Calculus Volume 3 · Chapter 4 · Section 4.1
Functions of Several Variables
Two inputs, one output. The graph is a landscape; its level curves are the landscape's map.
Colour key
key idea
· 3D figures turn when dragged · page links open the textbook
Level 1 · see it What a function of two variables looks like before any formula
A profit landscape
A hardware maker sells x x x y y y f ( x , y ) = 16 − ( x − 3 ) 2 − ( y − 2 ) 2 f(x,y)=16-(x-3)^2-(y-2)^2 f ( x , y ) = 16 − ( x − 3 ) 2 − ( y − 2 ) 2 Example 4.3 ).
16
thousand dollars at ( 3 , 2 ) (3,2) ( 3 , 2 ) Two inputs make a surface, and every height on it has a curve on the ground below.
drop the contour map
Heights drawn at 0.4 scale. Drag to turn.
Two numbers in, one number out
A function of two variables z = f ( x , y ) z=f(x,y) z = f ( x , y ) gives each point
( x , y ) (x,y) ( x , y ) of a set
D D D exactly one number (
p302 ).
Domain D D D : the points allowed in.
A square root needs something ≥ 0 \ge 0 ≥ 0 > 0 >0 > 0 Range : every output that actually happens.
Drag P P P g ( P ) g(P) g ( P ) g = 1 g=1 g = 1 g = 2 g=2 g = 2
Example 4.1b (p303 ): g ( x , y ) = 9 − x 2 − y 2 g(x,y)=\sqrt{9-x^2-y^2} g ( x , y ) = 9 − x 2 − y 2 x 2 + y 2 ≤ 9 x^2+y^2\le 9 x 2 + y 2 ≤ 9
Slice it flat: level curves
Cut the surface with the plane z = c z=c z = c f ( x , y ) = c f(x,y)=c f ( x , y ) = c p308 ). Every level at once is a contour map .
function
hemisphere · Ex 4.2a
bowl · Ex 4.2b
profit · Ex 4.3
saddle · Ex 15
sin x cos y · Ex 4.5
level c 2
play c
Try: the hemisphere, c up to 3 — the circles shrink to a point. The saddle, c through 0 — the hyperbolas switch sides. A level curve is a height you can walk along without climbing.
One picture
Horizontal cuts give the map; upright cuts give the profiles. Both are ordinary curves.
A function of two variables is a landscape: read it from above, or cut it and look from the side.
Level 2 · compute it Domains, level curves, traces, level surfaces and three labs
The five definitions
Domain and range · p302
z = f ( x , y ) , ( x , y ) ∈ D ⊆ R 2 z=f(x,y),\qquad (x,y)\in D\subseteq\mathbb R^2 z = f ( x , y ) , ( x , y ) ∈ D ⊆ R 2 Range: every z z z D D D
Graph · p304
{ ( x , y , f ( x , y ) ) : ( x , y ) ∈ D } \{(x,y,f(x,y)) : (x,y)\in D\} {( x , y , f ( x , y )) : ( x , y ) ∈ D } A surface in space.
Level curve · p308
f ( x , y ) = c f(x,y)=c f ( x , y ) = c c c c
Vertical trace · p310
z = f ( a , y ) or z = f ( x , b ) z=f(a,y)\ \text{ or }\ z=f(x,b) z = f ( a , y ) or z = f ( x , b ) The cut by the plane x = a x=a x = a y = b y=b y = b
Level surface · p313
f ( x , y , z ) = c f(x,y,z)=c f ( x , y , z ) = c Three inputs: the level sets are surfaces.
Level curves slice horizontally, traces slice upright; each is a curve you already know how to draw.
Finding a domain
the formula has so we need example a square root inside ≥ 0 \ge 0 ≥ 0 Ex 4.1b: 9 − x 2 − y 2 ≥ 0 9-x^2-y^2\ge 0 9 − x 2 − y 2 ≥ 0 a fraction bottom ≠ 0 \ne 0 = 0 Ex 4.6b: x 2 − y 2 ≠ 0 x^2-y^2\ne 0 x 2 − y 2 = 0 y ≠ ± x y\ne\pm x y = ± x a square root in the bottom inside > 0 >0 > 0 Ex 4.6a: x 2 + y 2 + z 2 < 9 x^2+y^2+z^2<9 x 2 + y 2 + z 2 < 9 a logarithm inside > 0 >0 > 0 Exercise 7: y 2 − x > 0 y^2-x>0 y 2 − x > 0 x < y 2 x<y^2 x < y 2 only polynomials, cube roots nothing Exercise 37: all of space
Write each restriction as an inequality; the edge of the region is where the inside expression equals 0.
Examples 4.1–4.7
Example given result 4.1 · p302 f = 3 x + 5 y + 2 f=3x+5y+2 f = 3 x + 5 y + 2 g = 9 − x 2 − y 2 g=\sqrt{9-x^2-y^2} g = 9 − x 2 − y 2 f f f R 2 \mathbb R^2 R 2 R \mathbb R R g g g x 2 + y 2 ≤ 9 x^2+y^2\le 9 x 2 + y 2 ≤ 9 [ 0 , 3 ] [0,3] [ 0 , 3 ] 4.2 · p304 graph g g g x 2 + y 2 x^2+y^2 x 2 + y 2 a hemisphere of radius 3; a paraboloid 4.3 · p306 16 − ( x − 3 ) 2 − ( y − 2 ) 2 16-(x-3)^2-(y-2)^2 16 − ( x − 3 ) 2 − ( y − 2 ) 2 disk of radius 4 about ( 3 , 2 ) (3,2) ( 3 , 2 ) 16 − z \sqrt{16-z} 16 − z ( 3 , 2 ) (3,2) ( 3 , 2 ) 4.4 · p308 8 + 8 x − 4 y − 4 x 2 − y 2 \sqrt{8+8x-4y-4x^2-y^2} 8 + 8 x − 4 y − 4 x 2 − y 2 c = 0 c=0 c = 0 ( x − 1 ) 2 4 + ( y + 2 ) 2 16 = 1 \frac{(x-1)^2}{4}+\frac{(y+2)^2}{16}=1 4 ( x − 1 ) 2 + 16 ( y + 2 ) 2 = 1 [ 0 , 4 ] [0,4] [ 0 , 4 ] 4.5 · p310 sin x cos y \sin x\cos y sin x cos y x , y = − π 4 , 0 , π 4 x,y=-\frac\pi4,0,\frac\pi4 x , y = − 4 π , 0 , 4 π x = ∓ π 4 x=\mp\frac\pi4 x = ∓ 4 π z = ∓ 2 2 cos y z=\mp\frac{\sqrt2}{2}\cos y z = ∓ 2 2 cos y x = 0 x=0 x = 0 z = 0 z=0 z = 0 y = ± π 4 y=\pm\frac\pi4 y = ± 4 π z = 2 2 sin x z=\frac{\sqrt2}{2}\sin x z = 2 2 sin x y = 0 y=0 y = 0 z = sin x z=\sin x z = sin x 4.6 · p312 3 x − 4 y + 2 z 9 − x 2 − y 2 − z 2 \frac{3x-4y+2z}{\sqrt{9-x^2-y^2-z^2}} 9 − x 2 − y 2 − z 2 3 x − 4 y + 2 z 2 t − 4 x 2 − y 2 \frac{\sqrt{2t-4}}{x^2-y^2} x 2 − y 2 2 t − 4 open ball x 2 + y 2 + z 2 < 9 x^2+y^2+z^2<9 x 2 + y 2 + z 2 < 9 t ≥ 2 t\ge 2 t ≥ 2 y ≠ ± x y\ne\pm x y = ± x 4.7 · p313 4 x 2 + 9 y 2 − z 2 = 1 4x^2+9y^2-z^2=1 4 x 2 + 9 y 2 − z 2 = 1 a hyperboloid of one sheet
* The book prints the range of Example 4.4 as [ 0 , 4 ) [0,4) [ 0 , 4 ) f ( 1 , − 2 ) = 16 = 4 f(1,-2)=\sqrt{16}=4 f ( 1 , − 2 ) = 16 = 4 [ 0 , 4 ] [0,4] [ 0 , 4 ]
Lab 1 · domains you can drag
function
Ex 4.1b
CP 4.1
Exercise 7
Exercise 10
Exercise 33
Ex 4.4
Shaded: the domain. A solid edge belongs to it; a dotted edge does not.
Try: Exercise 7 — drag across the parabola x = y 2 x=y^2 x = y 2 Every edge here is a level curve of the expression under the root or inside the log.
Lab 2 · vertical traces
function
Ex 4.5
CP 4.3
Exercise 31
cut by
x = a
y = b
where to cut
Try: Example 4.5 at x = − π 4 , 0 , π 4 x=-\frac\pi4,\ 0,\ \frac\pi4 x = − 4 π , 0 , 4 π z = 0 z=0 z = 0 Fixing one input leaves a function of one variable: its graph is the trace.
Lab 3 · level surfaces
function
Ex 4.7
Exercise 49
CP 4.5
level c 1
play c
Three inputs leave no room to draw the graph; instead, draw the set where the output is c c c Figure 4.13 ).
Try: Example 4.7 from 3 down to −3 — one sheet, a cone at 0, then two sheets. Nested level surfaces fill space the way nested level curves fill the plane.
Common mistakes
Mistake Result Fix Squaring 9 − x 2 − y 2 = − 2 \sqrt{9-x^2-y^2}=-2 9 − x 2 − y 2 = − 2 a "level curve" x 2 + y 2 = 5 x^2+y^2=5 x 2 + y 2 = 5 a square root is never negative: levels need c ≥ 0 c\ge 0 c ≥ 0 p308 ) ≤ \le ≤ the sphere x 2 + y 2 + z 2 = 9 x^2+y^2+z^2=9 x 2 + y 2 + z 2 = 9 the bottom cannot be 0: strict < < < x 2 ≠ y 2 x^2\ne y^2 x 2 = y 2 y ≠ x y\ne x y = x the line y = − x y=-x y = − x y ≠ ± x y\ne\pm x y = ± x Range read off the domain g = 9 − x 2 − y 2 g=\sqrt{9-x^2-y^2} g = 9 − x 2 − y 2 [ − 3 , 3 ] [-3,3] [ − 3 , 3 ] range is outputs: heights 0 to 3 4 ( x 2 − 2 x + 1 ) 4(x^2-2x+1) 4 ( x 2 − 2 x + 1 ) + 1 +1 + 1 4 ( x − 1 ) 2 + ( y + 2 ) 2 = 13 4(x-1)^2+(y+2)^2=13 4 ( x − 1 ) 2 + ( y + 2 ) 2 = 13 add 4 ⋅ 1 4\cdot 1 4 ⋅ 1 A trace drawn in the x y xy x y the curve z = f ( a , y ) z=f(a,y) z = f ( a , y ) it lives in the plane x = a x=a x = a y y y z z z Level surface named without the sign of c c c 4 x 2 + 9 y 2 − z 2 = − 1 4x^2+9y^2-z^2=-1 4 x 2 + 9 y 2 − z 2 = − 1 c < 0 c<0 c < 0 c = 0 c=0 c = 0
Practice and answers
Exercises from p315 ; the book's key: p861 , p864 –866 .
Checkpoints 4.1–4.5
# task answer 4.1 domain and range of 36 − 9 x 2 − 9 y 2 \sqrt{36-9x^2-9y^2} 36 − 9 x 2 − 9 y 2 x 2 + y 2 ≤ 4 x^2+y^2\le 4 x 2 + y 2 ≤ 4 [ 0 , 6 ] [0,6] [ 0 , 6 ] 4.2 level curve of x 2 + y 2 − 6 x + 2 y x^2+y^2-6x+2y x 2 + y 2 − 6 x + 2 y c = 15 c=15 c = 15 ( x − 3 ) 2 + ( y + 1 ) 2 = 25 (x-3)^2+(y+1)^2=25 ( x − 3 ) 2 + ( y + 1 ) 2 = 25 ( 3 , − 1 ) (3,-1) ( 3 , − 1 ) 4.3 trace of − x 2 − y 2 + 2 x + 4 y − 1 -x^2-y^2+2x+4y-1 − x 2 − y 2 + 2 x + 4 y − 1 y = 3 y=3 y = 3 z = 3 − ( x − 1 ) 2 z=3-(x-1)^2 z = 3 − ( x − 1 ) 2 y = 3 y=3 y = 3 4.4 domain of ( 3 t − 6 ) y − 4 x 2 + 4 (3t-6)\sqrt{y-4x^2+4} ( 3 t − 6 ) y − 4 x 2 + 4 { ( x , y , t ) : y ≥ 4 x 2 − 4 } \{(x,y,t): y\ge 4x^2-4\} {( x , y , t ) : y ≥ 4 x 2 − 4 } 4.5 level surface of x 2 + y 2 + z 2 − 2 x + 4 y − 6 z x^2+y^2+z^2-2x+4y-6z x 2 + y 2 + z 2 − 2 x + 4 y − 6 z c = 2 c=2 c = 2 ( x − 1 ) 2 + ( y + 2 ) 2 + ( z − 3 ) 2 = 16 (x-1)^2+(y+2)^2+(z-3)^2=16 ( x − 1 ) 2 + ( y + 2 ) 2 + ( z − 3 ) 2 = 16 ( 1 , − 2 , 3 ) (1,-2,3) ( 1 , − 2 , 3 )
Exercises
# task answer key step 1 W = 4 x 2 + y 2 W=4x^2+y^2 W = 4 x 2 + y 2 W ( 2 , − 1 ) W(2,-1) W ( 2 , − 1 ) W ( − 3 , 6 ) W(-3,6) W ( − 3 , 6 ) 17, 72 16 + 1 16+1 16 + 1 36 + 36 36+36 36 + 36 3 V = π x 2 y V=\pi x^2y V = π x 2 y V ( 2 , 5 ) V(2,5) V ( 2 , 5 ) 20 π ≈ 62.83 20\pi\approx 62.83 20 π ≈ 62.83 a cylinder of radius 2, height 5 7 domain of 4 ln ( y 2 − x ) 4\ln(y^2-x) 4 ln ( y 2 − x ) x < y 2 x<y^2 x < y 2 Lab 1 9 domain of y 2 − x 2 y^2-x^2 y 2 − x 2 all of R 2 \mathbb R^2 R 2 a polynomial 11 range of 16 − 4 x 2 − y 2 \sqrt{16-4x^2-y^2} 16 − 4 x 2 − y 2 [ 0 , 4 ] [0,4] [ 0 , 4 ] largest at ( 0 , 0 ) (0,0) ( 0 , 0 ) 13 range of y 2 − x 2 y^2-x^2 y 2 − x 2 R \mathbb R R y 2 y^2 y 2 − x 2 -x^2 − x 2 15 y 2 − x 2 = 4 y^2-x^2=4 y 2 − x 2 = 4 a hyperbola the saddle at c = 4 c=4 c = 4 17 4 − x − y 4-x-y 4 − x − y c = 0 , 4 c=0,4 c = 0 , 4 x + y = 4 x+y=4 x + y = 4 x + y = 0 x+y=0 x + y = 0 lines 19 2 x − y 2x-y 2 x − y c = 0 , − 2 , 2 c=0,-2,2 c = 0 , − 2 , 2 2 x − y = 0 , − 2 , 2 2x-y=0,\ -2,\ 2 2 x − y = 0 , − 2 , 2 three parallel lines 23 e x y e^{xy} e x y c = 1 2 , 3 c=\frac12,3 c = 2 1 , 3 e x y = 1 2 e^{xy}=\frac12 e x y = 2 1 e x y = 3 e^{xy}=3 e x y = 3 hyperbolas x y = − ln 2 xy=-\ln 2 x y = − ln 2 x y = ln 3 xy=\ln 3 x y = ln 3 27 ln ( y / x 2 ) \ln(y/x^2) ln ( y / x 2 ) c = − 2 , 0 , 2 c=-2,0,2 c = − 2 , 0 , 2 y = e − 2 x 2 y=e^{-2}x^2 y = e − 2 x 2 y = x 2 y=x^2 y = x 2 y = e 2 x 2 y=e^2x^2 y = e 2 x 2 parabolas 29 ( y + 2 ) / x 2 (y+2)/x^2 ( y + 2 ) / x 2 c c c y = c x 2 − 2 y=cx^2-2 y = c x 2 − 2 parabolas through ( 0 , − 2 ) (0,-2) ( 0 , − 2 ) 31 3 x + y 3 3x+y^3 3 x + y 3 x = 1 x=1 x = 1 z = 3 + y 3 z=3+y^3 z = 3 + y 3 Lab 2 33 domain of 100 − 4 x 2 − 25 y 2 \sqrt{100-4x^2-25y^2} 100 − 4 x 2 − 25 y 2 x 2 25 + y 2 4 ≤ 1 \frac{x^2}{25}+\frac{y^2}{4}\le 1 25 x 2 + 4 y 2 ≤ 1 Lab 1 35 domain of 1 / 36 − 4 x 2 − 9 y 2 − z 2 1/\sqrt{36-4x^2-9y^2-z^2} 1/ 36 − 4 x 2 − 9 y 2 − z 2 x 2 9 + y 2 4 + z 2 36 < 1 \frac{x^2}{9}+\frac{y^2}{4}+\frac{z^2}{36}<1 9 x 2 + 4 y 2 + 36 z 2 < 1 root in the bottom: strict 37 domain of 16 − x 2 − y 2 − z 2 3 \sqrt[3]{16-x^2-y^2-z^2} 3 16 − x 2 − y 2 − z 2 all of space cube roots take negatives 47 contours of x 2 + y 2 − 2 x − 2 y x^2+y^2-2x-2y x 2 + y 2 − 2 x − 2 y circles ( x − 1 ) 2 + ( y − 1 ) 2 = c + 2 (x-1)^2+(y-1)^2=c+2 ( x − 1 ) 2 + ( y − 1 ) 2 = c + 2 49 x 2 + y 2 + z 2 = 9 x^2+y^2+z^2=9 x 2 + y 2 + z 2 = 9 a sphere of radius 3 Lab 3 51 x 2 + y 2 − z 2 = 4 x^2+y^2-z^2=4 x 2 + y 2 − z 2 = 4 a hyperboloid of one sheet c > 0 c>0 c > 0 53 1 − 4 x 2 − y 2 1-4x^2-y^2 1 − 4 x 2 − y 2 P ( 0 , 1 ) P(0,1) P ( 0 , 1 ) 4 x 2 + y 2 = 1 4x^2+y^2=1 4 x 2 + y 2 = 1 f ( 0 , 1 ) = 0 f(0,1)=0 f ( 0 , 1 ) = 0 55 e x y ( x 2 + y 2 ) e^{xy}(x^2+y^2) e x y ( x 2 + y 2 ) P ( 1 , 0 ) P(1,0) P ( 1 , 0 ) 1 = e x y ( x 2 + y 2 ) 1=e^{xy}(x^2+y^2) 1 = e x y ( x 2 + y 2 ) g ( 1 , 0 ) = 1 g(1,0)=1 g ( 1 , 0 ) = 1 57 T T T T = k x 2 + y 2 T=\dfrac{k}{x^2+y^2} T = x 2 + y 2 k — 59 level curves T = 40 T=40 T = 40 T = 100 T=100 T = 100 x 2 + y 2 = k 40 x^2+y^2=\frac{k}{40} x 2 + y 2 = 40 k x 2 + y 2 = k 100 x^2+y^2=\frac{k}{100} x 2 + y 2 = 100 k circles of radius 10 k 20 \frac{\sqrt{10k}}{20} 20 10 k k 10 \frac{\sqrt k}{10} 10 k
Level 3 · why it works What level sets tell you short arguments
Why the range of g is [0, 3]
9 − x 2 − y 2 = c ⟺ x 2 + y 2 = 9 − c 2 and c ≥ 0 \sqrt{9-x^2-y^2}=c\quad\Longleftrightarrow\quad x^2+y^2=9-c^2\ \text{ and }\ c\ge 0 9 − x 2 − y 2 = c ⟺ x 2 + y 2 = 9 − c 2 and c ≥ 0 A circle of radius 9 − c 2 \sqrt{9-c^2} 9 − c 2 0 ≤ c ≤ 3 0\le c\le 3 0 ≤ c ≤ 3 p303 ). Squaring alone would accept c = − 2 c=-2 c = − 2 x 2 + y 2 = 5 x^2+y^2=5 x 2 + y 2 = 5
c c c
How far apart level curves sit
A point has one height, so it lies on one level curve: level curves never cross. Their spacing shows steepness (p307 ).
Bowl x 2 + y 2 = c x^2+y^2=c x 2 + y 2 = c c = 1 , 2 , 3 , 4 c=1,2,3,4 c = 1 , 2 , 3 , 4 1 , 1.41 , 1.73 , 2 1,\ 1.41,\ 1.73,\ 2 1 , 1.41 , 1.73 , 2 0.41 , 0.32 , 0.27 0.41,\ 0.32,\ 0.27 0.41 , 0.32 , 0.27
Equal steps in height, shrinking steps on the ground: the bowl gets steeper as you climb.
One function, three kinds of surface
4 x 2 + 9 y 2 − z 2 = c 4x^2+9y^2-z^2=c 4 x 2 + 9 y 2 − z 2 = c c c c
c < 0 · two sheets z 2 = 4 x 2 + 9 y 2 − c ≥ − c > 0 z^2=4x^2+9y^2-c\ge -c>0 z 2 = 4 x 2 + 9 y 2 − c ≥ − c > 0 z = 0 z=0 z = 0
c = 0 · a cone z = ± 4 x 2 + 9 y 2 z=\pm\sqrt{4x^2+9y^2} z = ± 4 x 2 + 9 y 2
c > 0 · one sheet At z = 0 z=0 z = 0 4 x 2 + 9 y 2 = c 4x^2+9y^2=c 4 x 2 + 9 y 2 = c
Different levels of one function never meet, so the cone is the wall between the two families.
Next: limits in two variables
§4.2 asks what f ( x , y ) f(x,y) f ( x , y ) ( x , y ) (x,y) ( x , y ) The contour map shows when the paths disagree: level curves crowding into one point.
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Calculus Volume 3 §4.1 Chapter 4 · Differentiation of Functions of Several Variables
Functions of Several Variables
Two inputs, one output: a surface you can walk on, and the map of its heights.
Read z = f ( x , y ) z=f(x,y) z = f ( x , y )
4.1.1 Recognize a function of two variables and identify its domain and range. 4.1.2 Sketch a graph of a function of two variables. 4.1.3 Sketch several traces or level curves of a function of two variables. 4.1.4 Recognize a function of three or more variables and identify its level surfaces.
Ask: "A hiking map shows height at every point. Input? Output?"Prereq: domain of √(4 − x) · x² + y² = 9 is a circle · complete the square Levels: L1 = pictures and definitions · L2 adds examples and checkpoints · L3 adds 2 why-slides
Functions of Two Variables bridge definition
A function of two variables sends each point ( x , y ) (x,y) ( x , y ) z z z
z = f ( x , y ) z=f(x,y) z = f ( x , y ) D ⊆ R 2 D\subseteq\mathbb R^2 D ⊆ R 2 Domain D D D Range : every z z z ( x , y ) ∈ D (x,y)\in D ( x , y ) ∈ D f ( x , y ) = z f(x,y)=z f ( x , y ) = z
g ( x , y ) = 9 − x 2 − y 2 g(x,y)=\sqrt{9-x^2-y^2} g ( x , y ) = 9 − x 2 − y 2
( 1 , 2 ) ↦ 9 − 5 = 2 (1,2)\mapsto\sqrt{9-5}=2 ( 1 , 2 ) ↦ 9 − 5 = 2
( 0 , 0 ) ↦ 3 (0,0)\mapsto 3 ( 0 , 0 ) ↦ 3 ( 0 , − 3 ) ↦ 0 \ (0,-3)\mapsto 0 ( 0 , − 3 ) ↦ 0
Values that happen: the range [ 0 , 3 ] [0,3] [ 0 , 3 ]
Geometry a region of the plane, each point sent to a height
Algebra · p302 a formula
z = f ( x , y ) z=f(x,y) z = f ( x , y ) in two letters
Ask: "Can one point give two values?" → no: that's what "function" meansAsk: "Can two points give the same value?" → yes: whole circles doFigure 4.2
Functions of Two Variables bridge Example 4.1
The domain is every point where the formula makes sense
Look for what breaks: square root needs inside ≥ 0 \ge 0 ≥ 0 ≠ 0 \ne 0 = 0
(a) f = 3 x + 5 y + 2 f=3x+5y+2 f = 3 x + 5 y + 2 R 2 \mathbb R^2 R 2 R \mathbb R R
(b) g = 9 − x 2 − y 2 g=\sqrt{9-x^2-y^2} g = 9 − x 2 − y 2 9 − x 2 − y 2 ≥ 0 ⟺ x 2 + y 2 ≤ 9 9-x^2-y^2\ge 0\iff x^2+y^2\le 9 9 − x 2 − y 2 ≥ 0 ⟺ x 2 + y 2 ≤ 9
rim → 0 0 0 3 3 3 [ 0 , 3 ] [0,3] [ 0 , 3 ]
Geometry closed disk of radius 3, rim included
Algebra · p303 x 2 + y 2 ≤ 9 x^2+y^2\le 9 x 2 + y 2 ≤ 9
Ask: "(a) Which z is impossible?" → none: set y = 0, x = (z − 2)/3Ask: "Is the rim in?" → yes, ≤: √0 = 0
Functions of Two Variables algebra Checkpoint 4.1
Factor out the 9: 36 − 9 x 2 − 9 y 2 \sqrt{36-9x^2-9y^2} 36 − 9 x 2 − 9 y 2
Inside ≥ 0 \ge 0 ≥ 0
Checkpoint 4.1 Domain and range of
f ( x , y ) = 36 − 9 x 2 − 9 y 2 f(x,y)=\sqrt{36-9x^2-9y^2} f ( x , y ) = 36 − 9 x 2 − 9 y 2
36 − 9 x 2 − 9 y 2 = 9 ( 4 − x 2 − y 2 ) ≥ 0 36-9x^2-9y^2=9(4-x^2-y^2)\ge 0 36 − 9 x 2 − 9 y 2 = 9 ( 4 − x 2 − y 2 ) ≥ 0
x 2 + y 2 ≤ 4 x^2+y^2\le 4 x 2 + y 2 ≤ 4 [ 0 , 6 ] [0,6] [ 0 , 6 ]
Pairs, 2 min Slip: range [0, 36] — forgot the root: √36 = 6
Graphing Functions of Two Variables geometry Example 4.2a
The graph of z = f ( x , y ) z=f(x,y) z = f ( x , y ) z z z
Surface : all points ( x , y , f ( x , y ) ) (x,y,f(x,y)) ( x , y , f ( x , y )) z > 0 z>0 z > 0 x y xy x y z < 0 z<0 z < 0
g ( x , y ) = 9 − x 2 − y 2 g(x,y)=\sqrt{9-x^2-y^2} g ( x , y ) = 9 − x 2 − y 2
radius 3 → height 0 · radius 2 2 2\sqrt2 2 2 5 \sqrt5 5
All the lifted circles: a hemisphere of radius 3.
Geometry upper half of the sphere of radius 3
Algebra · p304 z = 9 − x 2 − y 2 ⟺ z=\sqrt{9-x^2-y^2}\iff z = 9 − x 2 − y 2 ⟺ x 2 + y 2 + z 2 = 9 x^2+y^2+z^2=9 x 2 + y 2 + z 2 = 9 ,
z ≥ 0 z\ge 0 z ≥ 0
Ask: "Height over the origin?" → 3g sees only x² + y²: circles about 0 share a height Drag the figure
Graphing Functions of Two Variables bridge Example 4.2b
Fix one variable and a known curve appears: x 2 + y 2 x^2+y^2 x 2 + y 2
f ( x , y ) = x 2 + y 2 f(x,y)=x^2+y^2 f ( x , y ) = x 2 + y 2 y = 0 y=0 y = 0 z = x 2 z=x^2 z = x 2 x = 0 x=0 x = 0 z = y 2 z=y^2 z = y 2 c c c
Heights at half scale.
Cut by y = 0 y=0 y = 0 z = x 2 z=x^2 z = x 2
Cut by x = 0 x=0 x = 0 z = y 2 z=y^2 z = y 2
Spin the parabola about the z z z paraboloid .
Geometry bowl, lowest point at the origin
Algebra · p305 x 2 + y 2 ≥ 0 x^2+y^2\ge 0 x 2 + y 2 ≥ 0 ,
= 0 =0 = 0 only at
( 0 , 0 ) (0,0) ( 0 , 0 )
Ask: "Range?" → [0, ∞)These cuts return as vertical traces (p310)
Graphing Functions of Two Variables geometry Example 4.3
The profit function is a downward paraboloid, highest over ( 3 , 2 ) (3,2) ( 3 , 2 )
f ( x , y ) = 16 − ( x − 3 ) 2 − ( y − 2 ) 2 f(x,y)=16-(x-3)^2-(y-2)^2 f ( x , y ) = 16 − ( x − 3 ) 2 − ( y − 2 ) 2 ( x − 3 ) 2 + ( y − 2 ) 2 ≤ 16 (x-3)^2+(y-2)^2\le 16 ( x − 3 ) 2 + ( y − 2 ) 2 ≤ 16 f ( 3 , 2 ) = 16 f(3,2)=16 f ( 3 , 2 ) = 16
Heights at 0.4 scale.
x x x y y y z z z
Profit ≥ 0 \ge 0 ≥ 0 ( 3 , 2 ) (3,2) ( 3 , 2 )
Over it, the surface rises …
… to 16 at ( 3 , 2 ) (3,2) ( 3 , 2 )
Ask before clicking: "Best mix of nuts and bolts?"Book: domain = nonnegative profit, but prints range z ≤ 16; on that domain it is [0, 16]
Graphing Functions of Two Variables algebra Example 4.3
Every height below 16 is reached on a whole circle about ( 3 , 2 ) (3,2) ( 3 , 2 )
Set f = z f=z f = z ( x − 3 ) 2 + ( y − 2 ) 2 = 16 − z (x-3)^2+(y-2)^2=16-z ( x − 3 ) 2 + ( y − 2 ) 2 = 16 − z 16 − z \sqrt{16-z} 16 − z
z = 0 z=0 z = 0
z = 7 z=7 z = 7 z = 12 z=12 z = 12
z = 16 z=16 z = 16 ( 3 , 2 ) (3,2) ( 3 , 2 )
Geometry circles shrink as you climb
Algebra · p306 radius
16 − z \sqrt{16-z} 16 − z real only for
z ≤ 16 z\le 16 z ≤ 16
Ask: "z = 20?" → 16 − 20 < 0: no pointsPreview: these circles are level curves
Level Curves geometry Figure 4.7
A contour map joins points of equal height: bunched lines mean steep ground
Contour line = all points at one elevation. Equal height steps: close lines → steep, far apart → gentle.
A tower and a hill, seen from above.
Lines bunched: height changes fast — steep .
Lines far apart: gentle .
Contour lines never cross: a point has one height.
Show Devil's Tower map (Figure 4.7) in the panel Ask: "Easiest way up?" → where lines are far apartWeather maps: isobars, isotherms
Level Curves bridge definition
A level curve is a horizontal slice of the surface, dropped to the floor
Level curve for c c c ( x , y ) (x,y) ( x , y ) f ( x , y ) = c f(x,y)=c f ( x , y ) = c Contour map : several level curves drawn together.
Cut g = 9 − x 2 − y 2 g=\sqrt{9-x^2-y^2} g = 9 − x 2 − y 2 z = 2 z=2 z = 2
… drop the cut to the floor: x 2 + y 2 = 5 x^2+y^2=5 x 2 + y 2 = 5
c = 0 c=0 c = 0 1 1 1 2 2 2 3 3 3 3 , 2 2 , 5 , 0 3,\ 2\sqrt2,\ \sqrt5,\ 0 3 , 2 2 , 5 , 0
Geometry slice at height
c c c , seen from above
Algebra · p308 9 − x 2 − y 2 = c \sqrt{9-x^2-y^2}=c 9 − x 2 − y 2 = c ⟺ x 2 + y 2 = 9 − c 2 \iff x^2+y^2=9-c^2 ⟺ x 2 + y 2 = 9 − c 2
Ask: "c = 3?" → only the origin · "c = 4?" → nothing, 4 not in rangeFigure 4.8
Level Curves algebra why
Squaring can invent a level curve: c c c
9 − x 2 − y 2 = c ⟺ x 2 + y 2 = 9 − c 2 \sqrt{9-x^2-y^2}=c\iff x^2+y^2=9-c^2 9 − x 2 − y 2 = c ⟺ x 2 + y 2 = 9 − c 2 and c ≥ 0 c\ge 0 c ≥ 0 9 − x 2 − y 2 = c ⇒ 9 − x 2 − y 2 = c 2 ⇒ x 2 + y 2 = 9 − c 2 \sqrt{9-x^2-y^2}=c\ \Rightarrow\ 9-x^2-y^2=c^2\ \Rightarrow\ x^2+y^2=9-c^2 9 − x 2 − y 2 = c ⇒ 9 − x 2 − y 2 = c 2 ⇒ x 2 + y 2 = 9 − c 2 A real circle needs 9 − c 2 ≥ 0 9-c^2\ge 0 9 − c 2 ≥ 0 − 3 ≤ c ≤ 3 -3\le c\le 3 − 3 ≤ c ≤ 3
But c = − 2 c=-2 c = − 2 x 2 + y 2 = 5 x^2+y^2=5 x 2 + y 2 = 5 g = − 2 g=-2 g = − 2
Range [ 0 , 3 ] [0,3] [ 0 , 3 ] c c c
Book remark on p308: no extra solutions here because √ ≥ 0 Ask: "Which step lost information?" → squaring
Level Curves algebra Example 4.4
Complete the square: the level curves of 8 + 8 x − 4 y − 4 x 2 − y 2 \sqrt{8+8x-4y-4x^2-y^2} 8 + 8 x − 4 y − 4 x 2 − y 2 ( 1 , − 2 ) (1,-2) ( 1 , − 2 )
Set f = c f=c f = c 4 ( x − 1 ) 2 + ( y + 2 ) 2 = 16 − c 2 4(x-1)^2+(y+2)^2=16-c^2 4 ( x − 1 ) 2 + ( y + 2 ) 2 = 16 − c 2
4 ( x 2 − 2 x + 1 ) + ( y 2 + 4 y + 4 ) = 8 + 4 + 4 4(x^2-2x+1)+(y^2+4y+4)=8+4+4 4 ( x 2 − 2 x + 1 ) + ( y 2 + 4 y + 4 ) = 8 + 4 + 4
c = 0 c=0 c = 0 ( x − 1 ) 2 4 + ( y + 2 ) 2 16 = 1 \dfrac{(x-1)^2}{4}+\dfrac{(y+2)^2}{16}=1 4 ( x − 1 ) 2 + 16 ( y + 2 ) 2 = 1
c = 1 , 2 , 3 c=1,2,3 c = 1 , 2 , 3 c = 4 c=4 c = 4 ( 1 , − 2 ) (1,-2) ( 1 , − 2 )
Book prints range [ 0 , 4 ) [0,4) [ 0 , 4 ) f ( 1 , − 2 ) = 16 = 4 f(1,-2)=\sqrt{16}=4 f ( 1 , − 2 ) = 16 = 4 [ 0 , 4 ] [0,4] [ 0 , 4 ]
Slip: 4(x² − 2x + 1) balanced by +1 → add 4·1 Domain: the c = 0 ellipse and its inside Book erratum p310: range [0, 4) should be [0, 4]
Level Curves algebra Checkpoint 4.2
x 2 + y 2 − 6 x + 2 y = 15 x^2+y^2-6x+2y=15 x 2 + y 2 − 6 x + 2 y = 15 ( 3 , − 1 ) (3,-1) ( 3 , − 1 ) Level curve at c c c g = c g=c g = c x x x y y y
Checkpoint 4.2 Level curve at c = 15 c=15 c = 15
g ( x , y ) = x 2 + y 2 − 6 x + 2 y g(x,y)=x^2+y^2-6x+2y g ( x , y ) = x 2 + y 2 − 6 x + 2 y
( x 2 − 6 x + 9 ) + ( y 2 + 2 y + 1 ) = 15 + 9 + 1 (x^2-6x+9)+(y^2+2y+1)=15+9+1 ( x 2 − 6 x + 9 ) + ( y 2 + 2 y + 1 ) = 15 + 9 + 1
( x − 3 ) 2 + ( y + 1 ) 2 = 25 (x-3)^2+(y+1)^2=25 ( x − 3 ) 2 + ( y + 1 ) 2 = 25
Pairs, 2 min Slip: centre (−3, 1) — sign flips
Level Curves bridge definition · Example 4.5
A vertical trace is an upright slice: fix x x x y y y
Vertical trace : z = f ( a , y ) z=f(a,y) z = f ( a , y ) x = a x=a x = a z = f ( x , b ) z=f(x,b) z = f ( x , b ) y = b y=b y = b
Heights at 1.2 scale.
f ( x , y ) = sin x cos y f(x,y)=\sin x\cos y f ( x , y ) = sin x cos y
x = − π 4 x=-\tfrac\pi4 x = − 4 π z = − 2 2 cos y ≈ − 0.7071 cos y z=-\tfrac{\sqrt2}{2}\cos y\approx-0.7071\cos y z = − 2 2 cos y ≈ − 0.7071 cos y
x = 0 x=0 x = 0 z = 0 z=0 z = 0 x = π 4 x=\tfrac\pi4 x = 4 π z ≈ 0.7071 cos y z\approx 0.7071\cos y z ≈ 0.7071 cos y
Fix y y y z = sin x z=\sin x z = sin x y = 0 y=0 y = 0
Ask: "Trace at x = π/2?" → z = cos yLevel curves: in the xy-plane · traces: in xz- or yz-planes Tables 4.1, 4.2 · Figure 4.11 Same cut as the paraboloid slide, now named
Level Curves bridge Checkpoint 4.3
The trace of − x 2 − y 2 + 2 x + 4 y − 1 -x^2-y^2+2x+4y-1 − x 2 − y 2 + 2 x + 4 y − 1 y = 3 y=3 y = 3
Substitute y = b y=b y = b x x x z = 3 − ( x − 1 ) 2 z=3-(x-1)^2 z = 3 − ( x − 1 ) 2 y = 3 y=3 y = 3
Checkpoint 4.3 Vertical trace at y = 3 y=3 y = 3
g ( x , y ) = − x 2 − y 2 + 2 x + 4 y − 1 g(x,y)=-x^2-y^2+2x+4y-1 g ( x , y ) = − x 2 − y 2 + 2 x + 4 y − 1
g ( x , 3 ) = − x 2 − 9 + 2 x + 12 − 1 g(x,3)=-x^2-9+2x+12-1 g ( x , 3 ) = − x 2 − 9 + 2 x + 12 − 1 g ( x , 3 ) = − x 2 + 2 x + 2 \phantom{g(x,3)}=-x^2+2x+2 g ( x , 3 ) = − x 2 + 2 x + 2
z = 3 − ( x − 1 ) 2 z=3-(x-1)^2 z = 3 − ( x − 1 ) 2 x = 1 x=1 x = 1
Geometry the dome cut by the wall
y = 3 y=3 y = 3 Algebra · p311 substitute
y = 3 y=3 y = 3 into
g g g
Pairs, 2 min Slip: −y² at y = 3 is −9, not +9
Level Curves recap
One function, three views: the surface, its level curves, its traces
Horizontal cuts z = c z=c z = c x = a x=a x = a y = b y=b y = b
The profit hill again.
Horizontal cuts → level circles → the map.
Upright cut y = 2 y=2 y = 2 z = 16 − ( x − 3 ) 2 z=16-(x-3)^2 z = 16 − ( x − 3 ) 2
Sketch a surface from its map and its profiles.
Ask: "Which view shows steepness? which shows the maximum?"End of two-variable part
Functions of More Than Two Variables bridge
A function of three variables gives every point in space a number — too many dimensions for a graph
w = f ( x , y , z ) w=f(x,y,z) w = f ( x , y , z ) ( x , y , t ) (x,y,t) ( x , y , t ) f f f
f ( x , y , z ) = x 2 + y 2 + z 2 f(x,y,z)=x^2+y^2+z^2 f ( x , y , z ) = x 2 + y 2 + z 2
Every point carries a value: f ( 1 , 1 , 0 ) = 2 f(1,1,0)=2 f ( 1 , 1 , 0 ) = 2 f ( 1 , 2 , 2 ) = 9 f(1,2,2)=9 f ( 1 , 2 , 2 ) = 9
All points with f = 9 f=9 f = 9
Geometry a value at each point of space
Algebra · p312 f ( x , y , z ) f(x,y,z) f ( x , y , z ) , domain
⊆ R 3 \subseteq\mathbb R^3 ⊆ R 3
Ask: "Temperature in this room: how many inputs?" → 3 (or 4 with time)Book examples: a polynomial f(x, y, z), and g(x, y, t) with time
Functions of More Than Two Variables algebra Example 4.6
Same domain rules in space: the domain becomes a solid region
Root in the bottom: inside > 0 >0 > 0 x 2 ≠ y 2 x^2\ne y^2 x 2 = y 2 y ≠ x y\ne x y = x and y ≠ − x y\ne -x y = − x
(a) 3 x − 4 y + 2 z 9 − x 2 − y 2 − z 2 \dfrac{3x-4y+2z}{\sqrt{9-x^2-y^2-z^2}} 9 − x 2 − y 2 − z 2 3 x − 4 y + 2 z x 2 + y 2 + z 2 < 9 x^2+y^2+z^2<9 x 2 + y 2 + z 2 < 9
an open ball: the sphere itself is out
(b) 2 t − 4 x 2 − y 2 \dfrac{\sqrt{2t-4}}{x^2-y^2} x 2 − y 2 2 t − 4 t ≥ 2 t\ge 2 t ≥ 2 y ≠ ± x y\ne\pm x y = ± x
≤ 9 \le 9 ≤ 9
Ask: "Why < and not ≤ in (a)?"(b): two planes removed, everything below t = 2 removed
Functions of More Than Two Variables bridge Checkpoint 4.4
( 3 t − 6 ) y − 4 x 2 + 4 (3t-6)\sqrt{y-4x^2+4} ( 3 t − 6 ) y − 4 x 2 + 4 y ≥ 4 x 2 − 4 y\ge 4x^2-4 y ≥ 4 x 2 − 4 t t t A variable that appears in no restriction can take any value.
Checkpoint 4.4 Domain of
h ( x , y , t ) = ( 3 t − 6 ) y − 4 x 2 + 4 h(x,y,t)=(3t-6)\sqrt{y-4x^2+4} h ( x , y , t ) = ( 3 t − 6 ) y − 4 x 2 + 4
y − 4 x 2 + 4 ≥ 0 ⟺ y ≥ 4 x 2 − 4 y-4x^2+4\ge 0\iff y\ge 4x^2-4 y − 4 x 2 + 4 ≥ 0 ⟺ y ≥ 4 x 2 − 4
{ ( x , y , t ) ∈ R 3 : y ≥ 4 x 2 − 4 } \{(x,y,t)\in\mathbb R^3 : y\ge 4x^2-4\} {( x , y , t ) ∈ R 3 : y ≥ 4 x 2 − 4 }
Pairs, 2 min Slip: 3t − 6 ≥ 0 — it's outside the root, no restriction In (x, y, t)-space: this region extended in every t
Functions of More Than Two Variables bridge definition · Example 4.7
A level surface f ( x , y , z ) = c f(x,y,z)=c f ( x , y , z ) = c
Level surface for c c c ( x , y , z ) (x,y,z) ( x , y , z ) f ( x , y , z ) = c f(x,y,z)=c f ( x , y , z ) = c
f ( x , y , z ) = 4 x 2 + 9 y 2 − z 2 f(x,y,z)=4x^2+9y^2-z^2 f ( x , y , z ) = 4 x 2 + 9 y 2 − z 2
c = 1 c=1 c = 1
c = 2 , 3 c=2,\ 3 c = 2 , 3
Geometry nested hyperboloids, one per value
Algebra · p313 4 x 2 + 9 y 2 − z 2 = c 4x^2+9y^2-z^2=c 4 x 2 + 9 y 2 − z 2 = c
Ask: "Level curve : map :: level surface : ?" → onion layersFigure 4.13 shows c = 0, 1, 2, 3
Functions of More Than Two Variables algebra Checkpoint 4.5
Complete the square three times: the level surface is a sphere of radius 4 about ( 1 , − 2 , 3 ) (1,-2,3) ( 1 , − 2 , 3 )
Set g = c g=c g = c
Checkpoint 4.5 Level surface at c = 2 c=2 c = 2
g = x 2 + y 2 + z 2 − 2 x + 4 y − 6 z g=x^2+y^2+z^2-2x+4y-6z g = x 2 + y 2 + z 2 − 2 x + 4 y − 6 z
( x − 1 ) 2 + ( y + 2 ) 2 + ( z − 3 ) 2 (x-1)^2+(y+2)^2+(z-3)^2 ( x − 1 ) 2 + ( y + 2 ) 2 + ( z − 3 ) 2 = 2 + 1 + 4 + 9 = 16 \quad=2+1+4+9=16 = 2 + 1 + 4 + 9 = 16
sphere: centre ( 1 , − 2 , 3 ) (1,-2,3) ( 1 , − 2 , 3 )
Pairs, 2 min Slip: radius 16 — take the root
Functions of More Than Two Variables geometry why
Levels of one function never meet: as c c c
4 x 2 + 9 y 2 − z 2 = c 4x^2+9y^2-z^2=c 4 x 2 + 9 y 2 − z 2 = c c > 0 c>0 c > 0 c = 0 c=0 c = 0 c < 0 c<0 c < 0
Start at c = 1 c=1 c = 1
c = 0 c=0 c = 0
c = − 1 c=-1 c = − 1 z 2 = 4 x 2 + 9 y 2 + 1 ≥ 1 z^2=4x^2+9y^2+1\ge 1 z 2 = 4 x 2 + 9 y 2 + 1 ≥ 1 ∣ z ∣ < 1 |z|<1 ∣ z ∣ < 1
A point has one value of f f f
Ask: "Could two level surfaces of the same f touch?" → noSame argument: level curves never cross
§4.1 wrap-up
Read a function of several variables by its domain, its surface and its level sets
Domain = where the formula works · graph = surface · f = c f=c f = c
Objective You can now
4.1.1 domain and range9 − x 2 − y 2 \sqrt{9-x^2-y^2} 9 − x 2 − y 2 x 2 + y 2 ≤ 9 x^2+y^2\le 9 x 2 + y 2 ≤ 9 [ 0 , 3 ] [0,3] [ 0 , 3 ] 4.1.2 graphz = f ( x , y ) z=f(x,y) z = f ( x , y ) 4.1.3 traces, level curvesf ( x , y ) = c f(x,y)=c f ( x , y ) = c z = f ( a , y ) z=f(a,y) z = f ( a , y ) z = f ( x , b ) z=f(x,b) z = f ( x , b ) 4.1.4 three variablesdomain in space · level surface f ( x , y , z ) = c f(x,y,z)=c f ( x , y , z ) = c
Exit ticket: h(x, y) = √(16 − x² − y²): domain, range, level curve c = 2 (x² + y² ≤ 16 · [0, 4] · x² + y² = 12) Homework: 1, 3, 7, 9, 11, 13, 15, 17, 19, 23, 27, 29, 31, 33, 35, 37, 47, 49, 51, 53, 55, 57, 59