Introduction to Multivariable Calculus

Single-variable calculus studies one output as one input varies. Many relationships depend on several inputs at once. Surface height depends on two horizontal coordinates; a modelled quantity can depend on both position and time. Multivariable calculus keeps those inputs distinct and studies their combined effect.

With several inputs, limits must account for different directions of approach, derivatives can hold selected inputs fixed, and integrals can accumulate quantities over regions.

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Distinguish the input space from the graph

A scalar function $f(x,y)$ takes a pair of inputs and returns one number. Its domain lies in the input plane; its graph consists of points $(x,y,f(x,y))$ in three-dimensional space.

For

f(x,y)=x^2+2y^2,

the domain is all of $\mathbb R^2$ and the range is $[0,\infty)$ . Every nonnegative value $c$ occurs at $(\sqrt c,0)$ . The graph is an upward-opening elliptic paraboloid.

A function of three variables has a three-dimensional input domain, but its scalar graph would require four coordinates. Level surfaces and cross-sections are then more practical than a conventional graph.

Read level curves as equal-output sets

The level curve $f(x,y)=c$ consists of inputs with the same output. Here,

x^2+2y^2=c.

Positive levels are ellipses. At $c=8$ , the horizontal intercepts are $\pm2\sqrt2$ and the vertical intercepts are $\pm2$ . Level zero contains only the origin, and negative levels are empty.

These curves lie in the input plane. Their spacing can suggest rapid or gradual output change only when the level increments and axis scales are known.

Equal-output ellipses

Equal input-coordinate units.

Diagram explanation and exact data

For level eight, horizontal intercepts have magnitude 2√2 and vertical intercepts have magnitude two.

Equal-output ellipses: exact reference data
Level	Horizontal intercept magnitude	Vertical intercept magnitude
2	√2	1
4	2	√2
8	2√2	2

Change one input while holding another fixed

The partial derivative with respect to $x$ is

f_x(x,y)=\lim_{h\to0}\frac{f(x+h,y)-f(x,y)}h.

For the paraboloid, $f_x=2x$ and $f_y=4y$ . At $(1,1)$ , the coordinate-direction rates are two and four. They do not automatically describe every direction through that point.

Holding one variable fixed is an operation on the mathematical function. A physical experiment may not allow an input to change independently of the others. That interpretation needs modelling justification.

Approximate nearby values with a tangent plane

For a differentiable function,

f(a+h,b+k)\approx f(a,b)+f_x(a,b)h+f_y(a,b)k.

At $(1,1)$ , the paraboloid has tangent plane

z=3+2(x-1)+4(y-1)=2x+4y-3.

Take increments $h=0.1$ , $k=-0.05$ . The linear terms cancel, so the plane predicts value three. Direct calculation gives

f(1.1,0.95)=1.21+2(0.9025)=3.015.

The exact change is $0.015=h^2+2k^2$ . A zero first-order prediction is not an assertion that every finite change is zero. The tangent-plane reference distinguishes differentiability from the existence of isolated partial rates.

Combine changes along a path

If $x=x(t)$ and $y=y(t)$ , the chain rule for differentiable $f$ gives

\frac{d}{dt}f(x(t),y(t))=f_xx’+f_yy’.

For $x=t$ , $y=t^2$ , substitution gives $f=t^2+2t^4$ . Its derivative is $2t+8t^3$ .

Using partial derivatives gives the same result:

f_xx’+f_yy’=(2t)(1)+(4t^2)(2t)=2t+8t^3.

At $t=1$ , the output rate is ten per unit of parameter. Neither partial alone gives that answer because both inputs are changing.

Organise directional information with a gradient

The gradient is

\nabla f=\langle f_x,f_y\rangle.

For a differentiable function and a unit direction $\mathbf u$ , the rate per unit input-space distance is $D_{\mathbf u}f=\nabla f\cdot\mathbf u$ . At $(1,1)$ , the gradient is $(2,4)$ .

The unit-direction condition distinguishes a spatial rate from the rate along a faster or slower path. Differentiability justifies this formula in every direction. The mere existence of partial derivatives does not: additional hypotheses or a direct argument are needed.

A limit must accommodate every approach

In two variables, a point can be approached along infinitely many paths. For $xy/(x^2+y^2)$ away from the origin, both coordinate axes give zero, while $y=x$ gives $1/2$ . The limit therefore does not exist.

Agreement on two or several paths does not prove a limit. Disagreement on two paths can disprove it. Functions of Several Variables develops both counterexamples and bounds that control all approaches.

Accumulate over a region

A double integral adds contributions over a two-dimensional region. On $0\le x\le1$ , $0\le y\le2$ ,

\iint_R(x+2y)\,dA=\int_0^1\int_0^2(x+2y)\,dy\,dx.

First hold $x$ fixed and integrate in $y$ :

\int_0^2(x+2y)\,dy=[xy+y^2]_0^2=2x+4.

Then

\int_0^1(2x+4)\,dx=[x^2+4x]_0^1=5.

For a nonnegative height, the result can represent volume. For mass per unit area, it represents mass. Units and interpretation depend on the integrand and area element, not on the integral sign alone.

Continuity on a rectangle is sufficient for either integration order. Nonrectangular regions need bounds describing their actual shape; more general integrands require more careful hypotheses.

Recognise the boundaries of this introduction

The following pages develop domains, level sets, limits, partial derivatives and gradients. Multiple integration, vector fields, line integrals and the major vector-calculus theorems are substantial later subjects. Their entry points here should not be mistaken for full coverage.

A function with several inputs and several outputs has a derivative represented by a linear map or Jacobian matrix. The gradient is the scalar-output case, not a universal replacement for every multivariable derivative.

Practice questions

For $f=x^2+2y^2$ , find $f_x$ and $f_y$ at $(1,1)$ .
What is the zero level set? What happens at a negative level?
Along $x=t$ , $y=t^2$ , find the rate of change at $t=1$ .

Check your answers

The partial derivatives are $2x$ and $4y$ , giving two and four.
The zero level contains only $(0,0)$ . Negative levels are empty because the squared terms are nonnegative.
The path function is $t^2+2t^4$ , with derivative $2t+8t^3$ . Its value at one is ten.