Math B6C – Chapter 14 Quiz – Fall 2008

* Solutions* __________________________________________________________________________________ Problems #1-3 refer to the function

( )4 6

4 6

4 9,2 3

x yf x yx y−

=+

__________________________________________________________________________________ 1. The limit of f as x approaches 0 along the x-axis

=( ) ( )

4 6 4 4

4 6 4 4,0 0,0 0 0 0

4 9 4 9 0 4lim lim lim lim 2 22 3 2 3 0 2x x x x

x y x xx y x x→ → → →

− − ⋅= = = =

+ + ⋅

__________________________________________________________________________________ 2. The limit of f as y approaches 0 along the y-axis.

( ) ( )

4 6 6 6

4 6 6 60, 0,0 0 0 0

4 9 4 0 9 9 9lim lim lim lim 32 3 2 0 3 3 3y y y y

x y y yx y y y→ → → →

− ⋅ − − −= = = =

+ ⋅ += −

__________________________________________________________________________________

3. ( )

( )( )

( ), 0,0lim , ?

x yf x y

→=

( )

(4 by probl6

4 6,0 0

e

0

m

,

#14 9lim 22 3x

x yx y→

−=

+

) , but

( ) ( )

(4 by probl6

4 60, 0

e

0

m 2

,

#4 9lim 32 3y

x yx y→

−= −

+

).

Since these two limits aren’t equal, it follows that ( ) ( )

4 6

4 6, 0,0

4 9lim does not exist2 3x y

x yx y→

−+

__________________________________________________________________________________

4. If , then ( ) ( ) ( )( )2 2, ln cos sing x y x y y xπ= + π ( )2,1 ?xg =

( ) ( ) ( )( )( ) ( )( )

( ) ( )( ) ( )( ) ( )

2 22

2 22 2 2 2

cos sin 2 cos cos, ln cos sin

cos sin cos sinx

x y y x x y y xxg x y x y y xx x y y x x y y x

π π π π ππ π

π π π π

∂+ +∂ ∂= + = =

∂ + +

Hence,

( ) ( ) ( )( ) ( )

( )( )

( )( )

2

2 2

2 2cos 1 1 cos 2 4 1 1 4 42,12 cos 1 1 sin

112 4 1 4 0 4 4xg

π π π π π ππ π

⋅ ⋅ + ⋅ ⋅ ⋅ − + ⋅ − + −= = =

⋅ + ⋅ ⋅−−− + ⋅ −

⋅ =

__________________________________________________________________________________

5. If ( ) (, sin )xyh x y eπ= , then ( )1, ln 2 ?xyh =

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )( ) ( )( )

( ) ( ) ( )

, , , sin cos

cos cos cos

cos cos cos

xy xy xyxy x

xy xy xy xy xy xy

xy xy xy xy xy xy

h x y h x y h x y e e ey y x y x y x

e e xy e e y ye ey x y y

y e e y e e ye ey y y

π π

π π π π π π

π π π π

∂ ∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞ ⎛= = = =⎜ ⎟ ⎜ ⎟ ⎜∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝∂ ∂ ∂ ∂⎛ ⎞= ⋅ ⋅ = ⋅ ⋅ =⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂= + +⎜ ⎟ ⎜ ⎟ ⎜∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝

( ) ( ) ( ) ( )( )( )( ) ( ) ( ) ( )( )

1 cos cos sin

, cos cos sin

xy xy xy xy xy xy xy

xy xy xy xy xyxy

e e y xe e ye e e x

h x y e e xy e xye e

π π π π π

π π π π π

⎛ ⎞⎜ ⎟⎟

⎠⎝ ⎠

= ⋅ + + − ⋅ ⋅

= + −

π ⎞⎟⎠

Hence,

( ) ( ) ( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( ) ( )( ) ( )

ln 2 ln 2 ln 2 ln 2 ln 21, ln 2 cos ln 2 cos ln 2 sin

2 cos 2 ln 2 cos 2 ln 2 2sin 2 2 1 ln 2

xyh e e e e eπ π π π π

π π π π π π

= + −

= ⋅ ⋅ + ⋅ − ⋅ ⋅ = +, so

( ) ( ), sin xyh x y eπ=

__________________________________________________________________________________ 6. Find the linearization of ( ) 4 3 2, 3 2f x y x y x y= − − at ( )2,1− .

The linearization we seek is given by: ( ) ( ) ( )2

2,1 ' 2,11

,x

f fy

L x y⎛ − ⎞⎡ ⎤ ⎡ ⎤

= − + − −⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎝ ⎠

So now we collect all the pieces of this puzzle…

( )( ) ( ) ( )

( ) ( )

( ) ( )

4 3 2

4 3 2

4 3 2 4 3 2

3 3 4 2

3 3

, 3 2

2,1 2 1 3 2 2 1 16 6 2 20

'( , ) 3 2 3 2

'( , ) 4 3 3 4

' 2,1 4 2 1 3 3

x y

f x y x y x y

f

f x y x y x y x y x yx y

f x y x y x y y

f

f f

= ⋅ − ⋅ − ⋅

− = − ⋅ − ⋅ − − ⋅ = + − =

⎡ ⎤∂ ∂= − − − −⎢ ⎥∂ ∂⎣ ⎦

⎡ ⎤=

⎡ ⎤= ⎣

− −⎣ ⎦

− = ⋅ −

⎦

− ⋅ ⋅( ) [ ]4 22 1 4 1 35 44⎡ ⎤− ⋅ − ⋅ = −⎣ ⎦

Putting the pieces together, we get:

( ) ( ) ( ) [ ]

( ) ( ) ( )

2 22,1 ' 2,1 20 35 44

1 1

20 35 2 44 1 20 35 70 44 44

,

,

x xf f

y y

x y x y

L x y

L x y

⎛ − ⎞ +⎡ ⎤ ⎡ ⎤ ⎡ ⎤= − + − − = + −⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝ ⎠= − + + − = − − + −

Our desired linearization is thus: ( ) 35 44 94, x yL x y = − + −

A graph of ( ) 4 3 2, 3 2f x y x y x y= − − along with its linearization at ( )2,1− , : ( ) 35 44 94, x yL x y = − + −

__________________________________________________________________________________ 7. Find the linearization of ( ) ( )2 cosh, , yf x y z = xz at the point ( )1, 1, ln 2 . ( ) ?, ,L x y z =

The linearization we seek is given by: ( ) ( )1

( , , ) 1,1, ln 2 ' 1,1, ln 2 1ln 2

xL x y z f f y

z

⎛ ⎞⎡ ⎤ ⎡ ⎤⎜ ⎟⎢ ⎥ ⎢ ⎥= + −⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎝ ⎠

First we collect the pieces of this puzzle…

( ) ( ) ( )

( )

1lnln 2 ln 2 22

1lnln 2 ln 2 2

122 521,1, ln 2 1 cosh 1 ln 2 cosh ln 2

2 2 2

11 222 322 Similar sinh l

2

2

2n 2

2 2 2 2 2,

4ly

e e ef

e e e

−

−

⎛ ⎞+ ⋅⎜ ⎟+ + ⎝ ⎠= ⋅ ⋅ = = = = =⋅

⎛ ⎞− ⋅− ⎜ ⎟− − ⎝ ⎠= = = =⋅

4

=

( )( ) ( )( ) ( )( )

( ) ( ) ( ) ( )

( )

2 2 2

2 2

2

'( , , )

cosh cosh cosh

' , , sinh 2 cosh sinh

sinh

x y zf x y z f f f

y xz y xz y xzx y z

f x y z y xz z y xz y xz x

y z xz

⎡ ⎤= ⎣ ⎦

⎡ ⎤∂ ∂ ∂= ⎢ ⎥∂ ∂ ∂⎣ ⎦

⎡ ⎤= ⋅⎣ ⎦

= ( ) ( )

( ) ( ) ( ) ( )

2

2 2

2 cosh sinh

' 1,1, ln 2 1 ln 2 sinh 1 ln 2 2 1 cosh 1 ln 2 1 1 sinh 1

y xz xy xz

⋅

f z

⎡ ⎤⎣ ⎦

⎡ ⎤= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⎣ ⎦

( ) ( ) ( ) ( ) ( ) ( )

( )

' 1,1, ln 2 ln 2 2 ln 5cosh ln 24

3 3sinh ln 2 2

3ln 2 5 3' 1,1, ln 2

2 sinh ln 24

4

4

4 2

f

f

⎡ ⎤= =⎡ ⎤⎣ ⎦ ⋅ ⋅⎢ ⎥⎣ ⎦

⎡ ⎤= ⎢ ⎥⎣ ⎦

Putting these pieces together, we get:

( ) ( )1

( , , ) 1,1, ln 2 ' 1,1, ln 2 1ln 2

x xL x y z f f y y

z z

⎛ − ⎞⎡ ⎤ ⎡ ⎤⎜ ⎟⎢ ⎥ ⎢ ⎥= + − −⎜ ⎟⎢ ⎥ ⎢ ⎥⎜ ⎟⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎝ ⎠

( ) ( ) ( )1

5 3ln 2 5 3 5 3ln 2 5 3( , , ) 1 1 1 ln 24 4 2 4 4 4 2 4

ln 23ln 2 5 3 5 3ln 2 5 3ln 2( , , )

4 2 4 4 4 2 43ln 2 5 3 5 6ln 2( , , )

4 2 4 4

xL x y z y x y z

z

L x y z x y z

L x y z x y z

−⎡ ⎤⎡ ⎤ ⎢ ⎥= + − = + ⋅ − + ⋅ − + ⋅ −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥−⎣ ⎦

= ⋅ + ⋅ + ⋅ + − − −

+⋅ + ⋅ + ⋅ −=

The level surface, ( )2 5cosh4

y xz =

with tangent plane

5( , , )4

L x y z = ; i.e., the plane

( )3ln 2 10 3 10 6ln 2x y z+ + = + :

__________________________________________________________________________________

8. Suppose that and that ( , , )

( , , )( , , )

u x y zg x y z

v x y z⎡

= ⎢⎣ ⎦

⎤⎥

( , )( , )

( , )m u v

h u vn u v⎡ ⎤

= ⎢ ⎥⎣ ⎦

. Let f h g= .

'

xm m u u um mmy z xu vf

n n n n n v v vx y z u

xyz

v x y z

y z∂ ∂ ∂ ∂ ∂⎡ ⎤ ⎡∂ ∂⎡ ⎤ ⎤

⎢ ⎥ ⎢⎢ ⎥ ⎥∂ ∂ ∂∂ ∂ ∂ ∂⎢ ⎥ ⎢= = ⎢ ⎥ ⎥∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎢ ⎥ ⎢⎢ ⎥ ⎥⎢ ⎥ ⎢⎢ ⎥∂ ∂ ∂ ∂ ∂ ∂ ∂

⎡ ⎤⎢ ⎥⎢

∂⎣ ⎦⎣ ⎦

⎥⎢ ⎥⎣ ⎦

⎥⎣ ⎦

3 2

2

'

f

g h

mn

uxg

⎡ ⎤⎢ ⎥⎣ ⎦

⎯⎯

∂∂

=

→

'

u u m muv

y z u vhv v v n nx y z u v

⎡ ⎤⎢ ⎥⎣

∂ ∂⎡ ⎤ ∂ ∂⎡ ⎤⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂⎢ ⎥ = ⎢ ⎥∂ ∂ ∂ ∂ ∂⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂ ∂⎣ ⎦⎦

⎦⎣

Suppose further that and that ( )8

2, 2, 73

g ⎡ ⎤− − = ⎢ ⎥

⎣ ⎦

( )( )

( )( )

8,3 8,3 2 18,3 8,3 5 3

u v

u v

m mn n⎡ ⎤ −⎡ ⎤

=⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

and that

(2, 2, 7)(2, 2, 7) (2, 2, 7) 5 2 4(2, 2, 7)(2, 2, 7) (2, 2, 7) 1 1 2

yx z

yx z

uu uvv v

− −− − − − −⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥− −− − − − −⎣ ⎦⎣ ⎦

. Then ( )' 2, 2, 7 ?f − − =

Computing the given derivative matrices, we get:

( , , )( , , ) ( , , )'( , , )

( , , )( , , ) ( , , )yx z

yx z

u x y zu x y z u x y zg x y z

v x y zv x y z v x y z⎡ ⎤

= ⎢ ⎥⎣ ⎦

and ( , ) ( , )

'( , )( , ) ( , )

u v

u v

m u v m u vh u v

n u v n u v⎡ ⎤

= ⎢ ⎥⎣ ⎦

Applying the chain rule, we obtain:

( ) ( ) ( )

( ) ( )( )

( )( )

' , , ' ( , , ) ' , ,

( , , )( , , )( , , ) ( , , ) ( , , )' , ,

( , , )( , , )( , , ) ( , , ) ( , , )yxu v z

yxu v z

f x y z h g x y z g x y z

u x y zu x y zm g x y z m g x y z u x y zf x y z

v x y zv x y zn g x y z n g x y z v x y z

=

⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦

Thus,

( )( )( )( )( )

( )( )( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )

2, 2, 7 2, 2, 7 2, 2, 72, 2, 7 2, 2, 7' 2, 2, 7

2, 2, 72, 2, 7 2, 2, 72, 2, 7 2, 2, 7

8,3 8,3 5 2 48,3 8,3 1 1 2

2 1 5 2 45 3 1 1 2

9' 2, 2, 7

28

u v yx z

yx zu v

u v

u v

m g m g uu uf

vv vn g n g

m mn n

f

⎡ ⎤− − − − − −⎡ ⎤− − − −− − = ⎢ ⎥ ⎢ ⎥− −− − − −− − − −⎢ ⎥ ⎣ ⎦⎣ ⎦

⎡ ⎤ −⎡ ⎤= ⎢ ⎥ ⎢ ⎥−⎣ ⎦⎣ ⎦

− −⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

− − =

3 613 26−⎡ ⎤

⎢ ⎥−⎣ ⎦

__________________________________________________________________________________ 9. Same functions and situation as described in problem #8.

Suppose that ( )( , , )

, ,( , , )x y z

f x y zx y z

ϕψ⎡

= ⎢⎣ ⎦

⎤⎥ . Then at the point ( )2, 2, 7− − , ?

z x x zϕ ψ ϕ ψ∂ ∂ ∂ ∂

− =∂ ∂ ∂ ∂

By definition, ( )' , , x y z

x y z

f x y zϕ ϕ ϕψ ψ ψ⎡ ⎤

= ⎢⎣ ⎦

⎥

)

. But in problem #8 we found that:

. ( )9 3 6

' 2, 2, 728 13 26

f−⎡ ⎤

− − = ⎢ ⎥−⎣ ⎦

Thus evaluated at the point ( , we have: 2, 2, 7− −9 3 6

28 13 26x y z

x y z

ϕ ϕ ϕψ ψ ψ

−⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥−⎣ ⎦⎣ ⎦

.

Hence, 6 28 9 26 66z x x zz x x zϕ ψ ϕ ψ

ϕ ψ ϕ ψ∂ ∂ ∂ ∂

− = − = ⋅ − ⋅ =∂ ∂ ∂ ∂

− .

10. Let ( ), ,yz xz xy u

g x y zxyz v

− +⎡ ⎤= =⎢ ⎥⎣ ⎦

⎡ ⎤⎢ ⎥⎣ ⎦

and ( ) 4 2 2, 4f u v u u v v= − + , and let . h f g=

Then ( )' 1, 2,3 ?h = By the chain rule, ( ) ( )( ) ( )' , , ' , , ' , ,h x y z f g x y z g x y z= . So we need to do some differentiation:

( ) ( ) ( )4 2 2 4 4 2 2 4

3 2 2 3

' ,

4 2 2 4

f u v u u v v u u v vu v

u uv u v v

∂ ∂⎡ ⎤= − + − +⎢ ⎥∂ ∂⎣ ⎦⎡ ⎤= − − +⎣ ⎦

But ( ), ,yz zx xy u

g x y zxyz v

− +⎡ ⎤= =⎢ ⎥⎣ ⎦

⎡⎢⎣

⎤⎥⎦

( )2 3 3 1 1 2 5

1,2,31 2 3 6

ug

v⋅ − ⋅ + ⋅

, so ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⋅ ⋅⎣ ⎦ ⎣ ⎦ ⎣ ⎦

, hence:

( )( ) ( ) [ ]3 2 2 3' 1, 2,3 ' 5,6 4 5 2 5 6 2 5 6 4 6 140 564f g f ⎡ ⎤= = ⋅ − ⋅ ⋅ − ⋅ ⋅ + ⋅ =⎣ ⎦

Also,

( )( ) ( ) ( )

( ) ( ) ( )

( )

' , ,

2 3 1 3 2 1 1 4 1' 1,2,3

2 3 1 3 1 2 6 3 2

yz zx xy yz zx xy yz zx xyy z x z y xx y z

g x y zyz xz xyxyz xyz xyz

x y z

g

∂ ∂ ∂⎡ ⎤− + − + − +⎢ ⎥ − + −∂ ∂ ∂ ⎡ ⎤⎢ ⎥= = ⎢ ⎥∂ ∂ ∂⎢ ⎥ ⎣ ⎦⎢ ⎥∂ ∂ ∂⎣ ⎦

− + − −⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥⋅ ⋅ ⋅⎣ ⎦ ⎣ ⎦

Putting all this together, we get:

( ) ( )( ) ( ) [ ]

( )

( ) [ ]

1 4 1' 1,2,3 ' 1, 2,3 ' 1, 2,3 140 564

6 3 2

140 1 564 6 140 4 564 3 140 1 564 2

' 1,2,3 3244 2252 1268

h f g g

h

−⎡ ⎤= = ⎢ ⎥

⎣ ⎦

= ⋅ − + ⋅ ⋅ + ⋅ ⋅ + ⋅⎡ ⎤⎣ ⎦

=

__________________________________________________________________________________ 11. Find the gradient of

( )4 2

2 4 sin4

, x yx yg x y π⎛ ⎞⎜ ⎟⎝ ⎠

=

at the point ( . )1,2 What is the y-component of this gradient ?

( )

4 2 4 2 4 22 4 2 4 2 4

4 2 4 24 22 4 2 42 4

sin sin sin4 4 4

,sin sinsin

4 44

x

y

x y x y x yx y x y x yx x xg

g x yg x y x yx y x y x yx y

y yy

π π π

π ππ

⎡ ⎤⎛ ⎞⎛ ⎞ ⎡∂ ⎛ ⎞ ⎛ ⎞∂ ∂⎛ ⎞ +⎢ ⎥⎜ ⎟⎜ ⎟⎤

⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂⎡ ⎤ ⎝ ⎠⎝ ⎠⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢∇ = = =⎢ ⎥ ⎢ ⎥ ⎢⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂⎛ ⎞∂⎣ ⎦ ⎢ ⎥ ⎢ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎢ ⎥ ∂ ∂⎢∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎣⎝ ⎠⎣ ⎦ ⎦4 2 4 2 3 2

4 2 4

4 2 4 2 42 3 2 4

4 2 4 24 5 6

4 2 6 5 4 22 3

42 sin cos4 4 4

24 sin cos4 4 4

2 sin cos4 4

4 sin cos4 2 4

x y x y x yxy x y

x y x y x yx y x y

x y x yxy x y

x y x y x yx y

π π π

π π π

π ππ

π π π

⎥⎥⎥⎥

⎡ ⎤⎛ ⎞ ⎛ ⎞+ ⋅⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥= ⎢ ⎥⎛ ⎞ ⎛ ⎞⎢ ⎥+ ⋅⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦⎡ ⎛ ⎞ ⎛ ⎞

+⎢ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢=⎛ ⎞ ⎛ ⎞

+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣

( )

( ) ( )( )

4 2 4 24 5 6

4 2 6 5 4 22 3

1 2 1 22 1 2 sin 1 2 cos4 4

1, 21 2 1 2 1 24 1 2 sin cos4 2 4

0 64 132sin 64 cos1, 2 64 ,

0 16 132sin 16 cos

g

g

π ππ

π π π

ππ π ππ

ππ π π

⎤⎥⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎦⎡ ⎤⎛ ⎞ ⎛ ⎞⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ + ⋅ ⋅ ⋅⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥∇ = ⎢ ⎥⎛ ⎞ ⎛ ⎞⋅ ⋅ ⋅ ⋅ ⋅ ⋅⎢ ⎥⋅ ⋅ ⋅ +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

+ ⋅ −⎡ ⎤+⎡ ⎤∇ = = = −⎢ ⎥⎢ ⎥ + ⋅ −+⎣ ⎦ ⎣ ⎦

416 16

1π π

⎡ ⎤− = − ⎢ ⎥

⎣ ⎦

The y-component of this gradient is thus 16π− . __________________________________________________________________________________ 12. If ( ) 5 7 93, , y zf x y z x −= , then ? The x-component of this gradient is ___ . ( )1,1,1f∇ =

( ) ( ) ( ) ( )

( )

( )

5 7 9 5 7 9 5 7 9

4 7 5 6 8

, , 3 , 3 , 3

, , , 7 , 27

1,1,1 5, 7, 27

5

f x y z y z y z y zx y z

f x y z y y z

f

x x x

x x

∂ ∂ ∂∇ = − − −

∂ ∂ ∂

∇ = −

∇ = −

The x-component of this gradient is thus 5 .

The graph depicts the gradient field,

( ) 4 7 5 6 8, , , 7 , 275f x y z y y zx x∇ = −

which is always perpendicular to f ’s level surfaces, one of which is also shown:

( ) 5 7 93 2, , y zf x y z x − = −=

13. Find , for the function f featured in problem #12, with (ˆ 1,1,1uD f )1

1ˆ 13 1

u⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

.

( ) ( )

( )

( )

ˆ

ˆ

5 1 51 17 1 73 327 1 27 1

5 7 27

5 3

ˆ1,1,1 1,1,1

1 15 1533 3 3

1

3

, ,

3

1 1

u

u

D f f u

D f

11

15 3

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢= = ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢− − ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣

⋅+ −

⋅

−

= ∇

= = − = − =

=

i ii⎦

−

__________________________________________________________________________________ 14. In which direction is the temperature function

( ) ( )2 2

ln 2 sin3

,x y y x

T x yπ⎛ ⎞⎛ ⎞+

⎜ ⎟⎜ ⎟+⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

=

increasing most rapidly, at the point ? Find a vector (3 1, ) in this desired direction, and with a magnitude equal to the rate of increase of T in this direction of maximum increase.

The gradient points in the direction in which the scalar temperature field T is increasing most rapidly, and also has the desired magnitude. So we’ll calculate this gradient, and evaluate it at the given point.

T∇

( )

( )

( )

( ) ( )

( )

( ) ( )

2 2 2 2

2 2 2 2

2 2 2 22 2

2

cos3 3

ln 2 sin 2 sin3 3

,

cosln 2 sin3 33

2 sin

x

y

x y y x x y y xx

x y y x x y y xxT

T x yT x y y x x y y xx y y x

yy

x

π π

π π

π ππ

π

⎛ ⎞+ +∂⎜ ⎟ ⋅⎜ ⎟ ∂⎝ ⎠

⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞+ +∂⎢ ⎥⎜ ⎟⎜ ⎟+ ⎜ ⎟+⎜ ⎟⎜ ⎟ ⎜ ⎟∂⎢ ⎥⎡ ⎤ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎢ ⎥∇ = = =⎢ ⎥ ⎢ ⎥⎛ ⎞ ⎛ ⎞⎛ ⎞ + +⎣ ⎦ + ∂∂⎢ ⎥⎜ ⎟ ⎜ ⎟ ⋅⎜ ⎟+ ⎜ ⎟ ∂⎜ ⎟⎢ ⎥⎜ ⎟∂ ⎝ ⎠⎝ ⎠⎝ ⎠⎣ ⎦

+( )

( )

( )

( )

( )

( )

( )

2

2 2 2 22

2

2 2 2 2 22

3

2cos cos 23 332 22 sin 3 2 sin

3 3 3

3,

y y x

x y y x x y y xxy y xy y

x y y x x xy x y y x x xy

T

π ππ π

π π π

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎛ ⎞+⎢ ⎥⎜ ⎟

⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

⎛ ⎞ ⎛ ⎞+ +⎡ ⎤ ⎡ ⎤+⎜ ⎟ ⎜ ⎟ +⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥= ⋅ = ⋅⎢ ⎥ ⎢ ⎥⎛ ⎞ ⎛ ⎞⎛ ⎞+ + + +⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟+ ⎜ ⎟+⎜ ⎟ ⎢ ⎥ ⎜ ⎟ ⎣ ⎦⎜ ⎟⎣ ⎦⎝ ⎠ ⎝ ⎠⎝ ⎠

∇ ( )

( )

( )( )( )( )

2 2

2

2 22

3 1 1 3cos 2 3 1 1 7 73 7cos 4 6

15 563 2 sin 43 1 1 3 3 2 3 1 15 23 2 sin3

1

ππ ππ π π

πππ

⎛ ⎞⋅ + ⋅ ⎡ ⎤ ⎡ ⎤⎜ ⎟ ⋅ ⋅ + ⎡ ⎤⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎡ ⎤⎝ ⎠ ⎢ ⎥⎢ ⎥ ⎢ ⎥= ⋅ = ⋅ = ⎢ ⎥ = ⎢ ⎥⎢ ⎥ ⎢ ⎥+⎛ ⎞⎛ ⎞⋅ + ⋅ ⎣ ⎦+ ⋅ ⋅ ⎣ ⎦⎢ ⎥ ⎢ ⎥⎜ ⎟⎜ ⎟+⎜ ⎟ ⎣ ⎦ ⎣ ⎦⎜ ⎟⎝ ⎠⎝ ⎠

The gradient field,

( )

( )

( )

2 2

2

2 22

cos 23,

23 2 sin3

x y y xxy y

T x yx y y x x xy

ππ

π

⎛ ⎞+ ⎡ ⎤⎜ ⎟ +⎢ ⎥⎜ ⎟⎝ ⎠ ⎢ ⎥∇ = ⋅

⎢ ⎥⎛ ⎞⎛ ⎞+ +⎢ ⎥⎜ ⎟⎜ ⎟+⎜ ⎟ ⎣ ⎦⎜ ⎟⎝ ⎠⎝ ⎠

__________________________________________________________________________________

15. How fast is the temperature function ( ) ( )2 2

ln 2 sin3

,x y y x

T x yπ⎛ ⎞⎛ ⎞+

⎜ ⎜+⎜⎜ ⎟⎝ ⎠⎝ ⎠

= ⎟⎟⎟

)

increasing at the point

, in its direction of maximum increase? (3 1, The gradient points in the direction in which the temperature T is increasing most rapidly, and its magnitude is the rate of increase in this direction. We calculated this gradient in problem #14, so now we need but find this vector’s length:

T∇

( ) 2 27 2743 7 15156 6 6

,1T π π π⎡ ⎤∇ = = + =⎢ ⎥

⎣ ⎦

NOTE: If distance is measured in centimeters and temperature measured in degrees Celsius, then the instantaneous rate of change of the temperature in the direction of maximum increase from the point (3,1) we

just found is 274 8.66716

π≅ degrees Celsius per centimeter.

__________________________________________________________________________________ 16. Find the directional derivative of ( ) sin( ), , xz yzW x y z e= − at the point ( )0,1,2

in the direction of vector 1ˆ 2, 1, 23

v = − .

Next we need the gradient of W at the point ( )0,1,2 :

( )

( )( )( )

( )

( ) ( )

( ) ( )

( )

( ) ( )

( ) ( )

sin

sin

sin( )

sin

sin( )

sin 0 1

sin 0

cos, ,

cos

cos 0 2 2 20,1,2 2 2

1cos 0 2 0 1

xz

xzx

xzy

xzz

xz

z

e yzxW e

W x y z W e yz zy

W e xz xe yz

z

eW

e

⋅

⋅

⎡ ⎤∂−⎢ ⎥∂ xz z

y

⎡ ⎤⋅ ⋅⎢ ⎥⎡ ⎤⎢ ⎥∂⎢ ⎥⎢ ⎥∇ = = − = −⎢ ⎥⎢ ⎥⎢ ⎥ ∂ ⎢ ⎥⎢ ⎥⎢ ⎥ ⋅ ⋅ −⎣ ⎦ ⎣ ⎦∂⎢ ⎥−⎢ ⎥∂⎣ ⎦

⎡ ⎤⋅ ⋅ ⋅ ⎡ ⎤⎢ ⎥ ⎢ ⎥∇ = − = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⋅ ⋅ ⋅ − ⎣ ⎦⎣ ⎦

Now we can compute the desired directional derivative:

( ) ( ) ( )ˆ

2 240,1, 2 0,1, 2 2 2 4 2 23

1 1

2 21 1 1ˆ 1 13 3 3

2 2uD W W u

⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − + −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

= ∇ = − = − = =i i i

__________________________________________________________________________________ 17. Find the equation of the tangent plane to 2 3 4 3 233 2x z xy y z+ 4− = at the point . ( )1,1,1 Let ( ) 2 3 4 3 2, , 33 2x y zF x z xy y z= +−

( ), , 4x y zF =

, so we seek an equation for a tangent plane to the implicitly defined

surface . Since F∇ is normal to the level surface, it will be normal to the tangent plane as well.

( )

( )

( )

( )

( )

2 3 4 3 2

3 4

2 3 4 3 2 3 2 2

2 2 3

2 3 4 3 2

3 4

3 2 2

2 2 3

3

3 96

3

49 16 15

3 26 2

, , 3 2 89

3 2

6 1 1 2 1 1,1,1 8 1 1 1 1

9 1 1 1 1

x

y

z

x z xy y zxz y

F x y z x z xy y z xy y zx z y z

x z xy y z

F

⎡ ⎤∂+⎢ ⎥∂ ⎡ ⎤⎢ ⎥

∂ ⎢ ⎥⎢ ⎥+ +⎢ ⎥⎢ ⎥∂ ⎢ ⎥⎢ ⎥ +⎣ ⎦∂⎢ ⎥+⎢ ⎥∂⎣ ⎦

⎡ ⎤ ⎡⎢ ⎥ ⎢+⎢ ⎥⎢ ⎥+ ⎣⎣ ⎦

−−

∇ = − = −

−

⋅ ⋅ − ⋅∇ = − ⋅ ⋅ ⋅ ⋅ =

⋅ ⋅ ⋅ ⋅n

⎤⎥

⎢ ⎥⎢ ⎥⎦

=

The point of tangency, , is on both the implicit surface(1,1,1) ( ), , 4x y zF = and the tangent plane to the graph of F at that point. So the equation of the tangent plane is given by:

4 4 1

, , 1,1,1 1 1 1 4 15 4 1 1515 15 1

4 15 20

xn x y z n y x y z

z

x y z

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

→ →= = + +

+ + =

i i i i = + +

4

20

Note that although we could have obtained this same tangent plane using the linearization, by looking at its level surface . However, the above approach is quicker. ( ), , 4x y zL =

The implicitly defined surface,

2 3 4 3 233 2x z xy y z+− = ,

and it’s tangent plane at the point ( ) , 1,1,1

4 15x y z+ + = :

__________________________________________________________________________________

18. Find the equation of the tangent plane to the implicitly defined surface

sin sin sin 12 2 2yz xz xyπ π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ −⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠= at the point ( )1, 2,1 .

Let ( ), , sin sin sin2 2 2yz xzx y zF π π π⎛ ⎞ ⎛ ⎞ ⎛= + −⎜ ⎟ ⎜ ⎟ ⎜

⎝ ⎠ ⎝ ⎠ ⎝ ⎠xy ⎞

⎟ , so we seek an equation for a tangent plane to the

implicitly defined surface . Since ( ), , 1x y zF = F∇ is normal to the level surface, it will be normal to the tangent plane as well.

( )

sin sin sin cos cos2 2 2 2

sin sin sin2 2 2 2

sin sin sin2 2 2

, ,

yz xz xy xzz yx

yz xz xyy

yz xz xyz

F x y z

π π π π

π π π π

π π π

⎡ ⎤∂ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎢ ⎥⎢ ⎥∂ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎢ ⎥⎢ ⎥∂ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥+ −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦

∇ = =

( )

2

cos cos2 2

cos cos2 2

1 1 1 21 cos 2 cos2 22 1 1 21,2,1 1 cos 1 cos

2 2 22 1 1 12 cos 1 cos2 2

xy

yz xyz x

yz xzy x

F

π

π π

π π

π π

π π π

π π

⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥

⎢ ⎥⎛ ⎞ ⎛ ⎞−⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥

⎢ ⎥⎛ ⎞ ⎛ ⎞+⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠

⋅ ⋅ ⋅ ⋅⋅ ⋅

⋅ ⋅ ⋅ ⋅∇ = ⋅ ⋅

⋅ ⋅ ⋅ ⋅⋅ ⋅

0 2cos 2 11cos 1cos 0 0 0

2 22cos 0 2 1

nπ π

π ππ π ππ π

⎡ ⎤⎢ ⎥⎢ ⎥ −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢

⎤⎥− = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢+ − − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣⎢ ⎥⎢ ⎥⎟⎣ ⎦

= = = ⎥⎥⎦

The point of tangency, , is on both the implicit surface(1, 2,1) ( ), , 0x y zF = and the tangent plane to the graph of F at that point. So the equation of the tangent plane is given by:

1 1 1 1 1, , 1, 2,1 0 0 2 0 0 2

1 1 1 1 1

0

x xn x y z n y y

z z

x z

π π1

1

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

→

⎣ ⎦

→

⎣ ⎦

− =

i i i i i i

The implicitly defined surface,

sin sin sin 12 2 2yz xz xyπ π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ −⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠= ,

and it’s tangent plane at the point ( ) , 1, 2,1

0x z− =

19. Find the equation of the tangent line to the curve of intersection of the two surfaces implicitly

determined by the equations and 2 2 23 2 1yz x z xy− + = − 2 cos 04xyzπ⎛ ⎞ =⎜ ⎟

⎝ ⎠ at the point ( ) . 3, 2,1−

What is the x-coordinate of the point of intersection of the above tangent line with the xy-plane?

The first surface is the level surface defined by the equation:

( ) 2 2 2, , 3 2 12f x y z yz x z xy= − + = − ,

While the second surface is the level defined by the equation:

( ), , cos 04xyzg x y z π⎛ ⎞= =⎜ ⎟

⎝ ⎠

The direction of this tangent line is perpendicular to the normals to both surfaces at the point of tangency, and so their cross product is in the direction of the desired tangent line. So we first calculate the gradients of these functions (which are normal to their respective level surfaces).

( )

( )

( )

( )

( )( )( )

( )

2 2 2

2

2 2 2 2

2

2 2 2

2

2

2

3 24

, , 3 2 3 26 2

3 2

4 3 1 2 8 8 3, 2,1 3 1 2 3 2 9 9

6 2 1 2 3 30 30

yz x z xyx xz y

f x y z yz x z xy z xyy

yz xyz x z xy

z

f

⎡ ⎤∂− +⎢ ⎥∂ ⎡ ⎤− +⎢ ⎥

∂ ⎢ ⎥⎢ ⎥∇ = − + = +⎢ ⎥⎢ ⎥∂ ⎢ ⎥⎢ ⎥ −⎣ ⎦∂⎢ ⎥− +⎢ ⎥∂⎣ ⎦⎡ ⎤− ⋅ ⋅ + − −⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥∇ − = ⋅ + ⋅ ⋅ − = − = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⋅ − ⋅ − ⋅ −⎣ ⎦ ⎣ ⎦⎦

⇒

⎣

∇ ( )

cos sin4 4 4

, , cos sin sin4 4 4 4 4

sincos4 44

3,

xyz yz xyzx

yzxyz xz xyz xyzg x y z xz

yxy

xy xyzxyzz

g

π π π

π π π π π

π ππ

⎡ ⎤∂ ⎡ ⎤⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥∂ ⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥ ⎡ ⎤⎢ ⎥ ⎢ ⎥∂ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎢ ⎥= = − = −⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢ ⎥∂ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥∂ ⎛ ⎞⎛ ⎞ −⎢ ⎥ ⎢ ⎥⎜ ⎟⎜ ⎟∂ ⎝ ⎠⎣

⇒

⎝ ⎠ ⎣ ⎦⎦

∇ ( ) ( )

( )

2 1 2 23 2 1 32,1 sin 3 1 sin 3 3

4 4 4 2 43 2 6 6

ππ π π− ⋅ −⎡ ⎤ ⎡ ⎤ ⎡ ⎤

⋅ ⋅ − ⋅⎛ ⎞ ⎛ ⎞⎢ ⎥ ⎢ ⎥ ⎢ ⎥− = − ⋅ = − − = = −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎝ ⎠⎝ ⎠ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⋅ − −⎣ ⎦ ⎣ ⎦⎣ ⎦

π

The direction of the line is thus in the direction of the cross product of these two gradients:

( ) ( )

9 330 6

8 2 8 2 144 248 2 33, 2,1 3, 2,1 9 3 9 3 12 230 64 4 4 4 2

30 6 30 6 42 78 29 3

f g π π π π π

⎡ − ⎤⎢ ⎥⎢ ⎥⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥∇ − ×∇ − = − × − = − × − = − − = − = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦

So for our direction vector, we can select 2427

v⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥−⎣ ⎦

. Our tangent line goes through the point of tangency,

) , hence our tangent line is parametrized by ( )3 24 3 242 2 2 21 7 1 7

t xr t p tv t t y

t z

+⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢= + = − + = − + = ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢(3, 2,1− ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢− − ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

. When

this line intersects the xy-plane, we must have 0z = , hence 1 7 0t− = , which happens when 17

t = . So the

desired point of intersection is

1 453 24 37 71 1 122 27 7 7

1 11 7 07

247227

77

x

r y

z

⎡ ⎤ ⎡ ⎤⎡ ⎤⎛ ⎞+ ⎢ ⎥⎡ ⎤ ⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠ ⎢ ⎥+⎢ ⎥ ⎢ ⎥⎢ ⎥

⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎛ ⎞ ⎛ ⎞= − + = − =⎢ ⎥⎢ ⎥− + = ⎢ ⎥⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥

⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎛ ⎞− ⎢ ⎥⎢ ⎥− ⎢ ⎥⎢ ⎥⎜ ⎟⎝ ⎠ ⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

2

.

The implicitly defined surfaces,

2 2 23 2 1yz x z xy− + = − &

cos 04xyzπ⎛ ⎞ =⎜ ⎟

⎝ ⎠,

Along with the tangent line to their curve of intersection at the point ( ) , 3, 2,1−

( )3 24

2 21 7

tr t t

t

+⎡ ⎤⎢ ⎥= − +⎢ ⎥⎢ ⎥−⎣ ⎦

.

__________________________________________________________________________________

20. & 21. Find any relative extrema and saddle points of 2 2( , ) 2 3 3 2 1f x y x xy y x y= − + − + + . Extrema and saddle points occur at any points where '( , ) 0f x y = :

( )' , 2 2 3 2 6 2 0 0f x y x y x y⎡ ⎤⎣ ⎦= − − − + + = ⎡ ⎤⎣ ⎦

Solving this linear system (just add the two equations to start!) we get the critical point 7 14 4

,⎛ ⎞⎜ ⎟⎝ ⎠

.

To see whether this point is where f has an extreme value or saddle point, we invoke the multivariable second derivative test. We calculate determinant of the second derivative matrix, the Hessian:

2 22 6

12 4 8 0xx xy

yx yy

f fH

f f−

=−

= = − = >

The determinant of this Hessian matrix is positive, and 2 0xxf = > , hence our point 7 14 4

,⎛⎜⎝ ⎠

⎞⎟ is where f

attains a relative minimum.

The surface

2 22 3 3 2z x xy y x y= − + − + +1,

along with its relative minimum at the point

7 1 11,4 4 8

,⎛ ⎞−⎜ ⎟⎝ ⎠

.

__________________________________________________________________________________ 22. Find the absolute maximum of the function from #20,

2 2( , ) 2 3 3 2 1f x y x xy y x y= − + − + +

over the right triangular region in the xy-plane with its right angle at the origin and legs along the positive coordinate axes, both of length 3.

In #20 we found f’s only critical point at 7 14 4

,⎛⎜⎝ ⎠

⎞⎟

≤

)

. But this point is where f attains a minimum. Thus any

maximum values of f must be on the bounding triangle. The three sides of this triangle are parametrized by:

( ) ( ) ( )1 2 3

0, , , all with 0 3.

0 3t t

r t r t r t tt t

⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = = ≤⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Calculating the composition ( )( if r t , for each of these curves, we then differentiate to find where the derivatives are zero, thus locating any critical points.

( )( ) ( )

( )( ) ( )

2 2 21

21

2

,0 2 0 3 0 3 2 0 1 3 1

3 3 1 2 3 0 2

3 3 3 9 9 9 5,0 3 1 1 1 1.252 2 2 4 2 4 4

f r t f t t t t t t

d df r t t t t tdt dt

f

= = − ⋅ ⋅ + ⋅ − ⋅ + ⋅ + = − +

= − + = − = =

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + = − + = − + = − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⇒

⎝ ⎠

( )( ) ( )

( )( ) ( )

2 2 22

22

2

0, 0 2 0 3 3 0 2 1 3 2 1

1 3 2 1 6 2 0 3

1 1 1 1 2 20, 3 2 1 1 0.63 3 3 3 3 3

f r t f t t t t t t

d df r t t t t tdt dt

f

= = − ⋅ ⋅ + ⋅ − ⋅ + ⋅ + = +

= + + = + = = −

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − + − + = − + = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⇒

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

+

( )( ) ( ) ( ) ( ) ( )

( )

( )( ) ( )

223

2 2 2 2

23

2

, 3 2 3 3 3 3 2 3 1

6 2 3 9 6 3 6 2 1 6 29 34

29 6 29 34 12 29 0 12

29 29 29 29 25 ,3 6 29 3412 12 12 12 24

f r t f t t t t t t t t

t t t t t t t t t

d df r t t t t tdt dt

f

= − = − − + − − + − +

= − + + − + − + − + = − +

= − + = − = =

⎛ ⎞ ⎛ ⎞− = ⋅ − ⋅ + = − =⎜ ⎟

⇒

⎜ ⎟⎝ ⎠ ⎝ ⎠

1.0416−

Of the values we’ve found, the largest value occurs at 1 20, 0.63 3

f ⎛ ⎞− = =⎜ ⎟⎝ ⎠

. But we still have three last

points to check – the corners of the triangle:

2 2

2 2

2 2

(0,0) 0 2 0 0 3 0 3 0 2 0 1 1(3,0) 3 2 3 0 3 0 3 3 2 0 1 1(0,3) 0 2 0 3 3 3 3 0 2 3 1 34

fff

= − ⋅ ⋅ + ⋅ − ⋅ + ⋅ + == − ⋅ ⋅ + ⋅ − ⋅ + ⋅ + == − ⋅ ⋅ + ⋅ − ⋅ + ⋅ + =

←

We’ve found our maximum: (0,3) 34f = __________________________________________________________________________________ 23. Find the closest point in Octant I on the surface implicitly determined by 1yzx = to the origin. How far is this point from the origin ? We’ll use Lagrange multipliers to minimize the squared distance function,

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

subject to the constraint equation, 1( , , ) yzG x y z x= = .

Then f Gλ∇ = ∇ at the extreme values of f on the level curves of . Calculating the gradients, and setting up the Lagrange multiplier equations, we get:

G

( ) ( )222

&, ,x x

y y

z z

x yzy xz x

f Gf x y f G x y G

f G

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

∇ = ∇ = = zy

2

2 2

2

1

2 2

2 22 2

22

2

x yz x xyzy xz y xyz x y zz xy z xyz

λ λ λλλ λ λ

λ λ λ

= = == = = = = ==

⎫⇒ ⎪

⎪⇒ ⇒⎬⎪

⎭= ⎪=⇒

=

Then 2 2 2 32 2 2 2

x y z λ λ λ λ+ + = + + = , which means that the minimum we seek is on the sphere with radius

32λ centered at the origin, which implies that the minimum distance is 3

2λ . If only we knew what λ

equals! The squares of the coordinates are all equal, so they are all ± each other. But we seek a solution in

Octant I, so the coordinates are all positive and so 2

x y z λ= = = . Invoking the constraint equation:

3

1 1 22 12 2

yz x y zx λ λ⎛ ⎞

= = = = =⎜ ⎟⎜ ⎟⇒ ⇒⎝ ⎠

⇒ = =

Thus the closest point to the origin on the surface in Octant I is the point (1, 1, 1), and the minimum distance

from the part of the surface 1yzx = that dwells in Octant I to the origin is 3 . __________________________________________________________________________________

24. You need to construct a tank consisting of a right circular cylinder with height h and radius r, topped with a hemispherical top, and with a flat base, as shown in the figure. If the material for the hemispherical top costs $6/m2, and the material for the cylindrical side costs $3/m2, and material for the circular bottom costs $2/m2, find the value of r and h that minimize the cost of the materials for this tank, assuming that the volume must be 10π m3.

Then the ratio, ?2hr=

The area of the top hemisphere is 22 rπ , so it will cost ( ) [ ]2 22

2 $2 6 2 $1r rmm

π π⎡ ⎤⎡ ⎤⎣ ⎦ ⎢⎛ ⎞

=⎜ ⎟⎝ ⎠⎥⎣ ⎦

.

The area of the cylindrical side is 2 rhπ , so it will cost ( ) [ ]222 3 6$ $m

mrh rhπ π⎡ ⎤⎡ ⎤⎣ ⎦ ⎢

⎛ ⎞=⎜ ⎟

⎝ ⎠⎥⎣ ⎦.

The area of the circular bottom is 2rπ , so it will cost ( ) [ ]22 22

$2 $2mr rm

π π⎡ ⎤⎡ ⎤⎣ ⎦ ⎢⎛ ⎞

=⎜ ⎟⎝ ⎠⎥⎣ ⎦

.

We seek to minimize the overall cost function,

( )( ) ( )

2 2 212 6 2 14 6 2 7 3

, 2 7 3 ,

C r rh r r rh r r h

C r h r r h

π π π π π π

π

= + + = + = +

= +

subject to the constraint equation (which we first concoct, based on the volume constraint):

( ) ( )

3 2 3 2 3 2 3 2

3 2 2

1 4 2 210 10 30 2 32 3 3 3

, 2 3 2 3 30

V r r h r r h r r h r

G r h r r h r r h

π π π π π= = ⋅ + = + = + =

=

⇒

+

⇒ +

+ = =

r h

G

Then C λ∇ = ∇ at the extreme values of C on a level curve of . Calculating the gradients, and setting up the Lagrange multiplier equations, we get:

G

( )2

2

14 3 26 62 & 3

3 3r h r hr rh

C Gr rr

π+ ⎡ ⎤

r⎡ + ⎤+⎡ ⎤

∇ = ∇ = =⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

( ) ( ) ( ) ( )

2

2 14 3 6 14

3 32 2 214 3 3

6 3

2

r h r r h r h r r h

h hr r r

π λ π λπ ππ λπ λ λ λπ λ

λ

π

⇒ ⎫⎪⎪ ⇒⎬

+ = + + = +⎛ ⎞ ⎛ π ⎞⋅ + = ⋅ ⋅ +⎜ ⎟ ⎜⎝ ⎠ ⎝= = ⎪⇒

⎪⎭

⇒

⎟⎠

214 3 3hπ λλ

⎛ ⎞⋅ + =⎜ ⎟⎝ ⎠

2π⋅λ

2 28 12 28 123 6 6

16 1633

h h h

h h

π π π π πλ λ λ λ λ

π πλ λ

⎛ ⎞⋅ + + = + − = −⎜ ⎟⎝ ⎠

=

⇒

⇒=

⇒

⇒

3h h

We can already answer this question! The ratio of the height h to the diameter 2r can be easily calculated:

1616 43

22 3 42

hr

ππ λλ

π λ πλ

⎛ ⎞⎜ ⎟⎝ ⎠= = =⎛ ⎞⎜ ⎟⎝ ⎠

3

We are done, as far as the quiz problem is concerned. But we will finish this off anyway; substituting the expressions for r and h (in terms ofλ ) into the constraint equation allows us to solve for λ :

( )2 2 2

22 2 2 2 16 4 4 16 4 202 3 3 0 2 3 30 30

3r r h π π π π π π π π

λ λ λ λ λ λ λ λ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = ⋅ + ⋅ = + = ⋅ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠

⇒ ⇒⎠ ⎝ ⎠

⇒⎝

30

3 3 33

3 3 3

80 8 8 230 3 3

3

π π π λ λλ λ

⇒ ⇒ ⇒ ⇒= = =π

=

Now that we knowλ , we can find r and h:

3 3 33

3 3

& 2 2 3 16 16 16 3 82 32 22 3 3 23

3 3

r r hπ π π π πππ πλ π λ π

= = = ⋅ = = = = ⋅ =⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝

⇒

⎠

⇒

⎠ ⎝

33

h

So for the ratio of height h to diameter 2r, we get (though we already deduced this!):

3

3

3 3

8 33 8 3 1 4

2 32 3 2 3 h h

r r

⎛ ⎞⎜ ⎟⎝ ⎠= = ⋅ =⇒

2 3

__________________________________________________________________________________ 25. Suppose that p is on the level set defined by ( ) 1G x = . Assuming that the function G, , is

infinitely differentiable, and that the vector is tangent to its level set

n →

T ( ) 1G x = at p . Then ( ) ?T

D G p =

( ) ( ) 0T

D G p G p T= ∇ • = , since the gradient ( )G p∇ is perpendicular to the level set, and thus perpendicular to any tangent vector to the level set at p . Perpendicular vectors have dot product 0. __________________________________________________________________________________