High School Mathematics · Grades 10–12

12Differentiation: A First Course

How fast does a function change? The answer at a single point is the derivative, the single most useful number in all of applied mathematics. This chapter constructs it from average rates of change, computes it for the reference functions, and uses its sign to read the variations of a function. Everything here is extended and deepened in Chapter 22.

12.1 The average rate of change

Definition 12.1 (Average rate of change)

Let $f$ be defined on an interval $I$ , and $a \neq b$ in $I$ . The average rate of change of $f$ between $a$ and $b$ is

\frac{f(b) - f(a)}{b - a}.

It is the slope of the secant line through the points $\bigl(a, f(a)\bigr)$ and $\bigl(b, f(b)\bigr)$ of the curve.

Example 12.2

A car travels $140$ km between 2 p.m. and 4 p.m.: its average speed is $\frac{140}{2} = 70$ km/h. The speedometer, however, displays an instantaneous speed at every moment — that is exactly the notion that the derivative makes precise.

12.2 The derivative at a point

Definition 12.3 (Derivative at a point)

Let $f$ be defined on an interval containing $a$ . For $h \neq 0$ , form the average rate of change between $a$ and $a + h$ :

r(h) = \frac{f(a+h) - f(a)}{h}.

If $r(h)$ approaches a single fixed number as $h$ gets arbitrarily close to $0$ , we say that $f$ is differentiable at $a$ ; this number is the derivative of $f$ at $a$ , written $f'(a)$ :

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}.

Example 12.4

Let $f(x) = x^2$ and $a = 1$ . For $h \neq 0$ :

\frac{f(1+h) - f(1)}{h} = \frac{(1+h)^2 - 1}{h} = \frac{1 + 2h + h^2 - 1}{h} = \frac{h(2 + h)}{h} = 2 + h .

As $h$ approaches $0$ , this approaches $2$ : the function $x^2$ is differentiable at $1$ and $f'(1) = 2$ . Note the strategy: simplify the quotient until $h$ no longer appears in a denominator, then let $h \to 0$ .

Example 12.5 (A non-differentiable point)

Let $f(x) = \abs{x}$ and $a = 0$ . Then $\frac{f(h) - f(0)}{h} = \frac{\abs h}{h}$ , which equals $1$ for $h > 0$ and $-1$ for $h < 0$ : no single number is approached, and $\abs x$ is not differentiable at $0$ . Its curve has a corner there.

12.3 The tangent line

Definition 12.6 (Tangent)

If $f$ is differentiable at $a$ , the tangent to the curve at the point $A = \bigl(a, f(a)\bigr)$ is the line through $A$ with slope $f'(a)$ . Its equation is

y = f(a) + f'(a)(x - a).

The tangent is the limiting position of the secants through $A$ : as $h \to 0$ , the second point $\bigl(a+h, f(a+h)\bigr)$ slides along the curve towards $A$ , and the secant’s slope $\frac{f(a+h)-f(a)}{h}$ approaches $f'(a)$ .

The secant through A and B = (a+h, f(a+h) ) has slope f(a+h)-f(a)/h. As B slides towards A, the secant tilts into the tangent, whose slope is f'(a). — The secant through $A$ and $B = \bigl(a+h, f(a+h)\bigr)$ has slope $\frac{f(a+h)-f(a)}{h}$ . As $B$ slides towards $A$ , the secant tilts into the tangent, whose slope is $f'(a)$ .

Example 12.7

For $f(x) = x^2$ at $a = 1$ : $f(1) = 1$ and $f'(1) = 2$ (Example 12.4), so the tangent is

y = 1 + 2(x - 1) = 2x - 1 .

This is the line $d_2$ found by pure algebra in Exercise 10.11.

12.4 Derivatives of the reference functions

Proposition 12.8

On their domains of differentiability:

(c)' = 0, \quad (x)' = 1, \quad (x^2)' = 2x, \quad (x^3)' = 3x^2, \quad \left(\frac1x\right)' = -\frac{1}{x^2}, \quad (\sqrt x\,)' = \frac{1}{2\sqrt x} \ (x > 0).

More generally, $(x^n)' = n\,x^{n-1}$ for every integer $n \geq 1$ .

Proof. Each proof follows the strategy of Example 12.4: simplify the rate, then let $h \to 0$ . Fix $a$ in the domain.

Constant and identity. $\frac{c - c}{h} = 0$ and $\frac{(a+h) - a}{h} = 1$ for all $h$ : the limits are $0$ and $1$ .

Square. $\frac{(a+h)^2 - a^2}{h} = \frac{2ah + h^2}{h} = 2a + h \to 2a$ .

Cube. $(a+h)^3 = a^3 + 3a^2h + 3ah^2 + h^3$ , so the rate is $3a^2 + 3ah + h^2 \to 3a^2$ .

Reciprocal. For $a \neq 0$ and $h$ small enough that $a + h \neq 0$ :

\frac{1}{h}\left(\frac{1}{a+h} - \frac{1}{a}\right) = \frac{1}{h} \cdot \frac{a - (a+h)}{a(a+h)} = \frac{-1}{a(a+h)} \xrightarrow[h \to 0]{} -\frac{1}{a^2}.

Square root. For $a > 0$ , multiply by the conjugate:

\frac{\sqrt{a+h} - \sqrt a}{h} = \frac{(a + h) - a}{h\,(\sqrt{a+h} + \sqrt a)} = \frac{1}{\sqrt{a+h} + \sqrt a} \xrightarrow[h \to 0]{} \frac{1}{2\sqrt a}.

General power. The pattern $2a$ , $3a^2$ continues: expanding $(a+h)^n$ , the terms after $a^n + n\,a^{n-1}h$ all carry $h^2$ , so the rate tends to $n\,a^{n-1}$ . (A complete induction is given in Chapter 22.) ∎

12.5 Differentiation rules

Theorem 12.9 (Operations)

Let $u$ and $v$ be differentiable on $I$ and $\lambda \in \R$ . Then $u + v$ , $\lambda u$ , $uv$ are differentiable on $I$ , as is $\frac{u}{v}$ wherever $v \neq 0$ , and

(u+v)' = u' + v', \qquad (\lambda u)' = \lambda u', \qquad (uv)' = u'v + uv', \qquad \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}.

Proof of the first three rules. Fix $a \in I$ and abbreviate $\Delta u = u(a+h) - u(a)$ , $\Delta v = v(a+h) - v(a)$ .

Sum. The rate of $u + v$ is $\frac{\Delta u + \Delta v}{h} = \frac{\Delta u}{h} + \frac{\Delta v}{h} \to u'(a) + v'(a)$ .

Multiple. The rate of $\lambda u$ is $\lambda \frac{\Delta u}{h} \to \lambda u'(a)$ .

Product. Add and subtract the mixed term $u(a+h)\,v(a)$ :

\begin{align*} \frac{u(a+h)v(a+h) - u(a)v(a)}{h} &= \frac{u(a+h)v(a+h) - u(a+h)v(a) + u(a+h)v(a) - u(a)v(a)}{h}\\ &= u(a+h)\,\frac{\Delta v}{h} + v(a)\,\frac{\Delta u}{h}. \end{align*}

As $h \to 0$ : $\frac{\Delta v}{h} \to v'(a)$ , $\frac{\Delta u}{h} \to u'(a)$ , and $u(a+h) \to u(a)$ (its rate has a limit, so $\Delta u = h \cdot \frac{\Delta u}{h} \to 0$ ). The rate therefore tends to $u(a)v'(a) + v(a)u'(a)$ .

The quotient rule is proved along the same lines; it is admitted at this level and proved in Chapter 22. ∎

Example 12.10

Polynomial. $f(x) = 3x^3 - 5x^2 + 7x - 2$ :

f'(x) = 3 \times 3x^2 - 5 \times 2x + 7 = 9x^2 - 10x + 7 .

Quotient. $g(x) = \frac{2x+1}{x^2+1}$ : with $u = 2x + 1$ , $v = x^2 + 1$ ,

g'(x) = \frac{2(x^2+1) - (2x+1)\,2x}{(x^2+1)^2} = \frac{-2x^2 - 2x + 2}{(x^2+1)^2}.

Product. $k(x) = x\sqrt x$ for $x > 0$ : $k'(x) = 1 \times \sqrt x + x \times \frac{1}{2\sqrt x} = \sqrt x + \frac{\sqrt x}{2} = \frac{3\sqrt x}{2}$ .

12.6 Sign of the derivative and variations

Theorem 12.11 (Derivative and variations)

Let $f$ be differentiable on an interval $I$ .

If $f' \geq 0$ on $I$ , then $f$ is increasing on $I$ .
If $f' \leq 0$ on $I$ , then $f$ is decreasing on $I$ .
If $f' > 0$ on $I$ (except possibly at finitely many points where it vanishes), then $f$ is strictly increasing on $I$ .

Proof. The intuition is clear — a curve whose tangents all point upward must climb — but a rigorous proof needs the mean value theorem. The result is admitted at this level; see Theorem 22.7. ∎

Proposition 12.12 (Extrema)

If $f'$ changes sign at $a$ from $+$ to $-$ , then $f$ has a local maximum at $a$ ; from $-$ to $+$ , a local minimum. At such a point, $f'(a) = 0$ and the tangent is horizontal.

Proof. If $f' \geq 0$ on an interval ending at $a$ and $f' \leq 0$ on an interval starting at $a$ , then $f$ increases up to $f(a)$ and decreases afterwards (Theorem 12.11): $f(a)$ is largest nearby. Since $f'$ changes sign at $a$ and is defined there, $f'(a) = 0$ . ∎

Method 12.13 (Studying a function)

Compute $f'(x)$ and simplify it — ideally into a factored form.
Determine the sign of $f'(x)$ on each interval (for a quadratic $f'$ , use Theorem 10.11).
Record the results in a variation table: one row for the sign of $f'$ , one row of arrows for $f$ , with the values of $f$ at the turning points.
Conclude: extrema, number of solutions of $f(x) = k$ , inequalities.

Example 12.14

Study $f(x) = x^3 - 3x + 1$ on $\R$ . First, $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$ : a quadratic with roots $-1$ and $1$ , positive outside them. So $f$ is increasing on $\intoc{-\infty}{-1}$ , decreasing on $\intcc{-1}{1}$ , increasing on $\intco{1}{+\infty}$ , with a local maximum $f(-1) = -1 + 3 + 1 = 3$ and a local minimum $f(1) = 1 - 3 + 1 = -1$ .

The curve of f(x) = x3 - 3x + 1 (): horizontal tangents (orange) at the local maximum (-1, 3) and the local minimum (1, -1), where f' changes sign. — The curve of $f(x) = x^3 - 3x + 1$ (Example 12.14): horizontal tangents (orange) at the local maximum $(-1, 3)$ and the local minimum $(1, -1)$ , where $f'$ changes sign.

Example 12.15 (Optimization)

An open box is made from a $12 \times 12$ cm sheet by cutting a square of side $x$ from each corner and folding. Its volume is $V(x) = x(12 - 2x)^2$ for $0 < x < 6$ . Expanding, $V(x) = 4x^3 - 48x^2 + 144x$ , so

V'(x) = 12x^2 - 96x + 144 = 12(x^2 - 8x + 12) = 12(x - 2)(x - 6).

On $\intoo{0}{6}$ , the factor $x - 6$ is negative, so $V'(x)$ has the sign of $-(x-2)$ : positive before $2$ , negative after. The volume is maximal for $x = 2$ cm, giving $V(2) = 2 \times 8^2 = 128$ cm $^3$ .

12.7 Exercises

Exercise 12.1 ★

Using the definition (rate of change, then $h \to 0$ ), compute $f'(2)$ for $f(x) = x^2$ , then $g'(1)$ for $g(x) = x^2 + 3x$ .

Solution

Solution of Exercise 12.1.

For $f(x) = x^2$ at $a = 2$ :

\frac{(2+h)^2 - 4}{h} = \frac{4h + h^2}{h} = 4 + h \xrightarrow[h\to0]{} 4,

so $f'(2) = 4$ . For $g(x) = x^2 + 3x$ at $a = 1$ : $g(1) = 4$ and

\frac{(1+h)^2 + 3(1+h) - 4}{h} = \frac{5h + h^2}{h} = 5 + h \xrightarrow[h\to0]{} 5,

so $g'(1) = 5$ .

Exercise 12.2 ★

Differentiate:

f(x) = 4x^3 - 2x^2 + x - 7, \qquad g(x) = 5\sqrt{x} + \frac{3}{x}, \qquad h(x) = (2x+1)(x^2 - 3) .

Solution

Solution of Exercise 12.2.

$f'(x) = 12x^2 - 4x + 1$ .

$g'(x) = \frac{5}{2\sqrt x} - \frac{3}{x^2}$ (on $\intoo{0}{+\infty}$ ).

$h'(x) = 2(x^2 - 3) + (2x+1)(2x) = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6$ (product rule).

Exercise 12.3 ★

Give the equation of the tangent to the curve of $f$ at the point of abscissa $a$ :

f(x) = x^2 - 3x, \ a = 2; \qquad f(x) = \frac1x, \ a = 1; \qquad f(x) = \sqrt x, \ a = 4 .

Solution

Solution of Exercise 12.3.

$f(x) = x^2 - 3x$ , $a = 2$ : $f(2) = -2$ , $f'(x) = 2x - 3$ so $f'(2) = 1$ ; tangent $y = -2 + (x - 2) = x - 4$ .

$f(x) = \frac1x$ , $a = 1$ : $f(1) = 1$ , $f'(1) = -1$ ; tangent $y = 1 - (x - 1) = 2 - x$ .

$f(x) = \sqrt x$ , $a = 4$ : $f(4) = 2$ , $f'(4) = \frac{1}{2\sqrt4} = \frac14$ ; tangent $y = 2 + \frac14(x - 4) = \frac{x}{4} + 1$ .

Exercise 12.4 ★

Differentiate the quotients:

f(x) = \frac{x+2}{x-1}, \qquad g(x) = \frac{x^2}{x^2+1}, \qquad h(x) = \frac{1}{x^2 + x + 1} .

Solution

Solution of Exercise 12.4.

$f'(x) = \frac{(x-1) - (x+2)}{(x-1)^2} = \frac{-3}{(x-1)^2}$ .

$g'(x) = \frac{2x(x^2+1) - x^2 \times 2x}{(x^2+1)^2} = \frac{2x}{(x^2+1)^2}$ .

$h'(x) = -\frac{2x+1}{(x^2+x+1)^2}$ (quotient rule with $u = 1$ ).

Exercise 12.5 ★★

Study the variations of $f(x) = x^3 - 6x^2 + 9x - 2$ on $\R$ (variation table, local extrema).

Solution

Solution of Exercise 12.5.

$f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$ : positive outside $\intcc{1}{3}$ , negative inside. So $f$ is increasing on $\intoc{-\infty}{1}$ , decreasing on $\intcc{1}{3}$ , increasing on $\intco{3}{+\infty}$ , with a local maximum $f(1) = 1 - 6 + 9 - 2 = 2$ and a local minimum $f(3) = 27 - 54 + 27 - 2 = -2$ .

Exercise 12.6 ★★

Study the variations of $f(x) = \frac{x^2 + 3}{x + 1}$ on $\intoo{-1}{+\infty}$ .

Solution

Solution of Exercise 12.6.

On $\intoo{-1}{+\infty}$ :

f'(x) = \frac{2x(x+1) - (x^2+3)}{(x+1)^2} = \frac{x^2 + 2x - 3}{(x+1)^2} = \frac{(x+3)(x-1)}{(x+1)^2}.

For $x > -1$ , both $x + 3$ and $(x+1)^2$ are positive, so $f'(x)$ has the sign of $x - 1$ : $f$ is decreasing on $\intoc{-1}{1}$ and increasing on $\intco{1}{+\infty}$ , with minimum $f(1) = \frac{4}{2} = 2$ .

Exercise 12.7 ★★

Let $f(x) = x^2 - 4x + 5$ .

Find the point of the curve where the tangent is horizontal.
Find the point(s) where the tangent is parallel to the line $y = 2x + 1$ .

Solution

Solution of Exercise 12.7.

$f'(x) = 2x - 4$ .

1. Horizontal tangent: $f'(x) = 0$ at $x = 2$ ; the point is $(2, f(2)) = (2, 1)$ (the vertex of the parabola).

2. Parallel to $y = 2x + 1$ means slope $2$ : $2x - 4 = 2$ , so $x = 3$ , giving the point $(3, f(3)) = (3, 2)$ .

Exercise 12.8 ★★

A rectangular enclosure is built against a wall with $60$ m of fence (three sides). Express the area as a function of the width $x$ , and use the derivative to find the dimensions of maximal area. (Compare with the vertex method of Chapter 10.)

Solution

Solution of Exercise 12.8.

With width $x$ (two fenced sides) and length $60 - 2x$ : $A(x) = x(60 - 2x) = 60x - 2x^2$ on $\intoo{0}{30}$ . Then $A'(x) = 60 - 4x$ , positive before $x = 15$ and negative after: the area is maximal for $x = 15$ , giving a $15 \times 30$ m enclosure of area $450$ m $^2$ . The vertex formula $\alpha = -\frac{60}{2 \times (-2)} = 15$ of Chapter 10 gives the same answer without calculus.

Exercise 12.9 ★★

Show that for all $x > 0$ , $\sqrt{x} \leq \frac{x + 1}{2}$ , by studying the function $f(x) = \frac{x+1}{2} - \sqrt x$ on $\intoo{0}{+\infty}$ .

Solution

Solution of Exercise 12.9.

Let $f(x) = \frac{x+1}{2} - \sqrt x$ on $\intoo{0}{+\infty}$ . Then

f'(x) = \frac12 - \frac{1}{2\sqrt x} = \frac{\sqrt x - 1}{2\sqrt x},

which has the sign of $\sqrt x - 1$ : negative on $\intoo{0}{1}$ , positive on $\intoo{1}{+\infty}$ . So $f$ decreases then increases, with minimum $f(1) = 1 - 1 = 0$ . Hence $f(x) \geq 0$ for all $x > 0$ , that is $\sqrt x \leq \frac{x+1}{2}$ , with equality only at $x = 1$ .

Exercise 12.10 ★★

A cylindrical can must hold $500$ cm $^3$ . Its surface area (two disks plus the side) is $S = 2\pi r^2 + 2\pi r h$ with $\pi r^2 h = 500$ . Express $S$ as a function of $r$ alone, and show that $S'(r) = 4\pi r - \frac{1000}{r^2}$ . Deduce the radius that minimizes the surface, as an exact expression.

Solution

Solution of Exercise 12.10.

From $\pi r^2 h = 500$ , $h = \frac{500}{\pi r^2}$ , so

S(r) = 2\pi r^2 + 2\pi r \cdot \frac{500}{\pi r^2} = 2\pi r^2 + \frac{1000}{r}, \qquad S'(r) = 4\pi r - \frac{1000}{r^2}.

$S'(r) = 0$ when $4\pi r^3 = 1000$ , i.e. $r^3 = \frac{250}{\pi}$ , so $r = \sqrt[3]{250/\pi}$ ( $\approx 4.3$ cm). Since $S'(r) = \frac{4\pi r^3 - 1000}{r^2}$ is negative before this value and positive after, it is a minimum.

Exercise 12.11 ★★★

How many tangents to the parabola $y = x^2$ pass through the external point $P(1, -3)$ ? Determine their equations. (Let $a$ be the abscissa of the contact point and express that the tangent at $a$ passes through $P$ .)

Solution

Solution of Exercise 12.11.

The tangent to $y = x^2$ at the point of abscissa $a$ is $y = a^2 + 2a(x - a) = 2ax - a^2$ . It passes through $P(1, -3)$ when $-3 = 2a - a^2$ , i.e. $a^2 - 2a - 3 = 0$ , whose roots are $a = 3$ and $a = -1$ ( $\Delta = 16$ ). There are therefore two tangents through $P$ :

y = 6x - 9 \quad (a = 3), \qquad y = -2x - 1 \quad (a = -1).

Exercise 12.12 ★★★

Let $f(x) = x^3 - 3x + 1$ (see Example 12.14). Using the variation table, determine the number of solutions of the equation $f(x) = k$ according to the value of the real number $k$ .

Solution

Solution of Exercise 12.12.

From Example 12.14, $f$ increases up to the local maximum $f(-1) = 3$ , decreases to the local minimum $f(1) = -1$ , then increases again; for $x$ large positive (resp. negative), $x^3$ dominates and $f$ takes arbitrarily large (resp. small) values. Reading horizontal lines $y = k$ across the variation table:

$k > 3$ or $k < -1$ : the line meets the curve once — $1$ solution;
$k = 3$ or $k = -1$ : the line touches a turning point and crosses one other branch — $2$ solutions;
$-1 < k < 3$ : the line crosses all three branches — $3$ solutions.

12.8 Problem: The tangent’s three jobs

Problem 12.1

Weekend problem — the parabolic mirror proved, Newton’s root-finding machine, and the strongest beam in the log

A derivative draws tangents; that sounds like a modest talent. This problem shows the tangent holding three jobs at once: it proves the headlight theorem left pending in Problem 10.1 (every ray parallel to a parabola’s axis reflects through the focus), it powers the fastest equation-solving algorithm in common use — inside every calculator and neural network — and it finds the strongest rectangular beam that a round log can yield.

Part I — Tangent fluency.

Differentiate $f(x) = 3x^2 - 5x + 1$ and $g(x) = x^3 - 12x$ (Theorem 12.9); and recover $\left(x^2 \right)'$ at $x = 1$ from the definition (rate of change, then $h \to 0$ ).
Find the tangent to $y = x^3 - 12x$ at $a = 2$ . What is special about it, and what does that signal (Proposition 12.12)?
Establish the general tangent to the parabola $y = x^2$ at the point $(a, a^2)$ :

$y = 2ax - a^2 .$
Where does that tangent cross the $y$ -axis? Deduce the draftsman’s recipe: to draw the tangent at a point $P$ of height $h$ on the parabola, join $P$ to the point of the axis at depth … below the origin. State it.
Build the complete variation table of $g(x) = x^3 - 12x$ (sign of $g'$ , local extrema and their values, Method 12.13).

Part II — The mirror theorem. Let $P(a, a^2)$ be a point of the parabola $y = x^2$ (take $a \neq 0$ ), $F\left(0, \frac14\right)$ the focus, and $T$ the point where the tangent at $P$ crosses the $y$ -axis (question 4). Recall from Problem 10.1 that $PF = a^2 + \frac14$ .

Compute the distance $FT$ .
Conclude that the triangle $FPT$ is isosceles at $F$ . Which two angles are therefore equal?
The incoming ray through $P$ , parallel to the axis, is parallel to the line $(TF)$ (both vertical). Using alternate angles, show that the tangent at $P$ makes equal angles with the incoming vertical ray and with the segment $[PF]$ .
Physics: light reflects off a curve as off its tangent, leaving at the same angle it arrived (the law of reflection). Conclude the mirror theorem: every ray parallel to the axis reflects through the focus — and, reversing time, a bulb at the focus beams out parallel light. The dish and the headlight of Problem 10.1 are now theorems.
Check the isosceles triangle numerically at $a = 1$ : compute $T$ , $FP$ and $FT$ .

Part III — Newton’s machine. To solve $f(x) = 0$ , Newton proposed: start from a guess $x_0$ , follow the tangent at $x_0$ down to the axis, and let its $x$ -intercept be the next guess.

Derive the formula of the method:

$x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)} .$
Run it on $f(x) = x^2 - 2$ from $x_0 = 1$ : compute $x_1$ , $x_2$ , $x_3$ as exact fractions. Which two-thousand-year-old sequence — and which machine from Problem 3.1 — have you just rebuilt? Prove the identity behind the coincidence:

$x - \frac{x^2 - 2}{2x} = \frac12\left(x + \frac2x\right).$
Solve $x^3 = 100$ (the savings crossover of Problem 4.1): apply two Newton steps to $f(x) = x^3 - 100$ from $x_0 = 5$ (four decimals), and compare with the calculator’s $\sqrt[3]{100}$ .
Sabotage: try to launch the method on $f(x) = x^3 - 12x$ at $x_0 = 2$ . What goes wrong, and why (question 2)? State the practical caveat.
In question 12 the correct decimals went roughly $1 \to 3 \to 6$ . Compare with bisection (halving an interval each step) and say in one sentence why calculators, GPS receivers and neural-network training all run on Newton’s idea.

Part IV — The strongest beam. A rectangular beam of width $w$ and depth $d$ is sawn from a round log of diameter $30$ cm, so $w^2 + d^2 = 900$ . A beam’s stiffness is proportional to $w d^2$ (depth counts twice: joists stand on edge!).

Express the stiffness as a function of $w$ alone: $S(w) = 900w - w^3$ , on which interval?
Differentiate, build the variation table, and find the optimal $w$ and $d$ (exact values, then to the millimeter).
Show that the optimal proportions satisfy $d = w\sqrt2$ — the old carpenters’ ratio. (Bonus lore: dividing the log’s diameter in three equal parts and raising perpendiculars to the circle constructs exactly this rectangle.)
The naive choice is the square beam ( $w = d$ ). Compute its stiffness and the optimal beam’s, and give the carpenter’s gain in percent.
Finale — the tangent’s three jobs, one sentence each: geometry (the mirror theorem), numerics (Newton’s machine), optimization (the beam), all powered by the same object. And the forward pointer: grade 12 adds convexity — the tangent’s side of the curve — to sharpen all three.

Solution

Solution of Problem 12.1.

1. $f'(x) = 6x - 5$ ; $g'(x) = 3x^2 - 12$ . Definition: $\frac{(1 + h)^2 - 1}{h} = 2 + h \to 2$ : the derivative of $x^2$ at $1$ is $2$ .

2. $g(2) = 8 - 24 = -16$ and $g'(2) = 12 - 12 = 0$ : the tangent is the horizontal line $y = -16$ . A horizontal tangent signals a critical point — here a local minimum, as the variation table of question 5 confirms.

3. Slope $2a$ at $(a, a^2)$ : $y = a^2 + 2a(x - a) = 2ax - a^2$ .

4. At $x = 0$ : $y = -a^2$ : the tangent crosses the axis at depth $a^2$ below the origin — exactly as far below zero as $P$ sits above it. Recipe: to draw the tangent at a point of height $h$ , join it to the axis point at depth $h$ below the origin. (No derivative needed at the drawing board.)

5. $g'(x) = 3(x^2 - 4)$ : positive on $\intoo{-\infty}{-2}$ , negative on $\intoo{-2}{2}$ , positive beyond. $g$ increases to a local maximum $g(-2) = 16$ , falls to a local minimum $g(2) = -16$ , then increases.

6. $F\left(0, \frac14\right)$ and $T(0, -a^2)$ : $FT = \frac14 + a^2$ .

7. $FP = a^2 + \frac14 = FT$ : isosceles at $F$ , so the base angles are equal: $\widehat{FTP} = \widehat{FPT}$ .

8. The incoming ray at $P$ is vertical, hence parallel to the line $(TF)$ (the $y$ -axis). The tangent $(TP)$ crosses both parallels, so the angle between the ray and the tangent equals $\widehat{FTP}$ (alternate angles) — which equals $\widehat{FPT}$ , the angle between $[PF]$ and the tangent. Equal angles on the two sides of the tangent at $P$ .

9. By the law of reflection, the ray arriving vertically at $P$ leaves along the line making the equal angle on the other side — by question 8, along $[PF]$ : through the focus, whatever $a$ is. Reversed, light born at $F$ leaves every point of the mirror parallel to the axis. Dishes collect, headlights beam: proved.

10. $a = 1$ : tangent $y = 2x - 1$ , $T(0, -1)$ ; $FP = \sqrt{1 + \left(1 - \frac14\right)^2} = \sqrt{\frac{25}{16}} = \frac54$ ; $FT = \frac14 + 1 = \frac54$ : isosceles, as promised.

11. The tangent at $x_n$ is $y = f(x_n) + f'(x_n)(x - x_n)$ ; it crosses $y = 0$ where $x = x_n - \frac{f(x_n)}{f'(x_n)}$ .

12. $x_1 = 1 - \frac{-1}{2} = \frac32$ ; $x_2 = \frac32 - \frac{1/4}{3} = \frac{17}{12}$ ; $x_3 = \frac{577}{408}$ : Heron’s approximations of $\sqrt2$ , i.e. the fixed-point machine of Problem 3.1. Identity: $x - \frac{x^2 - 2}{2x} = \frac{2x^2 - x^2 + 2}{2x} = \frac{x^2 + 2}{2x} = \frac12\left(x + \frac2x\right)$ — Heron’s recipe is Newton’s method applied to $x^2 - 2$ , two millennia early.

13. $x_1 = 5 - \frac{125 - 100}{75} = \frac{14}{3} \approx 4.6667$ ; $x_2 \approx 4.6667 - \frac{1.63}{65.33} \approx 4.6417$ . Calculator: $\sqrt[3]{100} = 4.6416\ldots$ — four correct decimals in two steps.

14. $f'(2) = 0$ : the tangent at the starting point is horizontal (question 2) and never crosses the axis — the formula divides by zero. Caveat: start Newton away from critical points (and, more generally, close enough to the root you want).

15. Bisection gains one binary digit per step; Newton roughly doubles the number of correct digits each step ( $1 \to 3 \to 6$ here). When each step is costly — millions of parameters, real-time navigation — doubling beats inching, which is why the tangent line runs inside so much silicon.

16. $d^2 = 900 - w^2$ , so $S(w) = w(900 - w^2) = 900w - w^3$ , for $0 < w < 30$ .

17. $S'(w) = 900 - 3w^2 = 0$ at $w = \sqrt{300} = 10\sqrt3 \approx 17.3$ cm ( $S' > 0$ before, $< 0$ after: a maximum). Then $d = \sqrt{600} = 10\sqrt6 \approx 24.5$ cm.

18. $\frac{d^2}{w^2} = \frac{600}{300} = 2$ , so $d = w\sqrt2$ : depth is width times $\sqrt2$ — the carpenter’s ratio (and the diameter-in-thirds construction delivers it with a compass, no algebra on site).

19. Square beam: $w = d = \sqrt{450} \approx 21.2$ cm, stiffness $w^3 = 450^{3/2} \approx 9\,546$ . Optimal beam: $S = 10\sqrt3 \times 600 = 6\,000\sqrt3 \approx 10\,392$ . Gain: about $8.9\,\%$ stiffer from the same log — the derivative pays for the sawmill.

20. Geometry: the tangent’s equal angles turned a paraboloid into a light-gatherer. Numerics: sliding down tangents solves equations at digit-doubling speed. Optimization: a vanishing derivative found the log’s strongest rectangle. One object, three trades — and grade 12’s convexity will tell, in addition, on which side of every tangent the curve lies.