1. Consider the function f(x, y) = 2xy2 � y3 + xpy � 4.

(a) [7 points] Write the linearization L(x, y) for f at the point (3, 4).

(b) [3 points] Approximate a value of y such that f(3.01, y) = 33.95.

fx ( x, y ) = 2y2 + of

fy ( x, g) = 4xy - 3y2 + ¥

f( 3,4) = 34

f×( 3,

4) = 34

fy ( 3,4) = ¥

L ( x, g) = f ( 3

,4) + fx ( 3

,4) ( x - 3) + fy ( 3. 4) ( y

- 4)

L(x,y)=34+34(×-3)t3(y-4)@y33

.

95=34+34 ( 3.01 - 3) t ¥ (y- 4)

- 0.39 = Fly - 4)

y=3.@

2. [3 points per part] Here are the level curves of the surface z = f(x, y).

x

y

1 2 3 4 5 6 7 80

1

2

3

4

5

z = 8

z = 8

z = 8

z = 8

z = 8

z = 8

z = 7

z = 7 z = 6

z = 6z = 6

z = 6z = 5 z = 5

z = 5 z = 5

z = 7

z = 7

z = 8

z = 8

z = 9

z = 9

(a) Name three points at which fy(x, y) = 0.

(b) Write the equation for the tangent plane to the surface z = f(x, y) at the point (5, 3, 6).

(c) Estimate the value ofZ 5

3

Z 4

2

f(x, y) dy dx. (Circle one answer.)

Less than 0 Between 0 and 10 Between 10 and 20

Between 20 and 30 Between 30 and 40 Greater than 40

. . .

. . . .:.

Any of the red.points

in the graph above.

This is a saddlepoint! f×(5. 3) =fy( 5,31=0,

sothe tangent plane Is just 2¥

#volume of a solid

base is a square of area 4.

height is between z=5 and z=7.

So volume is between 4.5 and 4.7.

3. [12 points] Consider the function f(x, y) = x3 + xy � y2.

Find the absolute maximum and minimum values of f(x, y) on the triangle below:

x

y

(2, 0)

(2, 4)Critical points

:# 3(2y)2+y=°f×( x. y )=3x2ty=O

fylx,y)=x - 2y=O→×=2y l2y2ty=0y(l2y+D=O

One critical point :( goj .

y=o°ry=##×=0 not in domain

Bottom edge :y=o

f( x. D= ×3, increasing ,

no local extreme.

Right edge :x=2 Also,

check all corners.

f( 2. y )=8+2y -

y2

of '

so , f(

0,7=02¥,Y=0_ check ( 2

.' )

f(

2,4=8Top . left edge :y=2× f( 2,1)=9D←ma×flx ,2D=¥×f's 2×2-4×2

f( 2,4)=0

3×2-4×-0 f( ¥,§)= -3¥ ← min

x (3×-4)=0×=0 ,

x=¥

So also check ( ¥§).

4. [7 points per part] Evaluate each integral.

(a)Z 1

0

Z 3

0

yp

1 + xy dy dx

(b)Z 3

0

Z 6

2x

ey2dy dx

- -

f3fjyFxydxay-f3Mududy-f33fuBMdy-2sMHyY2-DdydYIjYf.osCHyI.yD3yHsC3D-D-2sEt.gy@Yn6yjtI6-6ffqeWdxdysx-f6CxeYJfgMdy-f.6hateY3dy-fYte4du-ytleFf36dIYyg.t

.iq

5. [7 points] Oh wow, another integral! Nice!

EvaluateZ p

2

0

Z p4�x2

x

sin(x2 + y2) dy dx.

6. [8 points] Let D be the region pictured below, bounded by the two circles x2 + y2 = 4,(x� 1)2 + y2 = 1, and the x-axis.

A lamina in the shape of D has density function ⇢(x, y) = y.

Compute the mass of the lamina.

x

y

" *IIIt¥÷.to?jItdotf?ftssinasaa.=fH#asHDlYdo=Fqftacosc4stfdo=Cfasl4)+tdoIf=fsasl

4) + 's)( ¥ - F) = ylang

^ ^

> >

in T2

= §£ | isinodrdo + f) ↳ issinodrdo

2 Cose

=fo÷t( is

.io#g.o+fgttCr3sinoDo2do=t(l?C8-8aos3Dsinodotfgt8sinodo)uaIIEinodo=s(sffu3Hdu+f8asoDg"

)= 's fruit . up,

:sfiD=F(F+ ' ) #