Math 211 * Final Exam * Dec 20, 2017 * Solution * Victor Matveev

(a) Since the line belongs to both planes, it is normal to both normal vectors, therefore its direction vector is:

1

1 2

2

4, 1, 24 1 2 3, 6, 9 3 1, 2, 3

1, 2, 11 2 1

i j kn

v n nn

A scalar factor does not change the vector’s direction, so for the sake of simplicity use v = 1, 2, 3

To find a point on this line, consider its intersection with any coordinate plane, for instance x = 0:

000 2

2 31

10

2, ,

2

y z yx

y z z

r

Thus, the line of intersection is: 0 2 1 20, 1, 1, 2, 3

2 3

t t t t

x t

y

z t

r r v

Check your answers!!

4 2 4 ( 1 2 ) 2( 2 3 ) 3 :)

2 2 1 2 2 3 0 :)

x y z t t t

x y z t t t

(b) The distance between a point and a line is illustrated below:

0 0 ˆsind P P r r v

where 0 2, 1, 1 1, 0, 1 1, 1, 2r r and

2, 2, 1 2, 2, 1ˆ

34 4 1v

0

1 1 25 9 16ˆ 1 13

2 5, 3, 43 3

50 5 2

332 2 1

d

i j k

r r v

Note: there is a typo in this problem, which however does not affect the results: the constant should be e – 3, not 2e – 3

2 3

3

21, ,1

2

21, ,1

( ) Denote , , ln 2 2 3 0 :

2 ln 2 2

6

2

66

/ 1

4

1

6

/ 1

66

x

ez

y

z e

F x y z x y x

e

z yz e

Fz x y z z

x F x exz y

Fz x y z z

e

ee

y F y ey exz

a

----------------------------------------------------------------------------------------------------------------------------------

0 10 2

20 2

( ) 0, 1 1, 0 cos cos 1, sin 0

cos sin 2 cos cos

cos sin 2 cos cos

2

u v x y xy y y

z z x w yy xy y u x xy x y v

u x u y u

z z x w yy xy y v x xy x y

v x v y v

b

0

0u

1 11, ln 2, 1, ln 2,2 2

11, ln 2,

2

(

3

) , , 2 2, 2, 1

2, 2, 1 1, 2, 21ˆ1, ln 2, 2, 2, 1

2 | | 1 4 4

4

y y

u

f e xe z

D f f

a

uu

u

Function increases most rapidly in the direction of the gradient: 3

2, 2, 1ˆ

| | 4 4

2 2 1, ,

3 31

f

f

u

----------------------------------------------------------------------------------------------------------------------------------

2 2 2

3,2,5

Denote , , 4 0

Normal to the tangent plane: 3, 2, 5 2 , 8 , 2 6, 16, 10 2 3, 8, 5

We can remove factor of 2 to simplify, since a scalar factor doesn't affect direction: 3, 8, 5

F x y z z x y

F x y z

(b)

n

n

0

0 0

Equation of tangent plane to the surface : 3 8 5 3, 8, 5 3, 2, 5 9 16 25 0

Equation of normal line to the su :

3 8 5 0

3 3

23 8

5 5

rface at P 3 , 2 8 , 5 5

x y z

t t

x y z

x t

y t

z

t

t

t t

n r n r

r r n

2

2 2

Find the critical points: 12 6 6 , 6 6 0

12 6 6 6 6 6 1 0 2 critic l0, 0

s1

a poin1

: ,

t

y x

f x x y y x

x x y x x x x

Second derivative test: check the signs of the determinant of the Hessian matrix and the curvature:

0,0 : local min

1, 1 sa

12(1 ) 6det 72 1 36 36 72 36 1 2

6 6

36 0, 36 0 (therefore 6 0, a

d

lso: curves upward)

: 36 0 dle

xx yy

xD H x x x

D f f

D

2 2 22 3 , 25

1 2 22 2

33

2

32

f x y z g x y z

yxxf g y

y x

z xzz

x

2 2 2 2 2 2 22 25 5 5 10 154 9 , ,

14 14 14 14 14

C1

ompare function values at these point

14 25

5 10 15 70 14 5, , 5 14 Maximum

14 14 14 14 14

5 0 15, , 5 14 Minimum

14 14 14

s

g x y z x x x xx x

f

f

f = const

f = const g = const

2 242 2 2 42 2

2 2 2 3

0 0 0 0 0 0 0

242 4 2

yr

x y dx dy r r dr d d r dr

Intersection of top and bottom surfaces defines the projection:

2 2bottom2 2 0 2 1 0 1topz r z r r r r r r

22 1 2 2 1 1

2 3 2

0 0 0 0 0

14 32

0

2

6

2 2

3 12 2

4

5

3 4 3

r

r

V dz rdr d d r r r dr r r r dr

r rr

2

2

r = 1

ztop = 2 – r2

zbottom = r

2 4 4 2

2 4

3

2 4 4 3 2 2 4 5

112 4 5 3 5 6

00

, , , ,, ,

1, 2 , 4

2

3

4 3 5 6

3 5 6

C

y z z x x y t t t t t tt t t t

d t t dt

d t t t t t t t t dt t t t dt

d t t t dt t t t

Fr

r

F r

F r

------------------------------------------------------------------------------------------------------------------------------------------------------

0, 0, cos cos cos cos Conservative!

sin cos cos sin

x y y x

x y z

y x y x z

i j k

F 0

2

2

s

sin

c

sii nc ,

,2

s n os

os in ,sin sin sin sin

,2

,

dx

dy

dz

fy x f

x

fy x f f x y z C

x y

y x x

y

x

z

g y z

f

zh x y

x

z

zw xf

, ,2 2 22 2 2, ,22 20,1,0

(0,1,0)

32

sin sin sin2 2 2

B B B B

AA A Adf

zd f d df f x y CF r r

2 24 , M N

x y x yF

1 1

0 0

2 21 1 1 1

0 0 0 0

34

3

R R

N Md dA dx dy

x y

x y x ydx dy dx dy

x y

F r F k

1 1

0 0

2 21 1 1 1

0 0 0 0

1 11 12 2

0 00 0

2

42 2

2 1 2

R R

x x

x x

M Nd dA dxdy

x y

x y x ydx dy x y dx dy

x y

x xy dy y dy y y

F n r F

1

1 x

y