constraint: the equation that lets a two-variable problem be written in one variable.
Linearisation: $L(x) = f(a) + f'(a)(x - a)$, the tangent line used as a stand-in for the function.
Related rates. Two quantities are tied by a relation, and both change with time. Differentiate the relation with respect to $t$ — every variable picks up a $\dfrac{d}{dt}$ by the chain rule — and then substitute the values at the instant in question. The order is the whole discipline: an instantaneous value put in early stops being a variable.
Extrema and optimisation. A critical point is where $f' = 0$ or fails to exist, and every local maximum or minimum is one — though not every critical point is an extremum. To classify, either check the sign of $f'$ on each side, or use the second derivative test: $f'' > 0$ means a minimum, $f'' < 0$ a maximum, and $f'' = 0$ means the test says nothing. For an optimisation problem, write the quantity to be optimised, use the constraint to reduce it to one variable, differentiate, and check the endpoints of the domain as well as the critical points — on a closed interval the extreme value theorem guarantees a maximum and a minimum exist, and either can be at an end.
Linearisation. Near $x = a$, a differentiable function is almost its tangent line: $f(x) \approx f(a) + f'(a)(x - a)$. This is what a derivative is good for in practice — and its error has a known direction, because a concave-up curve lies above its tangent and a concave-down one below.
Another way: picture
Picture a ripple spreading on a pond. The radius grows steadily, but the area does not: each new ring is longer than the last, so the same outward speed adds more area every second. $\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}$ is that sentence in symbols — and the $r$ in it is why the answer depends on when you ask.
Another way: steps
Substituting the instantaneous values too early. In a related-rates problem the numbers go in after differentiating. Put them in first and the varying quantity becomes a constant, whose derivative is zero.
"Every critical point is a maximum or a minimum." $f(x) = x^3$ has $f'(0) = 0 and no extremum at all. A critical point is a candidate; classifying it is a separate step.
Optimising without the constraint. An optimisation problem always has two equations: the thing to be optimised and the restriction. Differentiating before using the constraint to get down to one variable leaves an expression with two unknowns and no way forward.
A ladder 13 m long leans on a wall; its foot slides out at 2 m/s. $x^2 + y^2 = 169$.
The relation first.
Differentiate in $t$: $2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$.
Both terms need the chain rule.
When $x = 5$, $y = 12$: $10(2) + 24\dfrac{dy}{dt} = 0$, so $\dfrac{dy}{dt} = -\tfrac{5}{6}$ m/s — the top slides down.
Numbers last; the sign is the answer.
A pen against a wall with 40 m of fence: $2x + y = 40$, area $A = xy$.
Quantity and constraint.
$A(x) = x(40 - 2x) = 40x - 2x^2$, so $A'(x) = 40 - 4x = 0$ at $x = 10$.
One variable, then differentiate.
$A'' = -4 < 0$, so it is a maximum: $y = 20$ and the area is 200 m².
Classify, then answer what was asked.
$V = x^3$, so $\dfrac{dV}{dt} = 3x^2\dfrac{dx}{dt}$.
Differentiate in $t$.
At $x = 5$: $3(25)(3) = 225$ cm³/s.
Numbers after.
And it would be four times that at $x = 10$ — the same steady growth in the edge produces an ever faster growth in volume.
A circular ripple's radius grows at $6$ cm a second. How fast is the enclosed area growing when the radius is $3$ cm? Give the answer as a multiple of $\pi$.
answer
Find the critical point of $f(x) = 4x^2 + 56x - 3$.
answer
A rectangular pen is built against a straight wall, so only three sides need fencing. With $80$ metres of fence, what is the largest area it can enclose?
answer
$f(x) = 4x^2 - 40x$ has a critical point at $x = 5$. Is it a maximum or a minimum?
You can find critical points and classify them with the second derivative. This section makes the second derivative do a job of its own — describing the bend of a curve — and adds the theorem that connects an average rate of change to an instantaneous one.
Mean value theorem: if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, some $c$ inside has $f'(c) = \dfrac{f(b)-f(a)}{b-a}$.
Increasing / decreasing: $f' > 0$ / $f' < 0$.
Concave up / down: $f'' > 0$ / $f'' < 0$.
Inflection point: where concavity changes.
The mean value theorem says that if $f$ is continuous on $[a, b]$ and differentiable inside it, then for some $c$ in $(a, b)$
$$f'(c) = \frac{f(b) - f(a)}{b - a}$$
— somewhere, the instantaneous rate equals the average rate. Drive 120 miles in two hours and at some instant your speedometer read exactly 60. It sounds obvious and it is the reason the rest of this section works: it is what proves that a positive derivative throughout an interval forces the function to increase across it.
The first derivative gives direction. $f' > 0$ on an interval means $f$ increases there; $f' < 0$ means it decreases. Extrema occur where $f'$ changes sign.
The second derivative gives bend. $f'' > 0$ means concave up — the curve holds water, and lies above its tangents. $f'' < 0$ means concave down — it spills, and lies below its tangents. Where the concavity changes there is an inflection point, and finding one means solving $f'' = 0$ and confirming the sign really changes.
The two are independent, and all four combinations occur. Increasing and concave up is growth accelerating; increasing and concave down is growth easing off; and the difference between those two is most of what a news report about an economy is actually saying.
Another way: picture
Picture a curve and the straight chord joining its endpoints. Now slide a line parallel to that chord until it just touches the curve. Where it touches is the $c$ the mean value theorem promises — and you can see there must be such a point, and equally that nothing tells you where it is without solving for it.
Another way: steps
| $f'$ | $f''$ | The graph | A familiar example |
|---|---|---|---|
| $+$ | $+$ | Rising, ever more steeply | Compound growth |
| $+$ | $-$ | Rising, flattening off | Approaching a ceiling |
| $-$ | $+$ | Falling, decelerating | Cooling toward room temperature |
| $-$ | $-$ | Falling, ever faster | A collapse |
The two middle rows are the ones people describe wrongly. Both are 'slowing down', and they are going in opposite directions.
"$f'' < 0$ means the function is decreasing." It means the function is bending downward. A function can be increasing and concave down at once — rising, but flattening — and that combination is most of economics.
"$f'' = 0$ means an inflection point." Concavity has to change. $f(x) = x^4$ has $f''(0) = 0$ and is concave up on both sides, so there is no inflection there.
"The mean value theorem tells you where." It tells you a $c$ exists. Finding it requires solving an equation, and for most functions that is harder than the original problem.
$f(x) = x^2$ on $[1, 3]$: the secant slope is $\dfrac{9 - 1}{3 - 1} = 4$.
Average rate first.
$f'(x) = 2x$, so $2c = 4$ and $c = 2$.
Solve for the instant.
$c = 2$ is the midpoint of $[1, 3]$ — true for every parabola, and for almost nothing else.
$f(x) = x^3 - 3x$: $f'(x) = 3x^2 - 3$, zero at $x = \pm 1$.
Critical points.
$f'$ is positive outside $[-1, 1]$ and negative inside: a maximum at $-1$, a minimum at $1$.
Direction.
$f''(x) = 6x$, negative left of zero and positive right of it: an inflection at the origin.
Bend.
$f'(x) = 3x^2 - 12x$, so $f''(x) = 6x - 12$.
Differentiate twice.
$f'' > 0$ when $x > 2$.
So it is concave up on $(2, \infty)$, with an inflection at $x = 2$ — and it is decreasing over part of that region, which is no contradiction at all.
For $f(x) = x^2$ on $[\,5, 6\,]$, find the $c$ guaranteed by the mean value theorem.
answer
At $x = 8$ a function has $f'(8) < 0$ and $f''(8) > 0$. What is happening there?
Find the $x$-coordinate of the inflection point of $f(x) = 5x^3 + 30x^2 + 1x$.
answer
A minister says: “Prices are still rising, but the rate of increase has fallen for the $4$th month running.” If $P(t)$ is the price level, what does that say about its derivatives?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $f(x) = 6x^2$, use the tangent line at $x = 2$ to estimate $f(2 + \tfrac{1}{3})$.
answer
$f(x) = 2x^3$. Say whether $f$ is increasing or decreasing at $x = -5$, and whether it is concave up or concave down there.
a and b
You can apply derivatives to rates, extrema and shape. Without looking: at what point in a related-rates problem do the numbers go in, and what does a negative second derivative say about a function that is increasing?
8. Your turn: a cube's edge grows at 3 cm/s. How fast is the volume growing when the edge is 5 cm?, step 3
20. Your turn: where is $f(x) = x^3 - 6x^2$ concave up?, step 3