M4 Change and accumulation: the ideas of calculus

understand what derivatives and integrals mean, and do simple ones.

Before you begin: P1 · Describing motion, P2 · Forces and Newton's laws, P3 · Energy and momentum, M2 · Algebra and graphs for physics.

Videos coming soon

Three videos (part 1: rates of change and the derivative; part 2: accumulation and the integral; part 3: small steps and Euler's method) and a worked-example video (constant acceleration, with calculus) are made, and will appear here once they're uploaded to YouTube.

Chapters: part 1, rates of change and the derivative (25 minutes)

0:00 Speed at a single instant
1:25 The slope of a curve
3:28 Achilles and the tortoise
4:59 Limits without fear
7:50 Tangents before calculus
8:35 The derivative
10:32 Two ways to write a rate
12:07 The power rule
14:08 Galileo's falling stone
15:39 Acceleration: the derivative of the derivative
17:11 Reading a graph with derivatives
18:35 Ghosts of departed quantities
20:02 Four easy mistakes
21:23 Practice
23:34 Recap and next

Chapters: part 2, accumulation and the integral (22 minutes)

0:00 The broken odometer
1:16 Distance is an area
2:38 Slicing a curve
4:21 Archimedes and the parabola
6:44 Graphs and slices before calculus
7:58 The integral
9:35 Work from a varying force
10:40 The fundamental theorem
12:13 Who saw it first?
14:08 Integrating with the power rule
15:41 The constant you must not forget
16:54 Areas below the axis
17:42 Four easy mistakes
18:50 Practice
20:51 Recap and next

Chapters: part 3, small steps and Euler's method (20 minutes)

0:00 The skydiver's problem
1:29 Step by step
2:52 Stepping a falling stone
4:14 Euler
4:54 Smaller steps, smaller errors
6:02 Air resistance and terminal velocity
8:04 From Aristotle to the Moon
9:27 Stepping the skydiver
11:26 The record jumpers
12:35 Forecasts and flight paths
13:52 Where Euler's method goes wrong
15:14 Four easy mistakes
16:17 Practice
19:03 Recap, and what's next

Chapters: worked example, constant acceleration with calculus (5 minutes)

0:00 The problem
0:51 Velocity from acceleration
1:38 Position from velocity
2:52 The third equation
3:56 Does it make sense?

108 steps about 111 min

Watch first

Notes available, video soon.

Notes

In this chapter

Explain the speed at an instant as a limit of average speeds over smaller and smaller gaps, and say what the slope of a tangent means.
Differentiate powers and sums of powers with the power rule, and read position, velocity and acceleration as successive derivatives.
Sketch a derivative from a graph, and say what it means where a graph peaks, dips or is flat.
Explain an integral as a limit of thin strips, and use it for distance from velocity and work from a force.
State the fundamental theorem of calculus, and use it (with the power rule backwards and a constant of integration) to evaluate simple integrals.
Step a motion forward with Euler's method, see how the step size controls the error, and find a terminal velocity.

Speed at a single instant

FIRST, MAKE A PREDICTION

TRY IT / FOLLOW THE REASONING

For the falling-stone model x = 4.9t², keep t at 2 s and shrink the forward gap. Watch the interval's average velocity approach a particular value.

Bring two moments of a fall together

Time t: 2 s · Gap h: 1 s

Average velocity over the gap: 24.5 m/s
Velocity at t: 19.6 m/s

Time t 2 sGap h 1 s

Model assumptions

Dropped from rest with no air resistance: x = 4.9t² metres, positive downwards, g = 9.8 m/s². Time t is in seconds; the forward gap h stays positive. The line joining the dots gives the interval average; the tangent at the first dot gives the instantaneous rate.

MAKE THE CONNECTION

Key idea

A photograph has no time in it, and an average needs two moments. Calculus answers "how fast, exactly now?"

A stone is dropped. How fast is it falling after exactly two seconds? The distance fallen is $x = 4.9t^2$ metres (from $\tfrac12 g t^2$, with $g$ rounded to 9.8 m/s²). At $t = 2$ s the stone has fallen 19.6 m.

The average velocity over a gap $h$ starting at 2 s is the slope of a chord (a straight line joining two points on the curve):

Speed at a single instant

gap $h$ (s)	1	0.1	0.01
average velocity (m/s)	24.5	20.09	19.649

The chords tilt towards one line, the tangent, and the averages home in on 19.6 m/s.

Speed at a single instant

Chords from t = 2 s for gaps of 1, 0.5 and 0.1 s have slopes 24.5, 22.05 and 20.09 m/s. They swing round towards the tangent, whose slope is 19.6 m/s.

Checkpoint: Speed at a single instant (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

Checkpoint: Speed at a single instant (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Limits

FIRST, MAKE A PREDICTION

TRY IT / FOLLOW THE REASONING

Use progressively smaller gaps at t = 2. Compare the average and instantaneous readouts and distinguish approaching zero from substituting zero in a quotient.

Bring two moments of a fall together

Time t: 2 s · Gap h: 1 s

Average velocity over the gap: 24.5 m/s
Velocity at t: 19.6 m/s

Time t 2 sGap h 1 s

Model assumptions

MAKE THE CONNECTION

Key idea

We can't use a gap of zero (that is $0 \div 0$), so we ask what the average is getting close to as the gap shrinks to zero. That is a limit.

Exactly:

Limits

\[ \begin{gathered} \frac{4.9(2+h)^2 - 4.9\cdot 2^2}{h} \\ = \frac{19.6h + 4.9h^2}{h} \\ = 19.6 + 4.9h \\ \longrightarrow\; 19.6 \quad (h \to 0) \end{gathered} \]

The $h$ cancels while it is still small but not zero; then what is left fades to its limit. The limit never has to be reached: it is the value the numbers head for. Two pictures help:

A speed camera timing you between two painted lines: move the lines together, and the reading settles on your speed at that spot.
Zooming in on a smooth curve until it looks like a straight line: the slope of that line is the slope of the curve.

Checkpoint: Limits (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

Checkpoint: Limits (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

The derivative

FIRST, MAKE A PREDICTION

TRY IT / FOLLOW THE REASONING

Set t to 3 and shrink the gap. Compare the secant with the tangent. Then change t and describe the derivative as a rule for every point.

Bring two moments of a fall together

Time t: 2 s · Gap h: 1 s

Average velocity over the gap: 24.5 m/s
Velocity at t: 19.6 m/s

Time t 2 sGap h 1 s

Model assumptions

MAKE THE CONNECTION

Key idea

Doing the same algebra at any time $t$ gives $9.8t + 4.9h \to 9.8t$. A formula for the slope at every point is the derivative.

For position $x$ and time $t$:

The derivative

\[ v = \frac{dx}{dt} = \lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} \]

$\Delta x/\Delta t$ is the average rate over a finite gap; $dx/dt$ is the rate at an instant, where $d$ means "a tiny change in". The unit comes free: metres per second.
The derivative is not literally a fraction, but it behaves like one often enough that Leibniz's notation won.
The same idea gives any rate of change: how fast tea cools ($dT/dt$), the flow rate of a tap ($dV/dt$).

Checkpoint: The derivative (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

Checkpoint: The derivative (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

The power rule

FIRST, MAKE A PREDICTION

TRY IT / FOLLOW THE REASONING

Write a table of t, t² and t³ with their derivative rules. Check the t³ rule numerically using a small forward change at t = 1.

MAKE THE CONNECTION

Key idea

Bring the power down in front, then lower the power by one.

The power rule

\[ \frac{d}{dt}\,t^n = n\,t^{\,n-1} \]

Three more facts finish the job:

a constant has derivative 0;
a constant multiplier stays ($4.9t^2 \to 4.9\cdot 2t = 9.8t$);
sums are differentiated term by term.

The power rule

Worked number

For $x = 2t^3 - 5t^2 + 7t + 3$: $v = 6t^2 - 10t + 7$, which is 11 m/s at $t = 2$ s.

Checkpoint: The power rule (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

Checkpoint: The power rule (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Acceleration, and reading graphs

FIRST, MAKE A PREDICTION

TRY IT / FOLLOW THE REASONING

Sketch one horizontal and one rising velocity graph. Read acceleration from their slopes, then sketch position graphs consistent with each.

MAKE THE CONNECTION

Key idea

Acceleration is the derivative of velocity, so it is the second derivative of position, $a = d^2x/dt^2$ (the 2s are labels, not powers).

For the stone: $4.9t^2 \to 9.8t \to 9.8$, which is $g$.
For a rocket sled with $x = t^3$: $v = 3t^2$, $a = 6t$.

Acceleration, and reading graphs

Worked number: the ball thrown up

A ball thrown up at 20 m/s has $x = 20t - 4.9t^2$ and $v = 20 - 9.8t$. The top is where the slope of $x$ is zero, so $v = 0$: $t = 20/9.8 \approx 2.04$ s, at a height of about 20.4 m.

Acceleration, and reading graphs

Reading a graph: where it climbs, its derivative is positive; where it is flat, zero; where it falls, negative; and a peak or dip is a zero of the derivative. A zero derivative does not mean the quantity is zero: at the top the ball is 20 m up.

Checkpoint: Acceleration, and reading graphs (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

Checkpoint: Acceleration, and reading graphs (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Accumulation

FIRST, MAKE A PREDICTION

TRY IT / FOLLOW THE REASONING

Draw the 20-by-5 rectangle with units on both axes. Divide it into five strips and add their distances, then imagine a speed that changes between strips.

MAKE THE CONNECTION

Key idea

A car with a broken odometer but a working speedometer: how far did it go? The area under the speed–time graph.

At constant speed, distance is the area of a rectangle under the speed–time graph (20 m/s for 5 s is 100 m), because (m/s) × s = m. A staircase of speeds is a sum of rectangles. A smooth curve is approximated by thinner and thinner strips.

Accumulation

For v = 9.8t from 0 to 4 s, eight strips cover 68.6 m, under the exact 78.4 m; thinner strips squeeze it towards the exact answer.

Accumulation

For $v = 9.8t$ from 0 to 4 s, the true distance is $4.9 \times 16 = 78.4$ m, and the strips squeeze it:

Accumulation

strips	4	8	40
undershoot (m)	58.8	68.6	76.44
overshoot (m)	98	88.2	80.36

The area is the limit of the sum of the strips as they get thinner (a Riemann sum).

Checkpoint: Accumulation (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

Checkpoint: Accumulation (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

The integral

FIRST, MAKE A PREDICTION

TRY IT / FOLLOW THE REASONING

Sketch a rising curve and estimate its area with two left-edge strips, then four. Mark the missing pieces and explain what happens as the widths shrink.

MAKE THE CONNECTION

Key idea

An integral adds up thin slices to give an exact total: the limit of the strips as they get thinner.

The integral

\[ \int_0^4 9.8t\,dt = 78.4\ \text{m} \]

The stretched S means "sum"; $dt$ is the thin width of each slice.
0 and 4 are the limits of integration; $9.8t$ is the integrand.
The unit of the answer is the integrand's unit times the slice's unit.

An integral is a total built from thin slices, so it is an area only as a picture:

under a speed graph it is a distance;
under a force graph a work;
under a power graph an energy;
under a flow-rate graph an amount of water.

The integral

Worked number: the spring

A spring with $k = 200$ N/m stretched 0.1 m has a force climbing from 0 to 20 N, and $\tfrac12 \times 0.1 \times 20 = 1$ J of work.

Checkpoint: The integral (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

Checkpoint: The integral (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

The fundamental theorem

FIRST, MAKE A PREDICTION

TRY IT / FOLLOW THE REASONING

Draw the thin strip added as an endpoint moves by a small time. Divide its approximate area by that time and identify what remains as the strip narrows.

MAKE THE CONNECTION

Key idea

The slope of the area curve is the height of the original graph. Differentiating and integrating undo each other.

Let the end time slide, and call the area so far $A(t)$. When $t$ moves on by $dt$, a thin strip of area $v(t)\,dt$ is added, so $dA/dt = v(t)$. For the stone, $A = 4.9t^2$ and its derivative is $9.8t = v$.

Checkpoint: The fundamental theorem (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

Checkpoint: The fundamental theorem (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Integrating with the power rule

FIRST, MAKE A PREDICTION

TRY IT / FOLLOW THE REASONING

Find an antiderivative of t² and check it by differentiating. Add different constants and explain why all those curves have the same slope rule.

MAKE THE CONNECTION

Key idea

Run the power rule backwards: raise the power by one, then divide by the new power.

Integrating with the power rule

\[ \begin{gathered} \int t^n\,dt = \frac{t^{\,n+1}}{n+1} + C \\[.6em] (n \neq -1) \end{gathered} \]

Evaluate between two times by subtracting the antiderivative's values: $\int_0^4 9.8t\,dt = [4.9t^2]_0^4 = 78.4$.

Integrating with the power rule

Worked number: a force doing work

A force $F = 3x^2$ N pushing from 0 to 2 m does $[x^3]_0^2 = 8$ J of work.

Integrating with the power rule

The constant $C$ is the constant of integration, the starting value. For a ball thrown up at 20 m/s from the ground, $a = -9.8$ gives $v = -9.8t + C$ with $C = 20$, and then $x = 20t - 4.9t^2 + C'$ with $C' = 0$.

Integrating with the power rule

Worked number: forward and back

Areas below the axis count as negative. A cart with $v = 4 - 2t$ moves forward for 2 s (+4 m) and back for 2 s (−4 m): the integral (displacement) is 0, but the distance travelled is 8 m.

Checkpoint: Integrating with the power rule (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

Checkpoint: Integrating with the power rule (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

Checkpoint: Integrating with the power rule (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

Euler's method

FIRST, MAKE A PREDICTION

TRY IT / FOLLOW THE REASONING

For velocity changing under gravity, do two short updates by hand. Then compare the chapter's changing-acceleration calculations at different time steps, keeping each scheme's update order consistent.

MAKE THE CONNECTION

Key idea

When a formula can't be integrated by hand, step. Pick a small time step $\Delta t$, and repeat, using the old values on the right.

Euler's method

\[ \begin{gathered} x_{\text{new}} = x_{\text{old}} + v\,\Delta t \\[.6em] v_{\text{new}} = v_{\text{old}} + a\,\Delta t \end{gathered} \]

This is dead reckoning, the navigator's trick. On a graph, each step goes along the tangent.

Euler's method

Worked number: the stone, stepped

For the stone with $\Delta t = 0.5$ s the distance after 2 s is 14.7 m, against an exact 19.6 m. With $\Delta t = 0.1$ s it is 18.62 m (error 0.98), and with 0.05 s it is 19.11 m (error 0.49): halve the step and the error halves. Euler's method is therefore first order.

Checkpoint: Euler's method (1 of 3)

Question 1 of 3 Commit to an answer first, then look.

Checkpoint: Euler's method (2 of 3)

Question 2 of 3 Commit to an answer first, then look.

Checkpoint: Euler's method (3 of 3)

Question 3 of 3 Commit to an answer first, then look.

Air resistance and terminal velocity

FIRST, MAKE A PREDICTION

TRY IT / FOLLOW THE REASONING

In a = g − kv², set acceleration to zero and solve for the terminal speed. Then increase speed slightly above that value and predict the direction of acceleration.

MAKE THE CONNECTION

Key idea

At everyday speeds, drag grows roughly with the square of the speed, so a skydiver's acceleration is $a = g - kv^2$. The terminal velocity is where $a = 0$.

Air resistance and terminal velocity

\[ v_t = \sqrt{g/k} \]

Air resistance and terminal velocity

A skydiver stepped forward in 0.1 s steps: with air the speed levels off, at about 52 m/s after 10 s; with no air it would keep climbing to 98 m/s.

Air resistance and terminal velocity

Worked number: the skydiver

With $k = 0.0032\ \text{m}^{-1}$, $v_t = \sqrt{9.8/0.0032} \approx 55.3$ m/s, about 200 km/h or 120 mph.

Stepping with $\Delta t = 1$ s: $v = 0,\ 9.8,\ 19.29,\ 27.90$ m/s at 0, 1, 2, 3 s. With a 0.1 s step the speed levels off at about 55 m/s (after 10 s, about 52 m/s with air, against 98 m/s without).

The rules don't care if conditions change: a parachute ($k$ a hundred times larger) drops the speed to about 5.5 m/s.

Air resistance and terminal velocity

Roughly speaking, drag $\propto v^2$ suits fast objects like skydivers; very small, slow things like dust feel a drag proportional to $v$, and $k$ depends on shape.

Air resistance and terminal velocity

Weak spots of Euler's method

A step that is too big gives nonsense (with $\Delta t = 20$ s the skydiver's speed goes 196, then about −2000 m/s), and errors build up, so a simulated planet spirals outward. Better methods (Runge–Kutta, around 1900) average several slopes within each step.

Checkpoint: Air resistance and terminal velocity (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

Checkpoint: Air resistance and terminal velocity (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Common mistakes

FIRST, MAKE A PREDICTION

TRY IT / FOLLOW THE REASONING

Compare the derivative of 7 with the derivative of 7t. Label units and sketch both starting functions before deciding their slopes.

MAKE THE CONNECTION

Derivatives: mixing up the average (19.6 m over 2 s is 9.8 m/s) with the instantaneous rate (19.6 m/s); differentiating a constant to get the constant (it is 0); half-remembering the power rule ($t^3 \to 3t^2$, not $3t^3$ or $t^2$); thinking a zero derivative means the quantity is zero.
Integrals: forgetting to divide by the new power ($\int t^2\,dt = t^3/3$); forgetting the constant (use the starting value); confusing displacement and distance; losing track of units.
Euler's method: trusting one run (halve the step and compare); mixing old and new values within a step; mixing seconds and minutes; believing a simulation is exact.

Checkpoint: Common mistakes (1 of 2)

Question 1 of 2 Commit to an answer first, then look.

Checkpoint: Common mistakes (2 of 2)

Question 2 of 2 Commit to an answer first, then look.

Key terms

Chord: A straight line joining two points on a curve. Its slope is an average rate of change.
Tangent: The line that just touches a curve at one point. Its slope is the rate of change at that point.
Limit: The value that something approaches (such as an average over a shrinking gap), without having to reach it.
Derivative: The rate of change of one quantity with respect to another, at every point: $dx/dt$ is the derivative of position with respect to time.

Key terms

Power rule: $\frac{d}{dt}t^n = nt^{n-1}$.
Second derivative: The derivative of the derivative; for position, the acceleration, $d^2x/dt^2$.
Riemann sum: A total of thin rectangles under a curve, as an approximation to an area.
Integral: The limit of such sums as the strips get thinner; a total built from thin slices. Written $\int f\,dt$.

Key terms

Limits of integration: The start and end values of the integral, for example the 0 and 4 in $\int_0^4$.
Integrand: The thing being added up, for example $9.8t$ in $\int 9.8t\,dt$.
Fundamental theorem of calculus: Integrating and differentiating undo each other: the slope of the running-total curve equals the height of the original graph.
Antiderivative: A function whose derivative is the one you started with.

Key terms

Constant of integration: The $+C$ of an antiderivative, fixed by a starting value.
Dead reckoning: Working out a new position from the old one, plus velocity times a time step.
Time step: The small interval $\Delta t$ used in a step-by-step calculation.
Euler's method: Repeating "new = old + rate × step" to trace out a motion; a first-order method.

Key terms

Drag: The force of a fluid (air) pushing back on a moving object.
Terminal velocity: The steady speed at which drag balances weight, so the acceleration is zero.

History and sources

History

Where a primary text wasn't read, the sources are encyclopedic or institutional pages, and the wording below is cautious. The full notes, with what each source does and doesn't support, are in the sources.md of each video folder.

Zeno of Elea (5th century BC)

Zeno of Elea (5th century BC). Zeno's paradoxes of motion, such as Achilles and the tortoise, are known from Aristotle (Physics VI.9), because Zeno's own writings are lost. Adding the infinitely many shrinking gaps gives a finite total, which is the limit idea; whether that fully settles Zeno's point is still argued by philosophers. Sources: Stanford Encyclopedia: Zeno of Elea; Zeno's paradoxes.

Fermat and Descartes (1630s)

Fermat and Descartes (1630s). Pierre de Fermat found tangents and maxima and minima by comparing a quantity with a very slightly changed one; it circulated by letter and manuscript and was not printed until after his death. Descartes attacked it, then admitted it was very good. Source: MacTutor: Fermat.

Newton and fluxions (1665–1736)

Newton and fluxions (1665–1736). The plague closed Cambridge in 1665 and Newton worked at home in Lincolnshire for about two years; by the end of 1666 he had the key techniques of fluxions (a dot over a letter for a rate), which he kept private. De analysi circulated in manuscript in 1669, and his book on the method (written 1671) was printed in 1736. Sources: MacTutor: Newton; The October 1666 Tract on Fluxions (Newton Project).

Leibniz and the notation (1675–86)

Leibniz and the notation (1675–86). Leibniz worked out his notation in private manuscripts in 1675 and printed his first paper on the differential calculus in 1684 (Acta Eruditorum, Leipzig). The long S for summa is in a manuscript of October 1675 and was first printed in 1686; the word "integral" first appears in print in 1690, by Jacob Bernoulli. Sources: MacTutor: Leibniz; MacTutor: the integral sign, 29 October; Leibniz's notation.

Galileo (1638)

Galileo (1638). In Two New Sciences (Leiden, 1638), printed while Galileo was under house arrest near Florence, he showed that distance fallen grows as the square of the time (distances in successive equal times go as 1 : 3 : 5 : 7). He described a bronze ball rolling down a grooved beam, timed with a water clock; historians debate whether he performed the experiments exactly so. Sources: Two New Sciences; MacTutor: Galileo.

Berkeley, Cauchy and Weierstrass (1734–19th century)

Berkeley, Cauchy and Weierstrass (1734–19th century). George Berkeley's The Analyst (1734) called Newton's vanishing quantities "the ghosts of departed quantities": the answers were right, he argued, but the logic was shaky. Cauchy built calculus on limits in his Cours d'analyse (1821); Weierstrass later gave the limit its modern epsilon–delta form. Sources: The Analyst; MacTutor: Cauchy; Stanford Encyclopedia: Continuity and infinitesimals.

Archimedes and Eudoxus (3rd century BC)

Archimedes and Eudoxus (3rd century BC). In the Quadrature of the Parabola, a letter to Dositheus, Archimedes showed that a parabolic segment has $\tfrac43$ the area of the triangle that fits inside it, adding triangles whose total grows as $1 + \tfrac14 + \tfrac1{16} + \dots = \tfrac43$. Eudoxus had made the squeezing method rigorous; historians call exhaustion an early form of integration. Sources: MacTutor: Archimedes; The Quadrature of the Parabola.

The Archimedes Palimpsest (1906, 1998)

The Archimedes Palimpsest (1906, 1998). The tenth-century copy of Archimedes' works, including The Method, was scraped and overwritten with a prayer book dated 13 April 1229. Johan Heiberg recognised the underlying text in Constantinople in 1906. It was sold in October 1998 for $2 million to an anonymous buyer, and has been studied at the Walters Art Museum since January 1999. Sources: The Archimedes Palimpsest project: history; Archimedes Palimpsest.

Merton College and Oresme (14th century), Cavalieri (1635)

Merton College and Oresme (14th century), Cavalieri (1635). Scholars at Merton College, Oxford, showed that a steadily speeding-up body covers the same distance as one moving at its average speed; Nicole Oresme pictured it as an area under a speed graph. Bonaventura Cavalieri's method of indivisibles (1635) treated shapes as stacks of thin slices. Sources: Mean speed theorem; MacTutor: Oresme; MacTutor: Cavalieri.

Barrow, and the priority dispute (1669–1716)

Barrow, and the priority dispute (1669–1716). Isaac Barrow, first Lucasian Professor at Cambridge, showed in geometric form that areas and tangents are inverse problems (1670), and in 1669 resigned his chair, and Newton succeeded him. From about 1708 Newton's supporters accused Leibniz of plagiarism; the Royal Society's report (1712–13), largely written by Newton, sided with Newton. Leibniz died on 14 November 1716. Historians today agree that the two invented calculus independently. Sources: MacTutor: Barrow; Leibniz–Newton calculus controversy; Edward Worth Library: Newton and Leibniz.

Riemann (1854)

Riemann (1854). Bernhard Riemann's habilitation work (Göttingen, 1854; published 1868) gave the area-as-limit-of-strips idea its precise definition. Source: MacTutor: Riemann.

Euler (1768)

Euler (1768). Leonhard Euler described the stepping method in his textbook on integral calculus, Institutionum calculi integralis (St Petersburg, 1768), when he was almost blind; he kept producing mathematics for another fifteen years. Sources: MacTutor: Euler; Euler method.

Aristotle, Galileo, Newton, Stokes, and Apollo 15

Aristotle, Galileo, Newton, Stokes, and Apollo 15. Aristotle said heavier things fall faster; Galileo's thought experiment (1638) leads to a contradiction. The Leaning Tower of Pisa tale comes from a biography written after Galileo's death and is probably a legend. Stokes (1851) worked out the drag on a very small sphere moving slowly through a fluid. On 2 August 1971, Dave Scott dropped a geologic hammer and a falcon feather on the Moon; with no air, they landed together as far as anyone could tell. Sources: The Leaning Tower experiment; Stokes' law; NASA NSSDC: Apollo 15 feather drop.

Kittinger and Baumgartner (1960, 2012)

Kittinger and Baumgartner (1960, 2012). Joseph Kittinger jumped from about 31 km (102,800 ft) on 16 August 1960; Felix Baumgartner jumped from about 39 km on 14 October 2012 and reached 1,357.6 km/h (Mach 1.25), the first human to break the sound barrier in freefall. A belly-down skydiver's terminal velocity is about 54 to 55 m/s. Sources: Red Bull Stratos; Joseph Kittinger.

Richardson (1922) and Katherine Johnson (1962)

Richardson (1922) and Katherine Johnson (1962). Lewis Fry Richardson hand-calculated a weather forecast for 20 May 1910 (published 1922); he wrote that it took the best part of six weeks, and it predicted a pressure change about a hundred times too big, because of bad starting data. He dreamed of 64,000 human calculators. Before John Glenn's orbital flight in February 1962, Glenn asked for Katherine Johnson to check the computer's numbers by hand. That she used Euler's method is shown in the film Hidden Figures, but is not documented by NASA. Sources: Lynch: Richardson's forecast; NASA: Katherine Johnson.

Runge and Kutta (around 1900)

Runge and Kutta (around 1900). Carl Runge and Wilhelm Kutta developed more accurate versions of the same idea; Euler's method is the simplest of the family. Source: Runge–Kutta methods.

Worked example: constant acceleration, with calculus

The problem

Problem. A car is travelling at $u = 10$ m/s and speeds up steadily at $a = 3$ m/s². After 4 s, how fast is it going, and how far has it travelled?

Solution

Solution. Integrate the acceleration to get the velocity, and the velocity to get the position, using the starting values to fix the constants.

Solution

\[ \begin{gathered} v = \int a\,dt = at + C \\[.6em] v(0) = u \;\Rightarrow\; v = u + at \\ = 10 + 3\times4 = 22\ \text{m/s} \end{gathered} \]

Solution

\[ \begin{gathered} s = \int (u + at)\,dt \\ = ut + \tfrac12 at^2 + D \\[.6em] s(0) = 0 \\ \Rightarrow\; s = 10\times4 + \tfrac12\times3\times4^2 \\ = 40 + 24 = 64\ \text{m} \end{gathered} \]

The third equation needs no calculus: from $t = (v-u)/a$, $s = u\frac{v-u}{a} + \frac{(v-u)^2}{2a} = \frac{(v-u)(v+u)}{2a}$, so $v^2 = u^2 + 2as$. Check: $22^2 = 484 = 100 + 2\times3\times64$. ✓

Sanity checks. The area under the $v$–$t$ line (10 to 22 over 4 s) is a trapezium: $\tfrac12(10+22)\times4 = 64$ m, which splits into a 40 m rectangle and a 24 m triangle ✓. With $a = 0$, $s = ut$; with $u = 0$ and $a = g$, $s = \tfrac12 gt^2$, the falling stone ✓. These equations hold only for constant acceleration.

Practice

Use $g = 9.8$ m/s² unless told otherwise.

★ Warm-ups

1.1 A particle's position is $x = 3t^5$ metres. Find its velocity at $t = 2$ s, in metres per second. Then find the general formula $v(t)$.

Problem 1.2 (★ Warm-ups)

1.2 Differentiate $x = t^3 - 4t^2 + 9t - 7$. What is the velocity at $t = 3$ s, in metres per second?

Problem 1.3 (★ Warm-ups)

1.3 A cyclist's position is $x = 5t^2 + 2t$ metres. Find her velocity at $t = 3$ s, in metres per second. Then find her acceleration.

Problem 1.4 (★ Warm-ups)

1.4 For a stone with $x = 4.9t^2$, find the average velocity between $t = 1$ s and $t = 1.1$ s, in metres per second, and compare it with the instantaneous velocity at $t = 1$ s.

Problem 1.5 (★ Warm-ups)

1.5 Find $\int_0^3 2t\,dt$ by running the power rule backwards, then check it as the area of a triangle.

★★ Standard

2.1 A runner's speed rises steadily from 0 to 8 m/s in 2 s, stays at 8 m/s for 5 s, and then falls steadily to rest over 3 s. How far does she run, in metres?

Problem 2.2 (★★ Standard)

2.2 A particle moves with $x = t^3 - 6t^2 + 9t$ metres. (a) When is it at rest? (b) Where is it then? (c) What is its acceleration at each of those times?

Problem 2.3 (★★ Standard)

2.3 A variable force $F = 12x^2$ newtons acts on an object as it moves from $x = 0$ to $x = 1.5$ m. How much work is done, in joules?

Problem 2.4 (★★ Standard)

2.4 A particle starts at $x = 5$ m with velocity $v = 3t^2 + 2$ m/s. Find its position at $t = 3$ s, in metres.

Problem 2.5 (★★ Standard)

2.5 A spring with $k = 300$ N/m is stretched by 0.2 m. Find the work done, in joules, by integrating the force $F = kx$.

★★★ Stretch

3.1 A ball falls with acceleration $a = g - kv^2$, where $k = 0.02\ \text{m}^{-1}$. (a) Find its terminal velocity, in metres per second. (b) Use Euler's method with $\Delta t = 0.5$ s, starting from rest, to find the speed at each half-second up to 3 s. (c) Repeat the last stage with $\Delta t = 0.25$ s, and compare both with the exact answer 19.23 m/s at 3 s.

Hint

For each step, first work out $a = 9.8 - 0.02v^2$ from the old speed, then $v_{\text{new}} = v + a\Delta t$. Terminal velocity is where $a = 0$.

Your answer m/s

Solution

(a) $a = 0$ when $v^2 = g/k = 9.8/0.02 = 490$, so $v_t = 22.14$ m/s. (b) Steps of 0.5 s (acceleration used, then new speed): at 0.5 s, $a = 9.8$, $v = 4.90$; at 1.0 s, $a = 9.32$, $v = 9.56$; at 1.5 s, $a = 7.97$, $v = 13.55$; at 2.0 s, $a = 6.13$, $v = 16.61$; at 2.5 s, $a = 4.28$, $v = 18.75$; at 3.0 s, $a = 2.77$, $v = 20.14$ m/s. (c) With $\Delta t = 0.25$ s (twelve steps), $v(3) = 19.67$ m/s. The exact value from the full solution is $v_t\tanh(gt/v_t) = 19.23$ m/s. The errors are $0.91$ m/s with the big step and $0.44$ m/s with the half step: halving the step halves the error ✓. Sanity check: the speed climbs towards 22.1 m/s and never passes it, and the acceleration shrinks as it approaches ✓. Without drag it would be 29.4 m/s at 3 s.

Problem 3.2 (★★★ Stretch)

3.2 Consider $dv/dt = -v$ with $v = 1$ at $t = 0$ (the exact answer at $t = 1$ is $e^{-1} = 0.3679$). Apply Euler's method up to $t = 1$. What does it give with $\Delta t = 0.25$ s? Then try 0.1 and 0.05 s, and show that the error is roughly proportional to the step.

Hint

Each step multiplies $v$ by $(1 - \Delta t)$, since $v_{\text{new}} = v - v\Delta t$. So the result after $1/\Delta t$ steps is $(1-\Delta t)^{1/\Delta t}$.

Problem 3.3 (★★★ Stretch)

3.3 A car travelling at 30 m/s brakes with constant deceleration $4\ \text{m/s}^2$. Use calculus (integrate twice) to find how long it takes to stop, in seconds, and how far it travels, and check with $v^2 = u^2 + 2as$.

Hint

$a = -4$. Integrate once for $v$ (with $v(0) = 30$), and set $v = 0$. Integrate again for $x$ (with $x(0) = 0$).

Check yourself

Answer these out loud, from memory. Then try again in a few days.

Check yourself

How do limits and derivatives make instantaneous rates of change precise?

Check yourself

How do integrals describe accumulated quantities, and how are differentiation and integration connected?

Check yourself

How do derivatives and integrals appear in motion, work, charge distributions, and heat flow?

Check yourself

How can analytical approximations and numerical solutions complement one another?

Summary

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Further learning

Khan Academy (found in Khan Academy's search index on 2026-10-12; Khan Academy blocks automated link checking, so tell us if one has moved):

Free textbook: OpenStax, Calculus Volume 1 (checked 2026-10-12): 3.2 The Derivative as a Function · 5.3 The Fundamental Theorem of Calculus · 5.4 Integration Formulas and the Net Change Theorem.

Free course: MIT OpenCourseWare, 18.01SC Single Variable Calculus (lecture videos, recitations, problem sets with solutions; checked 2026-10-12). It is a full university course, so take it slowly: this chapter covers only its ideas.

Simulations: PhET: Calculus Grapher (draw a curve and see its derivative and integral alongside it; checked 2026-10-12).

Next: P4 · Gravity and orbits (coming soon) · Progress