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MVC1 · 1

Functions of several variables

7 concepts · 22 questions

Graphs and level sets, limits and continuity in several variables (and how they fail), and partial derivatives.

Overview 1The plan of this section

Functions like temperature T(x,y,z)T(x, y, z) depend on several variables. Before differentiating them we need:

  1. Ways to picture them: graphs and level sets (Def. 1.1).
  2. Limits and continuity (Def. 1.2), which are subtler than in one variable because you can approach a point along many paths (Prop. 1.3).
  3. Partial derivatives (Def. 1.4): differentiate in one variable, holding the others fixed.

Picture with level curves; limits need every path to agree; partials freeze all but one variable.

Definition 1.1Graphs and level sets

For f:D⊆R2→Rf: D\subseteq\mathbb{R}^2\to\mathbb{R}, the graph is the surface z=f(x,y)z = f(x, y) in R3\mathbb{R}^3. The level curve (contour) at height cc is {(x,y):f(x,y)=c}\{(x, y): f(x, y) = c\}. For ff of three variables the level sets f(x,y,z)=cf(x, y, z) = c are surfaces. Examples: f=x2+y2f = x^2 + y^2 has circles as level curves (a paraboloid); f=x2−y2f = x^2 - y^2 has hyperbolas (a saddle); f=1−x2−y2f = \sqrt{1 - x^2 - y^2} is a hemisphere on the disc x2+y2≤1x^2 + y^2\le1.

Level curves = contour map: f(x,y)=cf(x, y) = c for several cc.

Close contours = steep slope.

x2+y2x^2 + y^2: circles (bowl). x2−y2x^2 - y^2: hyperbolas (saddle).

Definition 1.2Limits and continuity

lim⁡(x,y)→(a,b)f(x,y)=L\lim_{(x, y)\to(a, b)}f(x, y) = L means: for every ε>0\varepsilon > 0 there is δ>0\delta > 0 with ∣f(x,y)−L∣<ε|f(x, y) - L| < \varepsilon whenever 0<(x−a)2+(y−b)2<δ0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta. So ff must approach LL along every path to (a,b)(a, b). ff is continuous at (a,b)(a, b) if the limit equals f(a,b)f(a, b). Sums, products, quotients (non-zero denominator) and compositions of continuous functions are continuous; polynomials in x,yx, y are continuous everywhere.

Limit in 2D: f→Lf\to L however (x,y)(x, y) approaches, along all paths at once.

Continuous = limit equals value. Polynomials and their combinations are continuous.

To prove a limit: bound ∣f−L∣|f - L| by something in r=x2+y2r = \sqrt{x^2 + y^2} (polar coordinates).

Proposition 1.3The two-path test for non-existence

If f(x,y)→L1f(x, y)\to L_1 along one path to (a,b)(a, b) and f→L2≠L1f\to L_2\ne L_1 along another, then lim⁡(x,y)→(a,b)f\lim_{(x, y)\to(a, b)}f does not exist. Agreement along all straight lines is not enough: f=x2yx4+y2f = \frac{x^2y}{x^4 + y^2} tends to 00 along every line through the origin, but to 12\frac12 along y=x2y = x^2.

Two paths, two answers ⇒ no limit.

Lines aren't enough: x2yx4+y2\frac{x^2y}{x^4 + y^2} needs the parabola y=x2y = x^2.

Typical paths: axes, y=mxy = mx, y=x2y = x^2.

Proof

Definition 1.4Partial derivatives

∂f∂x(a,b)=fx(a,b)=lim⁡h→0f(a+h,b)−f(a,b)h,\frac{\partial f}{\partial x}(a, b) = f_x(a, b) = \lim_{h\to0}\frac{f(a + h, b) - f(a, b)}h,

the derivative in xx with yy held fixed; similarly fyf_y. Geometrically fxf_x is the slope of the graph in the xx-direction. All the one-variable rules apply, treating the other variables as constants.

fxf_x: differentiate in xx, treat yy as a constant.

Slope of the surface in the xx-direction.

Example: f=x2y3f = x^2y^3: fx=2xy3f_x = 2xy^3, fy=3x2y2f_y = 3x^2y^2.

Example 1.5Partial derivatives can exist where f is not continuous

Let f(x,y)=xyx2+y2f(x, y) = \frac{xy}{x^2 + y^2} for (x,y)≠(0,0)(x, y)\ne(0, 0) and f(0,0)=0f(0, 0) = 0. Then fx(0,0)=fy(0,0)=0f_x(0, 0) = f_y(0, 0) = 0 (since f=0f = 0 on both axes), yet ff is not continuous at the origin (Prop. 1.3 quiz). So in several variables, having partial derivatives is much weaker than being differentiable; the right notion is in Section 2.

Partials only look along the axes. A function can have partials and still jump.

So we need a stronger notion of differentiability (Section 2).

Check 1.6Check: limits in two variables

  1. Continuous at the point? Substitute.
  2. Suspect no limit? Try the axes, lines y=mxy = mx, and curves like y=x2y = x^2; different answers prove non-existence.
  3. Suspect a limit LL? Bound ∣f−L∣|f - L| by a function of r=x2+y2r = \sqrt{x^2 + y^2} alone (polar coordinates), or squeeze; it must tend to 00 regardless of θ\theta.
  4. Never conclude existence from a few paths.

Disprove with two paths; prove with a bound in rr.

Polar coordinates: x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta, then check the r→0r\to0 limit is uniform in θ\theta.