Calculus

Things change: a car’s position, the temperature of a cup of tea, the money in a savings account, the number of people with a cold. Calculus is the mathematics that measures how fast something changes, and how much it adds up to over time. You start with the simplest version of the first question: a car trip.

You drive 150 kilometres and it takes 2 hours. Your average speed is the distance divided by the time:

The car was almost never going exactly 75 km/h. It stopped at lights and sped up on the motorway. The average hides all of that. Calculus is what lets you talk about the speed at one exact moment, the number on the speedometer. That takes the whole of the next three modules to do properly, and this module shows why it’s hard.

Rate of change

How much one quantity changes for each unit of another: kilometres per hour, degrees per minute.

Average speed

Total distance divided by total time.

Calculus

The mathematics of rates of change (derivatives) and of adding up (integrals).

Speed is one rate of change. In general you have a function $f$: a rule that turns an input $x$ into an output $f(x)$. When the input moves from $a$ to $b$, the output moves from $f(a)$ to $f(b)$. The average rate of change is the change in output divided by the change in input. The Greek letter $\mathrm{\Delta }$ (delta) means "change in".

A worked example with $f(x)=x^2$ from $a=1$ to $b=3$:

Draw the graph of $f$ and mark the two points $(a,f(a))$ and $(b,f(b))$. The straight line through them is a secant line (from the Latin for "cutting"). Its slope, rise over run, is exactly the average rate of change. So averages are slopes, and you can see them.

Why $2+h$? The two points are $(1,1)$ and $(1+h,{(1+h)}^2)$, and ${(1+h)}^2-1=2h+h^2$. Divide by the run, $h$, and you get $2+h$. Remember this: you will need it in a moment.

From the next slide on, small programs compute what each module teaches: rates, limits, derivatives, integrals. They are in C++, which the Algebra course taught. Here is the C++ they use most, plus one thing calculus needs all the time: handing a function to a function, as when you give a derivative or an integral the $f$ to work on. Run it.

1#include <cmath>2#include <functional>3#include <iostream>4#include <vector>5
6double square(double x) {  // returns a double, given a double7  return x * x;8}9
10// f is any function from double to double: call it like one.11double twiceAt(std::function<double(double)> f, double x) {12  return 2 * f(x);13}14
15int main() {16  std::cout << 7 / 2 << " but " << 7.0 / 2 << "\n";17  std::cout << std::pow(2, 10) << " " << std::sqrt(2.0) << "\n";18
19  std::vector<double> xs = {0.5, 1, 2};20  for (double x : xs) {21    if (x >= 1) std::cout << square(x) << "\n";22  }23
24  auto cube = [](double x) { return x * x * x; };  // a lambda25  std::cout << twiceAt(cube, 2) << " " << twiceAt(square, 3) << "\n";26}

• A program starts in int main(), and std::cout << x prints x. A function names the type it returns and its parameters’ types: double square(double x).
• int holds whole numbers, double numbers with a fraction. Two ints divide to a whole number, dropping the rest: 7 / 2 is 3 and 1 / 2 is 0. Write 7.0 / 2 when you mean $3.5$.
• <cmath> has std::sqrt, std::pow, std::exp, std::log, std::abs and std::sin (in radians). There is no power operator: $x^3$ is std::pow(x, 3) or x * x * x. ^ means something else entirely: 2 ^ 3 is 1 (bitwise XOR).
• std::vector<double> is a list of numbers, and for (double x : xs) visits each one.
• std::function<double(double)>, from <functional>, is the type "a function from double to double". [](double x) { return x * x * x; } is a lambda: a function written right where it is needed, and auto lets the compiler work out its type. Named functions and lambdas can both be passed.

Drop a ball from a tall tower. Ignoring air, the distance it has fallen after $t$ seconds is about $d(t)=4.9t^2$ metres. The average speed between two times is the average rate of change of $d$, which is the slope of a secant on the distance–time graph.

1#include <functional>2
3double averageRate(std::function<double(double)> f,4                   double a, double b) {5  // TODO: the change in f over the change in x6  return 0;7}

How fast is the ball going at exactly $t=1$ second? Average speed needs two times. If you use the same time twice, you get distance $0$ over time $0$, and $\frac{0}{0}$ means nothing. This is the puzzle that took mathematicians centuries to settle.

The way out: take averages over shorter and shorter intervals starting at 1 s, and watch where they head. From $t=1$ to $t=1+h$ the average speed is $4.9(2+h)$ m/s, by the same algebra as the secant of $x^2$.

Here is the table as a program. Change the time t0 or the function d, and run it again. Watch the averages settle on one number.

1#include <iostream>2
3int main() {4  auto d = [](double t) { return 4.9 * t * t; };  // metres fallen5  double t0 = 1;  // the moment we care about6
7  for (double h : {1.0, 0.5, 0.1, 0.01, 0.001, 0.0001}) {8    double average = (d(t0 + h) - d(t0)) / h;9    std::cout << "h = " << h << ": " << average << " m/s\n";10  }11}

Now turn the question around. Suppose you know the speed at every moment and want the distance. At a steady 20 m/s for 5 s you go $20\times 5=100$ m. On a speed–time graph that is a rectangle 5 wide and 20 tall: the distance is the area under the speed graph.

That stays true when the speed changes. A car that speeds up steadily has speed $v(t)=2t$ m/s. Under that line, from $0$ to $T$, is a triangle with base $T$ and height $2T$, so the distance is $\frac{1}{2}\cdot T\cdot 2T=T^2$ metres.

Calculus has two halves, one for each question you just met. Both are about curves, and both need you to deal with "infinitely small" pieces.

Look at the last line of each side. Speed comes from distance, and distance comes from speed: the two questions undo each other. That fact is the fundamental theorem of calculus, the high point of the first half of this course (module 10).

Parts of calculus are ancient. Archimedes (around 250 BC) found the area of a parabola’s segment by filling it with ever smaller triangles. In the 1630s Pierre de Fermat found tangents and highest points of curves. What was missing was a general method, and the link between the two questions.

Isaac Newton worked it out in England in 1665–66, calling derivatives "fluxions", but published little of it for decades. Gottfried Wilhelm Leibniz found it independently in Germany and published first, in 1684. The notation you will use, $\frac{dy}{dx}$ and $\int$, is Leibniz’s. A bitter argument over who came first followed. Both deserve the credit.

Their calculus worked but rested on vague "infinitely small" quantities. In the 1820s Augustin-Louis Cauchy built it on limits, and Karl Weierstrass, around 1860, gave the precise ε-δ definitions mathematicians use today (module 20).

• The average rate of change of $f$ from $a$ to $b$ is $\frac{f(b)-f(a)}{b-a}$: the slope of the secant line.
• Average speed is the average rate of change of distance.
• The rate "right now" is what averages over shrinking intervals approach, not $\frac{0}{0}$.
• Distance is the area under the speed graph.
• Slopes (derivatives) and areas (integrals) are the two halves of calculus, and they undo each other.

Take $f(x)=\frac{x^2-1}{x-1}$. At $x=1$ it is $\frac{0}{0}$, so $f(1)$ has no value. But for every other $x$ you can cancel: $x^2-1=(x-1)(x+1)$, so $f(x)=x+1$. The graph is the line $y=x+1$ with one point missing, a hole at $(1,2)$.

You say "the limit of $f(x)$ as $x$ approaches $1$ is $2$", and write it like this. It is about the values near $1$, never the value at $1$: the function doesn’t even need to be defined there.

Limit

The value $f(x)$ gets as close as you like to, by taking $x$ close enough to $a$ (but not equal to it).

$x \to a$

Read "$x$ approaches $a$": $x$ takes values nearer and nearer to $a$, from either side.

Before any algebra, you can estimate a limit by computing the function at inputs that close in on $a$ from both sides. For ${\lim }_{x\to 1}\frac{x^2-1}{x-1}$:

1#include <cmath>2#include <functional>3#include <vector>4
5std::vector<double> valuesNear(std::function<double(double)> f,6                               double a, int n) {7  std::vector<double> values;8  // TODO: one value for each k from 1 to n9  return values;10}

Sometimes a function heads for different values from each side. The left-hand limit ${\lim }_{x\to a^{-}}f(x)$ uses only $x<a$; the right-hand limit ${\lim }_{x\to a^{+}}f(x)$ uses only $x>a$. The two-sided limit exists only when both exist and are equal.

• A jump: the one-sided limits differ, as on the last slide.
• Blowing up: $\frac{1}{x^2}$ grows without bound near $0$. You write ${\lim }_{x\to 0}\frac{1}{x^2}=\infty$, but $\infty$ is not a number: it says how the limit fails to exist.
• Oscillation: $\sin \left(\frac{1}{x}\right)$ swings between $-1$ and $1$ infinitely often as $x\to 0$, never settling.

If ${\lim }_{x\to a}f(x)=L$ and ${\lim }_{x\to a}g(x)=M$ both exist, limits behave the way you would hope:

Together they give direct substitution: for a polynomial, or a fraction of polynomials whose bottom isn’t $0$ at $a$, the limit is just the value at $a$. So ${\lim }_{x\to 2}(x^3-4x+1)=8-8+1=1$.

$\frac{0}{0}$ is not an answer. It is a sign that something on top and bottom cancels. Two tricks find it: factor and cancel, or rationalise, multiplying top and bottom by the conjugate to clear a square root.

1x^2 - 9 = (x - 3)(x + )2(x^2 - 9) / (x - 3) = x + 3     when x is not 33limit as x -> 3 = 3 + 3 = 

Some limits can’t be simplified, but can be trapped. If $g(x)\le f(x)\le h(x)$ near $a$, and $g$ and $h$ have the same limit $L$ at $a$, then $f$ has no room to go anywhere else: its limit is $L$ too.

Example: $x^2\sin \left(\frac{1}{x}\right)$ near $0$. The sine is between $-1$ and $1$, so the function is between $-x^2$ and $x^2$. Both of those go to $0$, so ${\lim }_{x\to 0}{x}^2\sin \left(\frac{1}{x}\right)=0$, even though $\sin \left(\frac{1}{x}\right)$ alone has no limit.

With $x$ in radians, $\frac{\sin x}{x}$ approaches $1$ as $x\to 0$. You can’t substitute ($\frac{0}{0}$) or factor. It is proved with the squeeze theorem: a picture of the unit circle shows $\cos x\le \frac{\sin x}{x}\le 1$ for small $x$, and $\cos x\to 1$.

${\lim }_{x\to \infty }f(x)=L$ means $f(x)$ gets as close as you like to $L$ once $x$ is large enough. The line $y=L$ is then a horizontal asymptote. For a fraction of polynomials, divide top and bottom by the highest power of $x$ on the bottom:

1#include <cmath>2#include <vector>3
4double limitAtInfinity(const std::vector<double>& top,5                       const std::vector<double>& bottom) {6  return 0;7}

• ${\lim }_{x\to a}f(x)=L$: $f(x)$ gets as close as you like to $L$ for $x$ near $a$, never looking at $x=a$ itself.
• A two-sided limit exists only if both one-sided limits exist and agree.
• Limits fail by jumping, blowing up or oscillating.
• Try substitution first; on $\frac{0}{0}$, factor or rationalise.
• Squeeze a hard function between two easy ones with the same limit.
• ${\lim }_{x\to 0}\frac{\sin x}{x}=1$, in radians.
• At infinity, the highest powers decide.

Informally, a function is continuous if you can draw its graph without lifting your pen: no holes, no jumps, no shooting off to infinity. Temperature through a day is continuous; the price of a parking ticket, which jumps at each hour, is not.

Limits make that precise. $f$ is continuous at $a$ when three things hold:

1. $f(a)$ is defined (there’s a point on the graph at $x=a$),
2. ${\lim }_{x\to a}f(x)$ exists (the graph heads for one height from both sides),
3. and the two are equal: ${\lim }_{x\to a}f(x)=f(a)$.

Continuous at a point

${\lim }_{x\to a}f(x)=f(a)$.

Continuous on an interval

Continuous at every point of it. At a closed end, only the one-sided limit from inside counts.

Discontinuity

A point where a function isn’t continuous.

A removable discontinuity is called that because you can remove it: define (or redefine) $f(a)$ to be the limit, and the hole is filled. No choice of $f(a)$ can fix the other three.

You rarely check the three conditions by hand. These are continuous at every point of their domain (the inputs where they’re defined):

• polynomials, everywhere;
• fractions of polynomials, except where the bottom is $0$;
• $\sqrt{x}$ for $x\ge 0$, $e^x$ everywhere, $\ln x$ for $x>0$;
• $\sin x$ and $\cos x$ everywhere, $\tan x$ except at odd multiples of $\frac{\pi }{2}$.

Sums, differences, products and compositions of continuous functions are continuous, and so are quotients where the bottom isn’t $0$. So $e^{\sin x}+x^2$ is continuous everywhere, with no work.

A function built from pieces is continuous where the pieces meet only if their heights match. Take $f(x)=kx+1$ for $x<2$ and $f(x)=x^2$ for $x\ge 2$. At $x=2$ the left piece arrives at $2k+1$ and the right piece is at $4$. They meet when $2k+1=4$, so $k=\frac{3}{2}$.

1left piece at x = 1:   3 - c2right piece at x = 1:  33 - c = 2, so c = 

If you walk up a hill from 200 m to 500 m above sea level without teleporting, at some moment you were at exactly 350 m. That is the intermediate value theorem (IVT):

Its most useful form: if $f(a)$ and $f(b)$ have opposite signs, then $f$ has a root (a zero) between them. For example $f(x)=\cos x-x$ has $f(0)=1>0$ and $f(1)\approx -0.46<0$, so the equation $\cos x=x$ has a solution in $[0,1]$, though no formula gives it.

Drop the word "continuous" and the promise breaks. $f(x)=\frac{1}{x}$ has $f(-1)=-1$ and $f(1)=1$, opposite signs, yet $\frac{1}{x}$ is never $0$. It jumps from $-\infty$ to $\infty$ at $x=0$ instead of crossing.

The IVT gives an algorithm. Start with $[a,b]$ where $f$ changes sign. Look at the midpoint $m$. Whichever half still has a sign change contains a root; keep that half. Each step halves the interval, so the root is trapped ever more tightly. Here it finds $\sqrt{2}$, the root of $x^2-2$:

The extreme value theorem: a function continuous on a closed interval $[a,b]$ has a highest value (a maximum) and a lowest value (a minimum) somewhere on it. This is what makes "find the best value" problems sure to have an answer (module 7).

Both conditions matter. On the open interval $(0,1)$, $f(x)=x$ never reaches a maximum: $0.9$, $0.99$, $0.999$ keep getting bigger, but $1$ is not allowed. And $\frac{1}{x}$ on $[-1,1]$ has no maximum because it isn’t continuous at $0$.

• $f$ is continuous at $a$ when ${\lim }_{x\to a}f(x)=f(a)$.
• Discontinuities are removable (a hole), jumps, infinite, or oscillating.
• Elementary functions are continuous on their domains, and so are their combinations.
• IVT: a continuous function takes every value between $f(a)$ and $f(b)$; a sign change means a root.
• Bisection traps a root by halving, one binary digit per step.
• EVT: continuous on a closed interval means a maximum and a minimum exist.

In module 1 you found the speed of a falling ball "right now" by taking averages over shorter and shorter intervals. With limits you can now say exactly what that is. The slope of the secant from $a$ to $a+h$ is the difference quotient; its limit as $h\to 0$ is the derivative of $f$ at $a$:

Difference quotient

$\frac{f(a+h)-f(a)}{h}$: the slope of a secant, an average rate.

Derivative at a

The limit of the difference quotient as $h\to 0$: the instantaneous rate of change, the slope of the tangent.

Tangent line

The line through $(a,f(a))$ with slope $f^{\prime }(a)$.

The recipe: write the difference quotient, simplify until the $h$ on the bottom cancels, then let $h\to 0$. For $f(x)=x^2$: