Calculus II

The pressure of a gas depends on its volume and on its temperature at the same time, and the calculus of one variable, built for a quantity that depends on a single number, has no way to ask how that pressure changes when both are free to move.

The previous course was about $y=f(x)$: one input, one output, a curve with one slope at each point. Almost nothing measured in a laboratory, a factory or a weather station behaves like that. This lesson sets up the objects the rest of this course differentiates and integrates, shows three ways of looking at them, and ends on the precise reason the old machinery breaks. It assumes Calculus I and nothing about vectors or three dimensional geometry.

One output, several inputs

A function of two variables is a rule that takes an ordered pair of real numbers $(x,y)$ and returns one real number, written $f(x,y)$. The pair is best thought of as a single input, a point in the plane, because the question will always be what happens as the point moves. A function of three variables takes a point $(x,y,z)$, and so on.

The ideal gas law is the standard example. For $n$ moles of gas at absolute temperature $T$ in a volume $V$, the pressure is

with $R=8.314$ joules per mole per kelvin. One mole at the freezing point of water, $T=273.15$ K, in a volume of $0.0224$ cubic metres, has $p=8.314\times 273.15/0.0224=101383$ Pa, which is within a tenth of a percent of the standard atmosphere of $101325$ Pa.

Economics has the same shape. In 1928 the mathematician Charles Cobb and the economist Paul Douglas fitted index numbers for American manufacturing from 1899 to 1922, labour $L$ and capital $K$, to a production function of the form

and found that it tracked the actual output index closely for all twenty four years. The exponents add to $1$, which has a checkable meaning: doubling both inputs gives $1.01(2L{)}^{3/4}(2K{)}^{1/4}=2^{3/4+1/4}Q=2Q$, so doubling the factory doubles its output.

Weather forecasts supply a third. Since November 2001 the weather services of Canada and the United States have reported a wind chill index computed from the air temperature $T$ in degrees Celsius and the wind speed $v$ in kilometres per hour measured at ten metres:

The output is the temperature which, in near-calm air, would draw heat from exposed skin at the same rate. It is a fitted model of heat loss from a human face, not a law of nature.

Example. Find the wind chill at $T=-5$ with a wind of $v=30$.

First ${30}^{0.16}=1.7232$. Then the four terms are $13.12$, $0.6215\times (-5)=-3.1075$, $-11.37\times 1.7232=-19.593$ and $0.3965\times (-5)\times 1.7232=-3.4163$. Their sum is $W=-13.0$. A thermometer reading of $-5$ °C in a 30 km/h wind feels like $-13$ °C.

Now you. Find the wind chill at $T=-10$ with $v=20$, to one decimal place.

Where a function is defined

A function of two variables is defined on a set of points in the plane, its domain, and domains in the plane come in far more shapes than intervals on a line.

When a formula is given without comment, the domain is taken to be every point where the formula makes sense. For $f(x,y)=\sqrt{9-x^2-y^2}$ the square root needs $x^2+y^2\le 9$, and since $x^2+y^2$ is the square of the distance from the origin, the domain is the disc of radius $3$ centred there, boundary circle included. For $g(x,y)=\ln (x-y)$ the logarithm needs $x-y>0$, which is every point strictly below the line $y=x$: a half-plane, with its edge excluded. And $h(x,y)=xy/(x^2+y^2)$ is defined everywhere except the single point $(0,0)$, a hole of a kind that has no counterpart on the real line, because a point in the plane can be approached from every direction at once. A later lesson shows that this particular hole cannot be filled.

Physical models add restrictions the algebra does not see. The gas law makes sense only for $V>0$ and $T>0$, but it is also only accurate where the gas is dilute: squeeze carbon dioxide hard enough at room temperature and it liquefies, while the formula carries on regardless. The wind chill formula is published with the conditions $T\le 10$ and $v\ge 4.8$ attached, because outside them the underlying experiment says nothing. The honest domain of a model is where it was tested.

Graphs are surfaces

The graph of $y=f(x)$ needs two axes, one for the input and one for the output. A function of two variables needs three. Its graph is the set of points $(x,y,z)$ with $z=f(x,y)$, drawn with the $x$ and $y$ axes flat on a table and the $z$ axis pointing up, so that above each point of the domain there is exactly one point of the graph, at height $f(x,y)$. Those points form a surface, lying over the domain the way a landscape lies over a map.

Take $f(x,y)=x^2+y^2$. The height at each point is its squared distance from the origin, so all points at distance $r$ sit at height $r^2$. The surface is a bowl called a paraboloid, flat at the bottom and steepening outwards. Take instead $f(x,y)=\sqrt{9-x^2-y^2}$, defined on the disc of radius $3$. Squaring $z=\sqrt{9-x^2-y^2}$ gives $x^2+y^2+z^2=9$, which says every point of the graph is at distance $3$ from the origin in space, and since $z\ge 0$ the graph is the upper half of a sphere, a dome sitting on its domain.

The most useful trick for handling a surface is to slice it. Fixing $y=b$ cuts the surface with a vertical plane, and what remains is the curve $z=f(x,b)$, an ordinary function of the single variable $x$ called a trace. The paraboloid sliced at $y=2$ gives $z=x^2+4$, a parabola lifted by $4$. The dome sliced at $y=0$ gives $z=\sqrt{9-x^2}$, a semicircle. Every trace is a problem from the previous course.

The wind chill surface shows how different the traces in two directions can be. Hold the wind at $v=30$ and the formula collapses to $W=-6.473+1.3048T$: a straight line, saying that in that wind every degree of real cooling feels like $1.3$ degrees. Hold the temperature at $-5$ instead and $W=10.0125-13.3525v^{0.16}$, a curve that falls steeply for light winds and flattens for strong ones, since going from $10$ to $20$ km/h costs $2.3$ degrees of wind chill and going from $40$ to $50$ costs only $0.9$. One surface, straight when walked one way and bent when walked the other.

Level curves

The second way of looking at a surface slices it horizontally instead. The level curve of $f$ at height $c$ is the set of points in the domain where $f(x,y)=c$: cut the surface at height $c$ and drop the cut onto the input plane. Drawing several level curves for equally spaced values of $c$, each labelled with its value, gives a picture of the whole function using only two dimensions.

For the paraboloid the level curve $x^2+y^2=c$ is a circle of radius $\sqrt{c}$. At $c=1,2,3,4$ the radii are $1$, $1.414$, $1.732$ and $2$, so the circles crowd together further out. That crowding is the steepening of the bowl: the height rises by the same step between each pair of circles, and it does so over shorter and shorter distances.

Level curves have been reinvented in every subject that needed them. Temperature as a function of volume and pressure, $T=pV/(nR)$, has level curves $pV=\text{constant}$, the hyperbolas Robert Boyle measured in 1662 and physicists still call isotherms. Meteorologists draw isobars of equal pressure. Economists draw isoquants, the combinations of labour and capital giving equal output. Cartographers draw contours. The mathematics is the same in every case: a curve along which the function does not change.

For the Cobb-Douglas function the isoquant at output $Q_0$ can be solved explicitly. Write $c=Q_0/1.01$, so that $L^{3/4}{K}^{1/4}=c$, and raise both sides to the fourth power to get $L^{3}K=c^4$, that is $K=c^4/L^3$. Along an isoquant, capital must rise as the cube of the fall in labour.

Example. A plant with Cobb-Douglas production $Q=1.01L^{3/4}{K}^{1/4}$ runs at $L=16$, $K=81$. If labour is halved, how much capital keeps output the same?

The current output is $1.01\times {16}^{3/4}\times {81}^{1/4}=1.01\times 8\times 3=24.24$, so $c=24$ and the isoquant is $K={24}^4/L^3=331776/L^3$. At $L=8$ that is $331776/512=648$. Halving labour needs eight times the capital, the factor $2^3$ that $K=c^4/L^3$ predicts without computing $c$ at all.

Now you. The same plant runs at $L=81$, $K=16$. If labour is cut to $27$, how much capital keeps output the same?

Reading a contour map

Contour lines on maps have a mathematical origin. Charles Hutton, reducing the survey of the Scottish mountain Schiehallion made for Nevil Maskelyne's 1774 experiment to weigh the Earth, needed the volume of the mountain and had a mass of spot heights to organise; joining points of equal height turned the numbers into a picture, and his report of 1778 is often credited with the first contours of land relief. A walker's map is a level-curve diagram of height above sea level.

The Ordnance Survey's 1:50,000 Landranger maps of Great Britain draw contours every $10$ metres, and at that scale one millimetre on the paper is $50$ metres on the ground. Count the contour gaps crossed along a route to get the rise, measure the route on the map to get the run, and divide. Closely packed contours mean a steep slope, widely spaced ones a gentle slope, and walking along a contour means not climbing at all.

Example. On a Landranger map a path crosses five contour gaps in $4$ mm of paper. What is its average gradient, and at what angle does it climb?

The rise is $5\times 10=50$ metres and the run is $4\times 50=200$ metres, so the gradient is $50/200=0.25$, a slope of one in four. The angle is $\arctan 0.25=14.0$ degrees.

Now you. A route crosses eight contour gaps in $5$ mm on the same map. Find the gradient and the angle.

Why one variable is not enough

The contour map already says what goes wrong. On a hillside the steepness depends on which way you walk: straight up the fall line it is greatest, along the contour it is zero, and every compass bearing in between gives something else. A function of one variable has one line through each point to move along; a function of two variables has a slope in every direction.

The paraboloid $f(x,y)=x^2+y^2$ at the point $(3,4)$, where $f=25$, makes this quantitative. Step a distance $0.001$ due east, to $(3.001,4)$, and the difference quotient is $6.001$. Step due north instead and it is $8.001$. Step directly away from the origin, $0.0006$ east and $0.0008$ north, and it is $10.001$. Step along the level circle through the point, $0.0008$ west and $0.0006$ north, and it is $0.001$, heading for zero. Four directions, four limits, $6$, $8$, $10$ and $0$, and there are infinitely many more.

The same thing happens with the gas. At $T=273.15$ K and $V=0.0224$ cubic metres, warming one mole by a kelvin raises the pressure by $371$ Pa, while adding a litre of volume lowers it by $4333$ Pa. Doing both at once lowers it by $3977$ Pa, close to the sum of the two separate changes, a hint that the separate rates may be enough to recover the rest.

So the derivative of the previous course cannot be reused as it stands. Its difference quotient divided a change in output by a change in input, and here the input change is a step in the plane with both a length and a direction. The limit is affected too: at the hole in $xy/(x^2+y^2)$ the origin can be approached along infinitely many lines and curves, not just from the left and from the right. And the graph is a surface, so whatever replaces the tangent line must be a plane.

Every one of those needs geometry the previous course never built: distance between points in space, directions described by numbers, and equations for planes. That is the work of the next lesson, which sets up vectors in three dimensions.

The graph of a function of two variables is a surface in three dimensional space, and nothing in the calculus of one variable says how far apart two points in space are, which way a direction points, or what equation a flat plane has.

The previous lesson met that gap at its end: the slope on a hillside depends on the direction walked, and whatever replaces the tangent line must be a tangent plane. This lesson builds the geometry those ideas need, entirely by computing with coordinates, and assumes only Pythagoras and school trigonometry. Readers who know vectors from Linear Algebra will recognise the dot product; the cross product, and the lines and planes it makes easy, are new.

Coordinates in space

Take three mutually perpendicular number lines meeting at an origin, with the $x$ axis towards you, the $y$ axis to the right and the $z$ axis up, so that, as in the previous lesson, heights are measured along $z$. A point is an ordered triple $(x,y,z)$. The arrangement is right-handed: curl the fingers of the right hand from the positive $x$ axis to the positive $y$ axis and the thumb points along the positive $z$ axis. Distance does not care about this convention, but the cross product later on does.

Distance comes from Pythagoras applied twice. Between $P=(x_1,y_1,z_1)$ and $Q=(x_2,y_2,z_2)$, the horizontal distance across the floor is $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$, and the vertical rise $z_2-z_1$ is perpendicular to it, so the two are the legs of a right triangle whose hypotenuse runs from $P$ to $Q$. Squaring and adding,

From $(1,2,3)$ to $(4,6,15)$ the differences are $3$, $4$ and $12$, and the distance is $\sqrt{169}=13$. The inside of a standard twenty foot shipping container measures $5.898$ by $2.352$ by $2.393$ metres, so the longest pipe that fits, corner to opposite corner, is $\sqrt{34.786+5.532+5.726}=6.786$ metres.

The points at distance $r$ from a centre $(a,b,c)$ satisfy $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$, the equation of a sphere. Faced with $x^2+y^2+z^2-2x+4y-6z=2$, complete the square in each variable to get $(x-1{)}^2+(y+2{)}^2+(z-3{)}^2=16$: radius $4$, centre $(1,-2,3)$.

Vectors and components

A point says where; a vector says how far and which way. The displacement from $(1,2,3)$ to $(4,6,15)$ is written $⟨3,4,12⟩$, with angle brackets to keep it apart from a point, and the same vector carries $(10,10,10)$ to $(13,14,22)$: a vector has length and direction but no fixed position. The vector from $P$ to $Q$ is head minus tail, $⟨x_2-x_1,y_2-y_1,z_2-z_1⟩$. Vectors are printed in bold, $𝐚=⟨a_1,a_2,a_3⟩$.

Addition and scaling are componentwise. $𝐚+𝐛$ is one displacement followed by the other, which is also how forces combine, and $c𝐚$ stretches the arrow by the factor $c$, reversing it when $c<0$. The length is the distance the arrow covers, $|𝐚|=\sqrt{a_1^2+a_2^2+a_3^2}$, and dividing a non-zero vector by its length gives a unit vector in the same direction: $⟨3,4,12⟩/13=⟨0.2308,0.3077,0.9231⟩$. A unit vector is a pure direction, which is how later lessons will say "walk this way".

The unit vectors along the axes are $𝐢=⟨1,0,0⟩$, $𝐣=⟨0,1,0⟩$ and $𝐤=⟨0,0,1⟩$, and every vector combines them: $⟨3,4,12⟩=3𝐢+4𝐣+12𝐤$. Engineering texts favour this second notation; the two say the same thing.

The dot product and angles

The dot product multiplies matching components and adds, giving a number:

Its meaning comes from the law of cosines. Put the tails of $𝐚$ and $𝐛$ together with angle $\theta$ between them; the third side of the triangle is $𝐚-𝐛$, and the law says $|𝐚-𝐛{|}^2=|𝐚{|}^2+|𝐛{|}^2-2|𝐚||𝐛|\cos \theta$. Compute the left side in components instead: each $(a_1-b_1{)}^2$ expands to $a_1^2-2a_1{b}_1+b_1^2$, and the three together give $|𝐚{|}^2+|𝐛{|}^2-2𝐚\cdot 𝐛$. Comparing,

So the angle between two non-zero vectors follows from components alone. A positive dot product means an acute angle, a negative one an obtuse angle, and zero means the vectors are perpendicular. That last test is the one used most: $⟨2,-1,3⟩$ and $⟨1,5,1⟩$ are perpendicular because $2-5+3=0$.

Example. A methane molecule has its carbon atom at the centre of a cube and its four hydrogen atoms at alternate corners, which with the carbon at the origin are $(1,1,1)$, $(1,-1,-1)$, $(-1,1,-1)$ and $(-1,-1,1)$. Find the angle between two carbon to hydrogen bonds.

Take $𝐚=⟨1,1,1⟩$ and $𝐛=⟨1,-1,-1⟩$. The dot product is $1-1-1=-1$ and each length is $\sqrt{3}$, so $\cos \theta =-1/3$ and $\theta =109.47$ degrees, the same for every pair by symmetry. Spectroscopy measures the H-C-H angle of methane as $109.5$ degrees.

Now you. Find the angle between the long diagonal $⟨1,1,1⟩$ of a unit cube and the diagonal $⟨1,1,0⟩$ of its floor, to two decimal places.

Projection and work

The dot product also splits a vector into a part along a direction and a part across it. If $𝐮$ is a unit vector, $𝐛\cdot 𝐮=|𝐛|\cos \theta$ is the length of the shadow $𝐛$ casts on the line of $𝐮$, called the component of $𝐛$ along $𝐮$. Multiplying it by $𝐮$ gives the projection, and what remains of $𝐛$ is perpendicular to $𝐮$.

A $20$ kg box rests on a ramp rising $3$ metres for every $4$ of run, so the uphill direction is $𝐮=⟨4,0,3⟩/5$. Gravity is $𝐅=⟨0,0,-196.2⟩$ newtons, taking $g=9.81$ metres per second squared, and its component along the ramp is $-196.2\times 3/5=-117.7$ newtons, the pull needed to hold the box. The familiar $mg\sin \alpha$ has fallen out of the arithmetic without a triangle of forces.

Work is the same computation: a constant force $𝐅$ moving an object through a displacement $𝐝$ does work $W=𝐅\cdot 𝐝$, since only the part of the force along the motion counts. For a $70$ kg walker, gravity is $⟨0,0,-686.7⟩$ newtons, and over a route with displacement $⟨300,400,120⟩$ metres the work is $W=-686.7\times 120$, which is 82,404 joules of negative work. The horizontal components are multiplied by zero, which is why the contour map of the previous lesson, counting only height, is enough to reckon the effort of a climb.

The cross product

A plane is fixed by a point and the direction perpendicular to it, so tangent planes will need this: given $𝐚$ and $𝐛$, find $𝐧=⟨x,y,z⟩$ perpendicular to both. That is two equations,

Multiply the first by $b_3$ and the second by $a_3$ and subtract: $z$ cancels, leaving $(a_1{b}_3-a_3{b}_1)x+(a_2{b}_3-a_3{b}_2)y=0$, which $x=a_2{b}_3-a_3{b}_2$ and $y=a_3{b}_1-a_1{b}_3$ satisfy. Putting these into the first equation, two terms cancel and $z=a_1{b}_2-a_2{b}_1$ remains, and a direct check confirms that all three satisfy both equations. The result is the cross product:

The indices cycle $1\to 2\to 3\to 1$: the first component uses the second and third entries, the second uses the third and first, the third uses the first and second. On the axes, $𝐢\times 𝐣=𝐤$, which is where right-handedness enters: the cross product points along the right thumb when the fingers curl from $𝐚$ to $𝐛$. Swapping the factors flips every sign, so $𝐛\times 𝐚=-𝐚\times 𝐛$, and parallel vectors have cross product $𝟎$.

The length carries geometry. Expanding both sides in components shows $|𝐚\times 𝐛{|}^2=|𝐚{|}^2|𝐛{|}^2-(𝐚\cdot 𝐛{)}^2$; for $⟨1,2,3⟩$ and $⟨4,5,6⟩$ the cross product $⟨-3,6,-3⟩$ has squared length $54$, and $14\times 77-{32}^2=54$. Writing the dot product as $|𝐚||𝐛|\cos \theta$ and using $1-{\cos }^2\theta ={\sin }^2\theta$,

which is base times height for the parallelogram with sides $𝐚$ and $𝐛$. A triangle is half of one: the triangle with corners $(1,0,0)$, $(0,2,0)$ and $(0,0,3)$ has edges $⟨-1,2,0⟩$ and $⟨-1,0,3⟩$, whose cross product $⟨6,3,2⟩$ has length $7$, so its area is $3.5$.

In mechanics, a force $𝐅$ applied at position $𝐫$ from a pivot produces the torque $\tau =𝐫\times 𝐅$, pointing along the axis it tries to turn about, with length equal to the force times its perpendicular lever arm.

Example. A bicycle crank is $175$ mm long. With $x$ forward and $z$ up, the pedal is at $𝐫=⟨0.14,0,0.105⟩$ metres from the axle, and the rider presses straight down with $𝐅=⟨0,0,-600⟩$ newtons. Find the torque.

As a check, $|𝐫|=\sqrt{0.0196+0.011025}=0.175$. The components of $𝐫\times 𝐅$ are $0\times (-600)-0.105\times 0=0$, then $0.105\times 0-0.14\times (-600)=84$, then $0.14\times 0-0\times 0=0$. The torque is $⟨0,84,0⟩$, of $84$ newton metres along the axle. Only the crank's horizontal reach acts as lever arm against a vertical push, and $600\times 0.14=84$ agrees.

Now you. A spanner runs from the bolt to $𝐫=⟨0.2,0.15,0⟩$ metres, and a hand pulls its end with $𝐅=⟨0,60,80⟩$ newtons. Find the torque and its magnitude, to two decimal places.

Lines in space

In space a single equation in $x$, $y$ and $z$ cuts out a surface, so a line needs another description: a motion. Start at $P_0=(x_0,y_0,z_0)$ and move in the direction $𝐯=⟨v_1,v_2,v_3⟩$; after time $t$ the position is

These parametric equations trace the whole line as $t$ runs over the real numbers. The line through $(1,2,3)$ and $(4,6,15)$ is $x=1+3t$, $y=2+4t$, $z=3+12t$, at the first point when $t=0$, the second when $t=1$ and the midpoint $(2.5,4,9)$ when $t=\frac{1}{2}$. Another starting point or a rescaled direction describes the same line, worth remembering when two answers look different.

Where a line meets anything is found by substitution. A pole $5$ metres tall stands at the origin and sunlight travels in the direction $⟨2,1,-4⟩$. The ray through the top is $(2t,t,5-4t)$, which reaches the ground $z=0$ at $t=1.25$, the point $(2.5,1.25,0)$. The shadow is $\sqrt{7.8125}=2.795$ metres long, and the sun stands $\arctan (5/2.795)=60.8$ degrees above the horizon.

Two lines in the plane meet or are parallel; in space they can also be skew. The $x$ axis $(t,0,0)$ and the line $(0,s,1)$ have directions $𝐢$ and $𝐣$, so they are not parallel, yet one lies at height $0$ and the other at height $1$ and they never meet, like a road and the railway bridge over it. To test two lines, give them separate parameters, set the coordinates equal, and see whether three equations in two unknowns have a solution.

Planes

A plane is fixed by a point $P_0=(x_0,y_0,z_0)$ on it and a normal vector $𝐧=⟨a,b,c⟩$ perpendicular to it. A point $(x,y,z)$ lies in the plane exactly when its displacement from $P_0$ is perpendicular to $𝐧$, which by the dot product is

Collecting constants gives $ax+by+cz=d$ with $d=a{x}_0+b{y}_0+c{z}_0$. So the normal can be read off the coefficients: $2x-y+3z=7$ is perpendicular to $⟨2,-1,3⟩$, and the floor $z=0$ has normal $𝐤$. The angle between two planes is the angle between their normals. Given three points not on a line instead, the cross product of two edges supplies the normal, the construction tangent planes will use.

Example. A roof has its eaves along the line from $(0,0,3)$ to $(8,0,3)$ and rises to $(0,4,6)$ at the ridge, in metres. Find its equation and its pitch, and the fraction of full sunlight a panel lying on it receives when the direction towards the sun is $⟨-2,-1,4⟩$.

The edges from the first point are $⟨8,0,0⟩$ and $⟨0,4,3⟩$, with cross product $⟨0\times 3-0\times 4,0\times 0-8\times 3,8\times 4-0\times 0⟩=⟨0,-24,32⟩$, or $⟨0,-3,4⟩$ after dividing by $8$, of length $5$. The plane is $-3y+4z=d$, and $(0,0,3)$ gives $d=12$; the ridge checks, since $-12+24=12$. The pitch is the angle between this normal and $𝐤$: $\cos \theta =4/5$, so $\theta =36.87$ degrees. A panel collects sunlight in proportion to the cosine of the angle between its normal and the sun's direction, here $(0+3+16)/(5\sqrt{21})=0.829$, so $83$ percent of what it would facing the sun squarely.

Now you. Find the plane through $(0,0,3)$, $(6,2,3)$ and $(1,-3,5)$, and its angle to the horizontal, to one decimal place.

The distance from a point $Q=(x_1,y_1,z_1)$ to a plane is a projection. The shortest route runs along the normal, so the distance is the component, along $𝐧$, of the displacement from any point of the plane to $Q$, which works out as $|a{x}_1+b{y}_1+c{z}_1-d|/|𝐧|$. The origin is $|0-12|/5=2.4$ metres from the roof plane above.

Distance, and what comes next

The formula this lesson began with is the one the next lesson needs first. In the plane the distance from $(x,y)$ to $(a,b)$ is $\sqrt{(x-a{)}^2+(y-b{)}^2}$, and "$(x,y)$ is close to $(a,b)$" now means that number is small, whatever direction the approach comes from. Limits in several variables are defined that way, and the function $xy/(x^2+y^2)$ from the previous lesson will show how badly such a limit can behave. The rest of the toolkit is booked: unit vectors name the direction of a directional derivative, the dot product turns a gradient into a slope, and a normal and a point give a tangent plane.

One limit should be stated plainly. Length, distance, the dot product and angles work in any number of dimensions, with extra terms in the same formulas. The cross product does not: in four dimensions the vectors perpendicular to two given ones form a whole plane, so there is no single answer to pick. That is why Linear Algebra, which works in every dimension, does without it, and why this course uses it only in the three dimensional space where graphs of two variable functions live.

A point in the plane can be approached from infinitely many directions and along infinitely many curves, so a limit in several variables has far more ways to fail than a limit on the real line, where there are only a left and a right.

The previous lesson ended with the tool this one needs: the distance from $(x,y)$ to $(a,b)$ is $\sqrt{(x-a{)}^2+(y-b{)}^2}$, whatever direction separates them. This lesson uses that number to define limits and continuity in two variables, shows how a limit can depend on the path, and gives a reliable way to prove that one exists. It assumes the one variable limit from Calculus I.

Closeness is distance

In one variable, $\underset{x\to a}{lim}f(x)=L$ is a tolerance game: name any tolerance $\epsilon$ on the output, and some tolerance $\delta$ on the input keeps $f(x)$ within it. The several variable version keeps the game and changes only how the input tolerance is measured. Instead of $|x-a|$, use the distance to the point:

means that for every $\epsilon >0$ there is a $\delta >0$ such that $0<\sqrt{(x-a{)}^2+(y-b{)}^2}<\delta$ forces $|f(x,y)-L|<\epsilon$.

Geometrically, the points within $\delta$ of $(a,b)$ form an open disc, and the condition $0<$ removes its centre. So the limit is a promise about a punctured disc: every point in it, in every direction, has a value within $\epsilon$ of $L$. The value at $(a,b)$ itself plays no part and need not exist, as at the hole of $xy/(x^2+y^2)$.

Producing a $\delta$ works the same way as before, with one new habit: bound everything by the distance $d$. Two facts do most of the work. Each coordinate difference is at most the distance, $|x-a|\le d$ and $|y-b|\le d$, since a leg of a right triangle is never longer than its hypotenuse. And a dot product is at most the product of lengths, $|𝐮\cdot 𝐯|\le |𝐮||𝐯|$, since $|\cos \theta |\le 1$.

Take $\underset{(x,y)\to (0,0)}{lim}(3x+4y)=0$. The quantity $3x+4y$ is the dot product of $⟨3,4⟩$ with $⟨x,y⟩$, so $|3x+4y|\le 5d$, and $\delta =\epsilon /5$ works. The bound is sharp: at distance $0.01$ in the direction $⟨3,4⟩$, the point $(0.006,0.008)$ gives $0.018+0.032=0.05$, exactly $5\times 0.01$.

Every path must agree

If the limit is $L$, then every point of the punctured disc has its value within $\epsilon$ of $L$, and in particular every point of any path that runs into $(a,b)$. So the values along every path must tend to $L$: along the $x$ axis, along any line, along a parabola, along a spiral. This gives the standard way to show that a limit does not exist. Find two paths into the point along which the function tends to different values, or one path along which it tends to nothing, and no $L$ can satisfy the definition.

This is the left and right limit test of one variable, with infinitely many sides. It runs in one direction only: agreement along two paths, or ten, or every line, proves nothing, because a path not yet tried may disagree. Paths disprove; they never prove.

A limit that depends on the direction

Take $f(x,y)={\displaystyle \frac{xy}{x^2+y^2}}$, defined everywhere except the origin. Along the $x$ axis, $y=0$ and the function is $0$ at every point, so it tends to $0$. Along the $y$ axis it is also $0$. Along the diagonal $y=x$ it is $x^2/2x^2=\frac{1}{2}$ at every point, however close to the origin. Two paths, two different values, and the limit does not exist.

All the lines at once tell the whole story. Along $y=mx$ with $x\ne 0$,

which does not depend on $x$ at all: the function is constant along each line through the origin, with the constant set by the slope. The slope $m=2$ gives $0.4$, $m=-1$ gives $-\frac{1}{2}$, and as $m$ runs over the real numbers the constant covers every value from $-\frac{1}{2}$ to $\frac{1}{2}$. So every punctured disc around the origin, however small, contains points where $f=\frac{1}{2}$ and points where $f=-\frac{1}{2}$, and no tolerance smaller than $\frac{1}{2}$ can be met around any candidate $L$.

Polar coordinates make the same point in one line. Writing $x=r\cos \theta$ and $y=r\sin \theta$, where $r$ is the distance to the origin and $\theta$ the angle of the direction, gives $f=\cos \theta \sin \theta =\frac{1}{2}\sin 2\theta$. The distance $r$ has cancelled completely, so the value depends only on the direction. The level curves of $f$ are the lines through the origin, all crowding into the hole with every height from $-\frac{1}{2}$ to $\frac{1}{2}$.

Example. Show that ${\displaystyle \frac{x^2-y^2}{x^2+y^2}}$ has no limit at the origin.

Along $y=mx$ the value is ${\displaystyle \frac{x^2-m^2{x}^2}{x^2+m^2{x}^2}}={\displaystyle \frac{1-m^2}{1+m^2}}$, again constant on each line. The $x$ axis, $m=0$, gives $1$. The line $y=x$ gives $0$. The $y$ axis, which is not of the form $y=mx$ and must be checked on its own, has $x=0$ and value $-y^2/y^2=-1$. Three different values along three paths, so there is no limit. In polar coordinates the function is ${\cos }^2\theta -{\sin }^2\theta =\cos 2\theta$, which again has lost $r$.

Now you. Find the value that ${\displaystyle \frac{2xy}{x^2+4y^2}}$ approaches along the line $y=mx$, and use it to show that the function has no limit at the origin.

Every line is not enough

After the examples above it is tempting to check all the lines through the point and, if they agree, declare the limit. The function

shows why that fails. On the $x$ axis and the $y$ axis it is $0$. Along $y=mx$ with $m\ne 0$,

and as $x\to 0$ the numerator goes to $0$ while the denominator goes to $m^2$, so the limit is $0$. Every line into the origin gives $0$. Numerically, along $y=x$ the values at $x=0.1$, $0.01$ and $0.001$ are $0.0990$, $0.0100$ and $0.0010$, heading steadily for zero.

Now approach along the parabola $y=x^2$ instead:

at every point, however close. At $(0.01,0.0001)$ the value is exactly $0.5$. So there are points with $g=\frac{1}{2}$ in every punctured disc around the origin, and the limit does not exist, even though it is $0$ along every straight line. Along $y=k{x}^2$ the value is $k/(1+k^2)$: $g$ is constant on parabolas the way $xy/(x^2+y^2)$ is constant on lines.

Straight lines are simply the wrong family of paths for this function, and no finite list of families is always the right one. To prove a limit exists, something must control every point of a punctured disc at once.

Polar coordinates and the squeeze

The tool that controls a whole disc is a bound that depends on the distance alone. Suppose that near $(a,b)$

where $d$ is the distance to $(a,b)$ and $B(d)\to 0$ as $d\to 0$. Given $\epsilon$, choose $\delta$ so that $B(d)<\epsilon$ whenever $d<\delta$, which the one variable limit of $B$ guarantees. Then every point of the punctured disc of radius $\delta$ has $|f-L|<\epsilon$, whatever its direction. This is the squeeze theorem of Calculus I, with $f$ trapped between $L-B(d)$ and $L+B(d)$, and both walls closing in on $L$ as the distance shrinks.

At the origin, polar coordinates are the natural way to find such a bound, because $r$ is the distance. Substitute $x=r\cos \theta$, $y=r\sin \theta$, simplify, and try to write $|f-L|$ as at most something in $r$ alone. Since $|\cos \theta |\le 1$ and $|\sin \theta |\le 1$, any leftover trigonometric factors can be bounded by $1$. The one requirement is that the final bound contain no $\theta$: a bound that shrinks for each fixed direction but not uniformly across all of them proves nothing.

Example. Prove that ${\displaystyle \frac{3x^{2}y}{x^2+y^2}}\to 0$ as $(x,y)\to (0,0)$, and find a $\delta$ for each $\epsilon$.

In polar coordinates the numerator is $3r^3{\cos }^2\theta \sin \theta$ and the denominator is $r^2$, so the function equals $3r{\cos }^2\theta \sin \theta$. Its size is at most $3r$, since the trigonometric factors are at most $1$. The bound $B(r)=3r$ contains no $\theta$ and tends to $0$, so the limit is $0$, and $\delta =\epsilon /3$ works. As a check, at $(0.001,0.001)$ the function is $0.003\times {10}^{-6}/(2\times {10}^{-6})=0.0015$, and the bound there is $3\times 0.001414=0.0042$.

Now you. Prove that ${\displaystyle \frac{x^2{y}^2}{x^2+y^2}}\to 0$ at the origin, and give a $\delta$ that works for $\epsilon =0.0001$.

Counting degrees predicts the outcome. In $xy/(x^2+y^2)$ the numerator and denominator both have degree $2$, the powers of $r$ cancel exactly, and a pure function of $\theta$ is left, which is what makes the value depend on the direction. In the example the numerator has degree $3$, one spare power of $r$ survives, and it drives the whole expression to zero. The rule is a guide, not a theorem, and $g=x^{2}y/(x^4+y^2)$ misleads it. In polar coordinates

and for each fixed $\theta$ with $\sin \theta \ne 0$ this tends to $0$ as $r\to 0$. But when $\sin \theta$ is as small as $r{\cos }^2\theta$, the denominator shrinks to the size of $r^2$ and the fraction reaches $\frac{1}{2}$; those are the points of the parabola $y=x^2$. No bound in $r$ alone exists, which is exactly the failure the requirement above rules out.

Where continuous functions come from

A function $f$ of two variables is continuous at $(a,b)$ when $f(a,b)$ is defined and $\underset{(x,y)\to (a,b)}{lim}f(x,y)=f(a,b)$: the limit exists and the function takes the value it predicts. It is continuous on a region when it is continuous at every point of it. Continuity is recognised on sight the same way as in one variable: by building functions from simple ones with operations that preserve it.

The building blocks are the constants and the two coordinate functions $x$ and $y$. The function $x$ is continuous everywhere because $|x-a|\le d$, so $\delta =\epsilon$ works; likewise $y$.

Sums, products and quotients follow from the limit laws, and their one variable proofs go through unchanged, because they only ever manipulate output tolerances and never mention how the input distance is measured. The product shows the pattern. If $f\to L$ and $g\to M$, write

Near the point $f$ stays within $1$ of $L$, so $|f|\le |L|+1$, and both terms can be made as small as wanted by making $|g-M|$ and $|f-L|$ small, each of which some $\delta$ achieves; take the smaller of the two $\delta$ values. So products of continuous functions are continuous, and repeating the product and sum rules shows that every polynomial in $x$ and $y$ is continuous everywhere. A quotient is continuous wherever its denominator is not zero, so every rational function is continuous except where the denominator vanishes. That includes $xy/(x^2+y^2)$ everywhere but the origin, where no value fills the hole.

Composition is the rule that reaches everything else. Suppose $u=g(x,y)$ is continuous at $(a,b)$, and $h$ is a function of one variable continuous at $c=g(a,b)$. Given $\epsilon$, continuity of $h$ provides an $\eta$ such that $|u-c|<\eta$ keeps $|h(u)-h(c)|<\epsilon$. Continuity of $g$ then provides a $\delta$ such that $d<\delta$ keeps $|g(x,y)-c|<\eta$. Chaining the two, $d<\delta$ keeps $|h(g(x,y))-h(g(a,b))|<\epsilon$. So $e^{xy}$, $\sin (x^2+y)$, $\ln (x-y)$ on the half-plane $x>y$, and $\sqrt{9-x^2-y^2}$ on its disc are all continuous, since each is a continuous function of one variable applied to a polynomial.

The consequence is the same as in Calculus I: most limits are evaluated by substitution. Where nothing divides by zero the limit is the value, so $\underset{(x,y)\to (1,2)}{lim}(x^{2}y+3)/(x+y)=5/3$ with no $\epsilon$ in sight. The effort goes into the points where the formula breaks, and composition often handles those too.

Example. What value at the origin makes $F(x,y)={\displaystyle \frac{\sin (x^2+y^2)}{x^2+y^2}}$ continuous there?

Write $F=h(u)$ with $u=x^2+y^2$ and $h(u)=(\sin u)/u$. Calculus I gives $\sin u/u\to 1$ as $u\to 0$, so setting $h(0)=1$ makes $h$ continuous at $0$. The polynomial $u$ is continuous with $u(0,0)=0$, so by the composition rule $F$ tends to $h(0)=1$, and defining $F(0,0)=1$ makes it continuous. At $(0.1,0.2)$, where $u=0.05$, the value is $0.99958$.

Now you. What value at the origin makes ${\displaystyle \frac{1-\cos (x^2+y^2)}{(x^2+y^2{)}^2}}$ continuous there?

One variable at a time, and what comes next

A tempting shortcut is to test continuity one variable at a time: freeze $y$, check the resulting function of $x$, then freeze $x$ and check the function of $y$. The function $xy/(x^2+y^2)$, given the value $0$ at the origin, shows that this is not enough. Freeze $y=b$ with $b\ne 0$ and the result $bx/(x^2+b^2)$ is a rational function of $x$ with a denominator that never vanishes, continuous everywhere. Freeze $y=0$ and the result is the constant $0$, also continuous. The same holds with the roles swapped. So the function is continuous in each variable separately at every point, including the origin, and yet it is not continuous at the origin, since along $y=x$ it stays at $\frac{1}{2}$ while its value there is $0$.

The explanation is the theme of this lesson. Freezing a variable looks only along the two axes, two paths out of infinitely many, and paths can disprove continuity but never prove it. In three variables nothing changes but the distance formula: a ball replaces the disc.

Where continuity does hold, the next step is to measure slopes, and the idea that failed here as a test of continuity works well as a way of taking them. Freeze every variable but one, and the surface becomes a curve with an ordinary slope that Calculus I can compute. That is the partial derivative, the subject of the next lesson. It inherits the weakness exposed in this section: two slopes along the axes see only two directions, and $xy/(x^2+y^2)$ will turn up again as a function that has them both at a point where it is not even continuous.

A surface has a different slope in every direction through a point, but along the two coordinate directions the slope is an ordinary derivative that the calculus of one variable already knows how to take.

The previous lesson defined limits and continuity in two variables and found that freezing one variable at a time is a poor test of continuity, because it looks along only two paths. As a way of measuring slopes, freezing works well. This lesson defines the partial derivative, computes it with the rules of Calculus I, reads it off real models, differentiates twice, and meets the theorem that makes second derivatives manageable, together with the function where it fails.

Freeze every variable but one

Let $f$ be a function of $x$ and $y$ and fix a point $(a,b)$. Holding $y$ at $b$ leaves $x\mapsto f(x,b)$, a function of one variable, and its ordinary derivative at $x=a$ is the partial derivative of $f$ with respect to $x$:

Only $x$ moves in the quotient; $y$ sits at $b$ throughout. Holding $x$ at $a$ and moving $y$ instead gives $f_y(a,b)$, the same limit with $f(a,b+k)$ in place of $f(a+h,b)$.

Two notations are in use and they mean exactly the same thing: $f_x$, compact enough for long calculations, and the Leibniz form $\frac{\partial f}{\partial x}$, whose curly $\partial$ warns that other variables are being held still. Adrien-Marie Legendre used the symbol in 1786 and Carl Jacobi made it standard in 1841.

Take the paraboloid $f(x,y)=x^2+y^2$ at $(3,4)$, where $f=25$. Holding $y=4$, the difference quotient is

which tends to $6$, so $f_x(3,4)=6$. The same work in $y$ gives $8+k$, so $f_y(3,4)=8$. At $h=0.001$ the quotients are $6.001$ and $8.001$, the east and north slopes found by stepping away from $(3,4)$ when this surface first appeared, in the lesson on functions of several variables.

Nobody evaluates that limit for every function. Since $f_x$ is an ordinary derivative, every rule of Calculus I applies, and the one new instruction is to treat every other variable as a constant. In $x^2+y^2$ the term $y^2$ is a constant as far as $x$ is concerned, so $f_x=2x$ and likewise $f_y=2y$, and $f_x(3,4)=6$ falls out by substitution. In $x^2{y}^3$ with $y$ frozen, $y^3$ is a coefficient like the $5$ in $5x^2$, so the $x$ partial is $2x{y}^3$ and the $y$ partial is $3x^2{y}^2$. The product rule returns only when both factors contain the moving variable, and the chain rule whenever the moving variable sits inside another function. In three or more variables nothing changes: $g_z$ freezes $x$ and $y$ and differentiates in $z$.

Example. Find $f_x$ and $f_y$ for $f(x,y)=x^2{y}^3+\sin (xy)$, and evaluate both at $(1,2)$.

With $y$ constant, $x^2{y}^3$ gives $2x{y}^3$, and $\sin (xy)$ needs the chain rule with inner function $xy$, whose $x$ derivative is $y$. So $f_x=2x{y}^3+y\cos (xy)$, and with $x$ constant the same steps give $f_y=3x^2{y}^2+x\cos (xy)$. At $(1,2)$, $\cos 2=-0.4161$, so $f_x=16-0.8323=15.168$ and $f_y=12-0.4161=11.584$. A central difference in $x$ with step $0.001$ also gives $15.168$: a partial derivative can always be checked by nudging one input.

Now you. Find $f_x$ and $f_y$ for $f(x,y)=x{e}^{xy}+y^2$ at $(1,1)$, to three decimal places.

The slope of a slice

Cut the surface $z=f(x,y)$ with the vertical plane $y=b$ and what remains is the curve $z=f(x,b)$, the trace of the surface in that plane. Its slope at $x=a$ is $f_x(a,b)$. So $f_x$ is the slope of the surface for a walker heading due east, in the direction of increasing $x$, and $f_y$ is the slope for a walker heading due north. Neither says anything about a walker heading north-east.

On the paraboloid, the slice $y=4$ is the parabola $z=x^2+16$, whose tangent line at $x=3$ has slope $6$. In space that line passes through $(3,4,25)$ and climbs $6$ units for each unit east, so it runs along $⟨1,0,6⟩$; the slice $x=3$ gives a tangent line along $⟨0,1,8⟩$. Two lines through one point determine a plane, and any tangent plane at $(3,4,25)$ must contain both. Whether the surface actually has one is a question for the next lesson.

Each partial derivative has its own meaning and its own units. The wind chill index $W(T,v)=13.12+0.6215T-11.37v^{0.16}+0.3965T{v}^{0.16}$, with $T$ in degrees Celsius and $v$ in km/h, has $W_T=0.6215+0.3965v^{0.16}$, which at $v=30$ is $1.305$: in that wind each degree of real cooling feels like $1.3$ degrees. It also has $W_v=0.16v^{-0.84}(-11.37+0.3965T)$, which at $T=-5$ and $v=30$ is $-0.123$ degrees per km/h. Raising the wind from $30$ to $31$ km/h lowers the index by $0.121$, slightly less than the rate because the curve flattens as the wind grows.

A rate with everything else fixed

In a model, a partial derivative answers the question every controlled experiment asks: change this one thing and hold the rest. The ideal gas law $p=nRT/V$, with $R=8.314$ joules per mole per kelvin, has three. With $n$ and $V$ held, $nR/V$ is a constant coefficient; with $n$ and $T$ held, $V$ appears as $V^{-1}$. So

and $\partial p/\partial n=RT/V$. For one mole at $T=273.15$ K in $V=0.0224$ cubic metres, where $p=101383$ Pa, these give $\partial p/\partial T=371.2$ Pa per kelvin and $\partial p/\partial V=-101383/0.0224=-4.526\times {10}^6$ Pa per cubic metre, or $-4526$ Pa per litre. Finite steps from the same state, taken in the lesson on functions of several variables, found that one extra kelvin raises the pressure by $371$ Pa and one extra litre lowers it by $4333$ Pa. The temperature figures agree exactly because $p$ is linear in $T$. The volume step falls short of the rate because $p$ falls less and less steeply as $V$ grows, so a whole litre averages a weakening slope.

Example. A plant has the Cobb-Douglas production function $Q=1.01L^{3/4}{K}^{1/4}$, fitted to American manufacturing data in 1928, and runs at $L=16$, $K=81$. Find the marginal products $Q_L$ and $Q_K$, the extra output per extra unit of each input with the other held fixed.

Treating $K$ as constant, $Q_L=1.01\times \frac{3}{4}{L}^{-1/4}{K}^{1/4}=0.7575(K/L{)}^{1/4}$, and the fourth root of $81/16$ is $\frac{3}{2}$, so $Q_L=1.136$. Treating $L$ as constant, $Q_K=0.2525(L/K{)}^{3/4}=0.2525\times \frac{8}{27}=0.0748$. An actual extra unit of labour raises output from $24.24$ to $25.37$, an increase of $1.128$, close to the rate. As a check, $L{Q}_L+K{Q}_K=18.18+6.06=24.24=Q$. That is no accident: the derivatives show $L{Q}_L=\frac{3}{4}Q$ and $K{Q}_K=\frac{1}{4}Q$, so if each input were paid its marginal product, labour would take three quarters of the output, close to the share of wages Douglas found in the data.