Math B6C – Chapter 14 Quiz – Spring 2010

* Solutions* __________________________________________________________________________________

Problems #1-3 refer to the function

( )3 5

5 3

35 6,2 7

x yf x yy x

+=

−

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

=( ) ( )

( ) ( ) ( )3 5 3

5 3 3,0 0,0 0 0

35 6 0 35lim , ,0 lim lim 5 52 0 7 7x x x

x xf x y f xx x→ → →

+ ⋅= = = = − = −

⋅ − −

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

( ) ( )( ) ( ) ( )

3 5 5

5 3 5, 0, 0 0

35 0 6 6lim , 0, lim lim 3 32 7 0 2x y y x x

y yf x y f yy y→ → →

⋅ + ⋅= = = = =

⋅ − ⋅

__________________________________________________________________________________

3. ( ) ( )

( ), 0,0lim , ?

x yf x y

→=

( ) ( )( )

( )b

,0 0,0

y problem #1

lim , 5x

f x y→

= − , but ( ) ( )

( )( )by problem #2

0, 0,0lim , 3y

f x y→

= .

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

( ), 0,0lim , does not exist

x yf x y

→

__________________________________________________________________________________

4. If ( ) ( ) ( )( )2 2, ln sin cosg x y x y xy= − ,

then 3 3, ?2 2yg π π⎛ ⎞

=⎜ ⎟⎜ ⎟⎝ ⎠

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

( ) ( )

( ) ( ) ( )( ) ( )

2 2

2 22 2

2 2 2

2 2

sin cos, ln sin cos

sin cos

cos 2 sin,

sin cos

y

y

x y xyyg x y x y xy

y x y xy

x x y xy xyg x y

x y xy

∂−

∂ ∂= − =∂ −

+=

−

Hence,

2 2 21 1 1 1 1 1 13 3 3 3 3 3 3

1 13 3

21 13 3

cos 2 sin2 2 2 2 2 2 2

,2 2

sin2 2

yg

π π π π π π π

π π

π π

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

21 13 3

2 2 1 2 1 23 3 3 3 3 3

2 1 1 23 3 3 3

cos2 2

cos 2 sin2 2 2 2 2 2

sin cos2 2 2 2

π π

π π π π π π

π π π π

π

⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟− ⋅⎜ ⎟ ⎜ ⎟⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

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

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

=

1

2 2 2 23 3 3 3

2 2213 333

2 23

2 23 3

3

3

cos 2 sin 0 2 12 2 2 2 2 2

1 0sin c

2

os2 2

2 22 2 22 2 2

π π π π π

π π

π π π

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

−−

⎛ ⎞= = ⋅ = ⋅ = ⋅⎜ ⎟⎝ ⎠

⋅23

23

2

π⋅ 3 22π=

__________________________________________________________________________________

5. If ( ) ( )sin, xyh x y e= , then , ?2 2xyh π π⎛ ⎞

=⎜ ⎟⎜ ⎟⎝ ⎠

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

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

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

sin sin

sin sin sin

sin sin sin

, , , cos

cos cos cos

cos cos sin cos 1

xy xyxy x

xy xy xy

xy xy xy

xy

h x y h x y h x y e e xy yy y x y x y

e xy y e xy y e xy yy y y

e xy x xy y e xy x y e xy

h

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

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

= ⋅ ⋅ ⋅ ⋅ + ⋅ − ⋅ ⋅ + ⋅ ⋅

( ) ( ) ( ) ( ) ( )( )sin 2, cos sin cosxyx y e xy xy xy xy xy= − +

Hence,

sin 2 12, cos sin cos 0 1 02 2 2 2 2 2 2 2 2 2xy

eh e eππ π π π π π π π π π⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= ⋅ − ⋅ + = ⋅ − ⋅ + = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠

.

( ) ( )sin, xyh x y e=

__________________________________________________________________________________ 6. Find the linearization of ( ) 5 2 3 3, 3f x y x y x y= − − at ( )1, 2− .

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

1, 2 ' 1, 22

,x

f fy

L x y⎛ − ⎞⎡ ⎤ ⎡ ⎤

= − + − −⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎝ ⎠

So now we collect all the pieces of this puzzle…

( )

( ) ( ) ( )

( ) ( )

( ) ( ) ( )

5 2 3 3

5 32 3

5 2 3 3 5 2 3 3

4 2 2 5 2

4 22

, 3

1, 2 1 2 1 3 2 4 1 24 27

'( , ) 3 3

'( , ) 5 3 2 9

' 1, 2 5 1 2 3 1

x y

f x y x y x y

f

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

f x y x y x x y

f

f

y

= ⋅ − − ⋅

− = − ⋅ − − − ⋅ = − + − = −

⎡ ⎤∂ ∂= − − − −⎢ ⎥∂ ∂⎣ ⎦

⎡ ⎤= − −⎣ ⎦

− = ⋅ − ⋅

⎡

⋅ −

=

−

⎤⎣ ⎦

( ) [ ]5 22 1 2 9 2 17 40⎡ ⎤⋅ − ⋅ − ⋅ = −⎣ ⎦

Putting the pieces together, we get:

( ) ( ) ( ) [ ]

( ) ( ) ( )

1 11, 2 ' 1, 2 27 17 40

2 2

27 17 1 40 2 27 17 17 40 80

,

,

x xf f

y y

x y x y

L x y

L x y

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

= − + + − − = − + + − +

Our desired linearization is thus: ( ) 17 40 70, x yL x y = − +

A graph of ( ) 5 2 3 3, 3f x y x y x y= − − along with its linearization at ( )1, 2− , ( ) 17 40 70, x yL x y = − + :

__________________________________________________________________________________ 7. Find the linearization of ( ) ( )2 sinh, , z xyf x y z = at the point ( )1, ln 2, 1− . ( ) ?, ,L x y z =

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

( , , ) 1, ln 2, 1 ' 1, ln 2, 1 ln 21

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

122 321, ln 2, 1 1 sinh 1 ln 2 sinh ln 2

2 2 2 4

31, ln 2, 14

11 222 522 cosh lSimilarly, n 22 2 2

2

22

2

2

4

e e ef

f

e e e

−

−

⎛ ⎞− ⋅⎜ ⎟− − ⎝ ⎠− = − ⋅ ⋅ = = = = =⋅

− =

⎛ ⎞+ ⋅+ ⎜ ⎟+ + ⎝ ⎠= = = = =⋅

( )( ) ( )( ) ( )( )2 2 2

'( , , )

sinh sinh sinh

x y zf x y z f f f

z xy z xy z xyx y z

f

⎡ ⎤= ⎣ ⎦

⎡ ⎤∂ ∂ ∂= ⎢ ⎥∂ ∂ ∂⎣ ⎦

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

2 2

2 2

2 2

' , , cosh cosh 2 sinh

cosh cosh 2 sinh

' 1, ln 2, 1 1 ln 2 cosh ln 2 1 1 co

x y z z xy y z xy x z xy

z y xy z x xy z xy

f

⎡ ⎤= ⋅ ⋅⎣ ⎦

⎡ ⎤= ⎣ ⎦

− = − ⋅ ⋅ − ⋅ ⋅ ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( )

sh ln 2 2 1 sinh ln 2

5 5 3' 1, ln 2, 1 ln 2 cosh ln 2 cosh ln 2 2sinh ln 2 ln 2 24 4 4

5ln 2 5 3' 1, ln 2, 14 4 2

f

f

⎡ ⎤⋅ − ⋅⎣ ⎦

⎡ ⎤− = ⋅ − = ⋅ − ⋅⎡ ⎤⎣ ⎦ ⎢ ⎥⎣ ⎦

⎡ ⎤− = −⎢ ⎥⎣ ⎦

Putting these pieces together, we get:

( ) ( )1

( , , ) 1, ln 2, 1 ' 1, ln 2, 1 ln 21

xL x y z f f y

z

−⎡ ⎤⎢ ⎥= − + − −⎢ ⎥⎢ ⎥+⎣ ⎦

( ) ( ) ( )1

3 5ln 2 5 3 3 5ln 2 5 3( , , ) ln 2 1 ln 2 14 4 4 2 4 4 4 2

1

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

5ln 2 5 3 3 10ln 2( , , )4 4 2 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 3sinh4

z xy =

with tangent plane

3( , , )4

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

( )5ln 2 5 6 6 10ln 2x y z+ − = + :

__________________________________________________________________________________

8. Suppose that ( , , )

( , , )( , , )

u x y zg x y z

v x y z⎡ ⎤

= ⎢ ⎥⎣ ⎦

and that ( , )

( , )( , )

m u vh u v

n u v⎡ ⎤

= ⎢ ⎥⎣ ⎦

. Let f h g= .

Chain Rule

'

m m m u u um mx y z x y zu vfn n n n n v v vx y z u v x y z

xyz

↓

∂ ∂ ∂ ∂ ∂ ∂⎡ ⎤ ⎡ ⎤∂ ∂⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥∂ ∂ ∂ ∂ ∂ ∂∂ ∂⎢ ⎥ ⎢ ⎥= = ⎢ ⎥∂ ∂ ∂ ∂

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

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

⎣ ⎦

⎦

3 2

2

'

f

g h

u u ux y zv v vx y z

u

mn

v

g∂ ∂ ∂⎡ ⎤⎢ ⎥∂ ∂ ∂⎢ ⎥=∂ ∂ ∂⎢ ⎥⎢ ⎥∂

⎡ ⎤⎢

∂ ∂⎣

⎥

⎡ ⎤

⎣ ⎦

⎦

⎦

⎥

→

⎢⎣

⎯⎯

'

m mu vhn nu v

∂ ∂⎡ ⎤⎢ ⎥∂ ∂= ⎢ ⎥∂ ∂⎢ ⎥

⎢ ⎥∂ ∂⎣ ⎦

Suppose further that ( )7

1,1,19

g ⎡ ⎤= ⎢ ⎥⎣ ⎦

and that ( )( )

( )( )

7,9 7,9 4 17,9 7,9 2 5

u v

u v

m mn n⎡ ⎤ −⎡ ⎤

=⎢ ⎥ ⎢ ⎥−⎣ ⎦⎣ ⎦ and that

( )( )

( )( )

( )( )

1,1,11,1,1 1,1,1 7 4 11,1,11,1,1 1,1,1 1 3 2

yx z

yx z

uu uvv v

⎡ ⎤ −⎡ ⎤=⎢ ⎥ ⎢ ⎥− −⎣ ⎦⎣ ⎦

. Then ( )' 1,1,1 ?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,

( )( )( )( )( )

( )( )( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )

1,1,1 1,1,1 1,1,11,1,1 1,1,1' 1,1,1

1,1,11,1,1 1,1,11,1,1 1,1,1

7,9 7,9 7 4 17,9 7,9 1 3 2

4 1 7 4 12 5 1 3 2

29 19 6' 1,1,1

19 23 12

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

⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

⎡ ⎤ −⎡ ⎤= ⎢ ⎥ ⎢ ⎥− −⎣ ⎦⎣ ⎦

− −⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦

−⎡=

− −⎣

⎤⎢ ⎥

⎦

__________________________________________________________________________________ 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 ( )1,1,1 , ?z y y 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:

( )29 19 6

' 1,1,119 23 12

f−⎡ ⎤

= ⎢ ⎥− −⎣ ⎦

.

Thus evaluated at the point ( )1,1,1 , we have: 29 19 619 23 12

x y z

x y z

ϕ ϕ ϕψ ψ ψ

−⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥− −⎣ ⎦⎣ ⎦

.

Hence, ( ) ( )6 23 19 12 90z y y zz y y zϕ ψ ϕ ψ ϕ ψ ϕ ψ∂ ∂ ∂ ∂

− = − = ⋅ − − ⋅ − = −∂ ∂ ∂ ∂

.

10. Let ( ) 2 3

2, ,

yz xz xy ug x y z

xy z v− +⎡ ⎤ ⎡ ⎤

= =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

and ( ) 5 3 3 5,f u v u u v v= + − , and let h f g= .

Then ( )' 0,1, 2 ?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:

( ) ( ) ( )5 3 3 5 5 3 3 5

4 2 3 3 2 4

' ,

5 3 3 5

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

u u v u v v

∂ ∂⎡ ⎤= + − + −⎢ ⎥∂ ∂⎣ ⎦⎡ ⎤= + −⎣ ⎦

But ( ) 2 3

2, ,

yz xz xy ug x y z

xy z v− +⎡ ⎤ ⎡ ⎤

= =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

, so ( ) 2 3

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

0 1 2 0u

gv

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

, hence:

( )( ) ( ) [ ]4 2 3 3 2 4' 0,1, 2 ' 2,0 5 2 3 2 0 3 2 0 5 0 80 0f g f ⎡ ⎤= = ⋅ + ⋅ ⋅ ⋅ ⋅ − ⋅ =⎣ ⎦

Also,

( )( ) ( ) ( )

( ) ( ) ( )

( )

2 3 3 2 22 3 2 3 2 3

2 3 3 2 2

2 2 22 2

' , ,2 3

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

1 2 2 0 1 2 3 0 1 2 8 0 0

yz xz xy yz xz xy yz xz xyz y z x y xx y z

g x y zy z xyz xy zxy z xy z xy z

x y z

g

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

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

Putting all this together, we get:

( ) ( )( ) ( ) [ ]

( )

( ) [ ]

3 2 1' 0,1,2 ' 0,1, 2 ' 0,1, 2 80 0

8 0 0

80 3 0 8 80 2 0 0 80 1 0 0

' 0,1,2 240 160 80

h f g g

h

−⎡ ⎤= = ⎢ ⎥

⎣ ⎦

= ⋅ − + ⋅ ⋅ + ⋅ ⋅ + ⋅⎡ ⎤⎣ ⎦

= −

__________________________________________________________________________________ 11. Find the gradient of

( )3 3

1tan4

, x yg x y − ⎛ ⎞⎜ ⎟⎝ ⎠

=

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

( )

3 31

3 31

3 3

2 323 3 6 6

3 33 2

6 623 3

tan4

,

tan4

1 1 34 121 44 411 3

4 41 44

x

y

x yxg

g x yg x y

y

x yx yxx y x y

x y x yx yyx y

−

−

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

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

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

( ) ( )

2 3

6 6 2 2

6 63 2

6 6

2 2 4

6 6 4 2

16 121612

16

31 1 112 2 1 2 3 3 52,12 2 2 616 2 1 52 1 2

5

x yyx y x yxx yx y

x y

g

⎡ ⎤⎢ ⎥+ ⎡ ⎤⎢ ⎥ = ⎢ ⎥⎢ ⎥ + ⎣ ⎦⎢ ⎥

+⎣ ⎦

⎡ ⎤⎢ ⎥⎡ ⎤ ⎡ ⎤ ⎡ ⎤⋅ ⋅ ⋅

∇ = = = = ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ ⋅ ⋅ +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎢ ⎥⎣ ⎦

The y-component of this gradient is thus 65

.

__________________________________________________________________________________ 12. If ( ) 3 5 72, , y zf x y z x −= , then ( )2, 1,1f∇ − = ? The y-component of this gradient is ___ .

( )

( )

( )

( )

( )

3 5 7

2 5

3 5 7 3 4

6

3 5 7

212

, , 2 2, 1, 1 4014 14

2

35

y zx y

f x y z y z y fy

zy z

z

xx

x x

x

⎡ ⎤∂−⎢ ⎥∂ ⎡ ⎤ −⎢ ⎥ ⎡ ⎤

∂ ⎢ ⎥⎢ ⎥ ⎢ ⎥∇ = − = ∇ − =⎢ ⎥⎢ ⎥ ⎢ ⎥∂ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦⎣ ⎦∂⎢ ⎥−⎢ ⎥∂⎣

⇒

⎦

The x-component of this gradient is thus 40 . The graph depicts the gradient field,

( )

2 5

3 4

6

, ,14

35

yf x y z y

z

xx

⎡ ⎤⎢ ⎥∇ = ⎢ ⎥⎢ ⎥−⎣ ⎦

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

( ) 3 5 72 1, , y zf x y z x − = −=

13. Find ( )ˆ 2, 1,1uD f − , for the function f featured in problem #12, with 1

1ˆ 335 5

u⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

.

( ) ( )

( ) ( ) ( )

ˆ

ˆ

12 1 6 11 22, 1,1 2, 1,1 40 3 20 335 3514 5 7 5

386 60 35 19 35

ˆ

2 2 38 2, 1,135 35 35

u

u

D f f u

D f

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

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

⋅ − + ⋅ ⇒−

=∇

− −= = = − =

i ii

14. In which direction is the temperature function

( ) ( )3 3cos6,

x y y xeT x y

π⎛ ⎞+⎜ ⎟⎝ ⎠=

increasing most rapidly, at the point ( )1 2, ? Find a vector in this desired direction, and with a magnitude equal to the rate of increase of T in this direction of maximum increase.

The gradient T∇ 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.

( )

( )

( )

( ) ( )( ) ( )

( )

3 33 3

3 3 3 3

3 3

coscos 3 366

cos cos6 3 36

cos6

cos6

,

cos6

sin6

x y y xx y y x

x

x y y x x y y xy

x y y x

e x y y xeT xxT x yT

e e x y y xy y

xe

ππ

π π

π

π

π

π

⎛ ⎞⎛ ⎞ ++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞+⎜ ⎟⎝ ⎠

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

−= ⋅

( ) ( )

( ) ( )

( ) ( )

( )( )

( )

3 3

3 3 3 3

3 3 3 3

2 3cos3 36

3 2

3cos 2 8 co6

2

6

sin6 6

3sin

6 6 3

3 2 2sin 2 8

6 6 61 3 21,2

x y y x

y y x x y y xx

x y y x x y y xy

x y ye x y y x

x y x

T e e

π

π

π

π π

π π

π π π

⎛ ⎞+⎜ ⎟⎝ ⎠

⎛ ⎞+⎜ ⎟⎝ ⎠

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

⎡ ⎤+⎛ ⎞= − ⋅ + ⋅ ⎢ ⎥⎜ ⎟ +⎝ ⎠ ⎣ ⎦

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

5s3

12

145sin133

7 314 143 3 613 136 2 12 13 3

12

eee

e

π π

ππ π

π

⎛ ⎞⎜ ⎟⎝ ⎠ ⎡ ⎤⎛ ⎞ ⋅⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

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

The gradient field,

( )( ) ( )

3 3 2 3cos3 36

3 2

3, sin

6 6 3

x y y x x y yT x y e x y y x

x y x

ππ π⎛ ⎞+⎜ ⎟⎝ ⎠

⎡ ⎤+⎛ ⎞∇ = − ⋅ + ⋅ ⎢ ⎥⎜ ⎟ +⎝ ⎠ ⎣ ⎦

__________________________________________________________________________________

15. How fast is the temperature function ( ) ( )3 3cos6,

x y y xeT x y

π⎛ ⎞+⎜ ⎟⎝ ⎠= increasing at the point ( )1 2, ,

in its direction of maximum increase? The gradient T∇ 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:

( ) 2 2143 3 109514 131312 12 12

1,2 e e eT 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 1095 14.28312

eπ≅ degrees Celsius per centimeter.

__________________________________________________________________________________ 16. Find the directional derivative of

( ) 2 2sin( ) cos( ), , xy z x yzW x y z −= at the point ( )1,1,π in the direction of vector

1 1, 2, 23

u = − .

First we need the gradient of W at the point ( )1,1,π :

( )

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

( ) ( )( )

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

( )( ) ( )

2 22 2 2

2 2 2 2 2

2 2 2 2

2 2

sin coscos sin 2

, , sin cos cos 2 sin

cos sinsin cos

cos sin1,1,

x

y

z

xy z x yzxy z y z x yz xyzxW

W x y z W xy z x yz xy z xyz x yz x zy

W xy z xy x yz x yxy z x yz

z

Wπ π π

π

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

⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥⋅ + ⋅∂⎢ ⎥ ⎣ ⎦−⎢ ⎥∂⎣ ⎦

∇ =⋅ +

( ) ( )( ) ( )

21

2cos 2 sin

cos 1 sin 1

ππ

ππ π π ππ π

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

⋅⋅ + ⋅⋅ + ⋅

Now we can compute the desired directional derivative:

( ) ( ) ( )ˆ3 21,1, 1,1, 2 2 4 2

31 1

1 11 1 1ˆ 2 23 3 3

2 2uD W W u

π πππ π π π π π

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

⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

= ∇ = − =− − = − =i i i

__________________________________________________________________________________ 17. Find the equation of the tangent plane to 3 2 4 422 3 1x y xz yz+− = at the point ( )1,1,1 . Let ( ) 3 2 4 4, , 22 3x y zF x y xz yz= +− , so we seek an equation for a tangent plane to the implicitly defined

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

( )

( )

( )

( )

( )

3 2 4 4

2 2 4

3 2 4 4 3 4

3 3

3 2 4 4

2

2 28

2

32 6

8 4

2 36 3

, , 2 3 412

2 3

6 3 1,1,1 4

12

x

y

z

x y xz yzx y z

F x y z x y xz yz x y zxz yz

x y xz yz

F n

⎡ ⎤∂+⎢ ⎥∂ ⎡ ⎤⎢ ⎥

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

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

−−

∇ = − =−

−

−∇ = = =

−

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

3 3 6 6

4 4

1, , 1,1,1 1 3 6 4 3 6 4

1

3 6 4 5

xn x y z n y x y z

z

x y z

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

→ →= = + − = + −

+ − =

i i i i

Note that we could have obtained this same tangent plane using the linearization, by looking at its level surface ( ), , 1x y zL = . However, the above approach is quicker.

The implicitly defined surface,

3 2 4 422 3 1x y xz yz+− = ,

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

3 6 4 5x y z+ − = :

__________________________________________________________________________________ 18. Find the equation of the tangent plane to the implicitly defined ellipsiod

2 2 2

15 3 4x y z

+ + = at the point 41, 2,30

⎛ ⎞⎜ ⎟⎝ ⎠

.

Let ( )2 2 2

, ,5 3 4x y zx y zF = + + , so we seek an equation for a tangent plane to the implicitly defined surface

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

( )

2 2 2

2 2 2

2 2 2

25 3 4 5

25 3 4 3

25 3 4

2 1 25 5

4 2 2 2 21, 2,3 3304 2

2

152

313030 30

, ,

2

x y z xx

x y z yy

zx y zz

F x y z

F

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

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎛ ⎞ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

⎡⎢

⎣

∇ = =

⋅

⋅∇ = = = 2n

⎤⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥

⎥⎢ ⎥

=

⎢⎦

The point of tangency, 41, 2,30

⎛ ⎞⎜ ⎟⎝ ⎠

, is on both the implicit surface ( ), , 1x 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 15 5 1

4 2 2 2 1 2 2, , 1, 2, 2 3 3 5 3 5 3 1530 3041 1

3030 30

2 15 3 30

xx y zn x y z n y

z

x y z

=

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

⋅+ + =

→ →i i i i

The implicitly defined ellipsoid,

2 2 2

15 3 4x y z

+ + = ,

and it’s tangent plane at the point 41, 2,30

⎛ ⎞⎜ ⎟⎝ ⎠

,

2 1

5 3 30x y z⋅+ + =

__________________________________________________________________________________ 19. Find the equation of the tangent line to the curve of intersection of the two surfaces implicitly determined by the equations

2 22 5 14yz xz xy+ + = − and sin 02xyzπ⎛ ⎞ =⎜ ⎟

⎝ ⎠

at the point ( )2, 1, 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 5 14f x y z yz xz xy= + + = − ,

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

( ), , sin 02xyzg 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

22

2 55

, , 2 5 2 22 10

2 5

5 1 1 6 3 2, 1, 1 2 1 2 2 1 6 2 3

2 1 10 2 1 22 11

yz xz xyx z y

f x y z yz xz xy z xyy

y xzyz xz xy

z

f

⎡ ⎤∂+ +⎢ ⎥∂ ⎡ ⎤+⎢ ⎥

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

⇒

( )

sin cos2 2 2

, , sin cos cos2 2 2 2 2

cossin2 22

2, 1

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

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

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

⇒

⎣ ⎦⎣ ⎦

∇ − −( ) ( ) ( )

( )

11 1 1 12 1 1

, 1 cos 2 1 cos 2 22 2 2 2

2 1 2 2

ππ π ππ=−− ⋅ −⎡ ⎤ ⎡ ⎤ ⎡ ⎤

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

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

( ) ( )

3 211 2

3 1 3 1 163 1

2, 1,1 2, 1,1 2 3 2 3 2 511 22

11 2 11 2 33 13 2

f g π π π π

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

So for our direction vector, we can select 16

53

v⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦

. Our tangent line goes through the point of tangency,

( )2, 1, 1− − , hence our tangent line is parametrized by ( )2 16 2 161 5 1 51 3 1 3

t xr t p tv t t y

t z

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

.

When this line intersects the xy-plane, we must have 0z = , hence 1 3 0t+ = , which happens when 13

t = − .

So the desired point of intersection is

1 22162 16 23 331 1 5 21 5 13 3 3 3

1 1 1 01 33

xr y

z

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

.

The implicitly defined surfaces,

2 22 5 14yz xz xy+ + = − &

sin 02xyzπ⎛ ⎞ =⎜ ⎟

⎝ ⎠,

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

( )2 8

11

tr t

t

− +⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥+⎣ ⎦

.

__________________________________________________________________________________

20. & 21. Find any relative extrema and saddle points of 2 2( , ) 2 5 2 3f x y x xy y x y= − + − − + . Extrema and saddle points occur at any points where '( , ) 0f x y = :

( )' , 4 5 1 5 4 1 0 0f x y x y x y⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= − − − + − =

Solving this linear system (I prefer Cramer’s rule!) we get the critical point ( )1 1,− − . 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:

4 55 4

16 25 9 0xx xy

yx yy

f fH

f f−

=−

= = − = − <

The determinant of this Hessian matrix is negative, so we’ve found a saddle point on this surface. Since ( )1, 1 2 5 2 1 1 3 4f − − = − + + + + = , the point ( )1 1 , 4,− − on the surface ( ),x yf z= is a saddle point .

The surface

2 22 5 2 3z x xy y x y= − + − − + ,

With it’s saddle point (in the middle of the “saddle”), the point

( )1 1, 4,− − .

__________________________________________________________________________________

22. & 23. Find any relative extrema and saddle points of 2 2( , ) 5 3 20 18g x y x y x y= + − − . Extrema and saddle points occur at any points where ( )' , 0g x y = :

( )' , 10 20 6 18 0 0g x y x y⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= − − =

Solving this linear system (this one’s as easy as can be!) we get the critical point ( )2 3, . 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:

10 00 6

60 0 0xx xy

yx yy

g gH

g g= = = − >

The determinant of this Hessian matrix is positive, so we’ve found a relative maximum or minimum on this surface. Since 0xxg > , , the point ( )2 3 , 47, − on the surface ( ),x yg z= is a relative minimum .

The paraboloid surface

2 25 3 20 18z x y x y= + − − ,

with it’s relative minimum at the point

( )2 3, 47, − . __________________________________________________________________________________ 24. & 25. Find the absolute maximum and the absolute minimum of the function from #20,

2 2( , ) 2 5 2 3f x y x xy y x y= − + − − +

over the right triangular region in the xy-plane with its corners at the points ( ) ( ) ( )3 0 , 2 0 , 2 5 ., , ,− − In #20 we found f’s only critical point at ( )1 1,− − . But this is a saddle point, which is thus neither a maximum nor a minimum. Thus any extreme values of f must be on the bounding triangle. The three sides of this triangle can be parametrized by:

( ) ( ) ( )1 2 3

2 3 with 3 2, with 5 0, with 0 5.

0t t

r t t r t t r t tt t

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

(…Your parametrizations my differ...)

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 21

21

2

1

,0 2 0 0 0 3 2 3

1 2 3 4 1 0 4

1 1 1 1 1 1 232 3 3 3 2.8754 4 4 8 4 8 8

f r t f t t t t t

d df r t t t t tdt dt

f r

⇒

= = − + − − + = − +

= − + = − = =

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

This value of t is within our parametrization interval 3 2t− ≤ ≤ , and corresponds to the point on the

bounding triangle 11 1 ,04 4

r ⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

.

( )( ) ( )

( )( ) ( )

2 2 22

22

2, 2 2 5 2 2 2 3 2 11 9

11 2 11 9 4 11 0 2.754

f r t f t t t t t t

d df r t t t t tdt dt

= = ⋅ − ⋅ ⋅ + − − + = − +

= − + = − = = =⇒

Since this last value of t isn’t in our parametrization interval 5 0t− ≤ ≤ , this value isn’t of interest to us, as it lies outside of our triangular region of interest. So we move on to the third side’s parametrization:

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

( ) ( )

( )( ) ( ) ( )

3

2 2 2 2 2

23

3

2 23, 3 3 3

2 6 9 5 15 2 3 3 9 27 24 3 3 9 8

9 3 3 3 9 8 3 6 9 0 1.56 2

3 33 32 2

2 5 2 3f r t f t t t t t t t t

t t t t t t t t t t t

d df r t t t t t tdt dt

f r

= − − = ⋅ − ⋅ − ⋅ − ⋅ − − −

= − + + − + − + + + = − + = − +

= − + = − = = = =

⎛ ⎞⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎜ ⎟⎝

⇒ ⇒

⎠ ⎝ ⎠⎝ ⎠

− + − − +

2 3 27 27 27 32 159 8 3 8 3 3.752 4 2 4 4

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

This value of t is within our parametrization interval 3 2t− ≤ ≤ , and corresponds to the point on the

bounding triangle ( )3 3

3 33 3 32 ,32 2 22

r r t

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

.

So far our potential extreme values are the two points along the edge we just found. But we also must look at the values of f at the 3 corners of our triangular region. Here are all the candidates, including the corners:

( ) ( )

( ) ( ) ( )

2

2

22

1 ,0 2.8754

3 3, 3.752 2

( 3,0) 2 3 0 0 3 0 3 24

(2,0) 2 2 0 0 2 0 3 9

(

min

m2, 5) 2 2 5 2 5 2 5 2 5 3 1 a1 x4

f

f

f

f

f

⎛ ⎞ =⎜ ⎟⎝ ⎠

⎛ ⎞− − =⎜ ⎟⎝ ⎠

− = ⋅ − − + − − − + =

= ⋅ − + − − + =

− = ⋅ − ⋅ ⋅ − + ⋅ − − − − + ←=

←

We’ve found our absolute maximum (2, 5) 114f − = ,

and our absolute minimum 1 ,0 2.8754

f ⎛ ⎞ =⎜ ⎟⎝ ⎠

,

over the given triangular region:

__________________________________________________________________________________

26. Find the closest point in Octant I on the surface implicitly determined by 2 1xy yz− = to the origin. How far is this point from the origin (approximately, to the nearest millionth) ? We’ll use Lagrange multipliers to minimize the squared distance function,

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

subject to the constraint equation, 2 1( , , ) xy yzG x y z = − = .

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

( ) ( )2 22 22

, & ,x x

y y

z z

x yy x zz y

f Gf x y f G x y G

f G

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

∇ = ∇ = =

( )

2 2

2 2 2 2

2 2

x y x y

y x z y x z

z y z y

λ λ

λ λ λ

λλ

↓

↑

= =

= − = −

= − = −

⎫⎪⇒ ⎪⎪

⇒ ⎬⎪⎪

⇒ ⎪⎭

*

Substituting the top and bottom of these equations (on the right) into the middle one, we get

( )

( )

22 2 2 2

2 2

2 2 2 2 4 4 4 52 2

2 4 5 0 4 5 0 0 or 5

y y y y y y y y y y y

y y y y

λ λλ λ λ λ λ λ λ

λ λ λ

⎛ ⎞= − − = + = + =⎜ ⎟⎝ ⎠

− =

⇒ ⇒ ⇒

⇒ ⇒ − = = = ±⇒

Notice from the equations at * that if 0y = , then so does x and z. However, the origin (0,0,0) doesn’t satisfy our constraint equation, so this possibility is nixed in the bud.

We are now able to express x, y, and z, all in terms of y, now that we’ve found λ :

2 2 5 5

21 15

2 2 5 5

x y y x y

y y y y

z y y y z y

λ

λ

= ⋅ = ± ⋅ = ± ⋅

= =

⎛ ⎞− ±⎜ ⎟⎝ ⎠= − = ⋅

→

⋅ = ⋅

→

→= ∓ ∓

Substituting these values into our constraint equation, 2 1xy yz− = , we get:

We must chose + in the ,( as well as the - in the ) since otherwis

2 2 2 2

2

2 1

2 1 4 1 5 52 1 1 1 55 5 5 5 5

x y y z

y y y y y y y y

y

±

↓

⋅ ⋅ − ⋅ =

⎛ ⎞ ⎛ ⎞⋅ ± ⋅ ⋅ − ⋅ ⋅ = ± ⋅ ± ⋅ = ± ⋅ = = ±⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⇒ ⇒ ⇒

⇒ =

∓

∓

e isn't a real number.

14

344

344

5 5 5 5 5 55 5 5 5

2 2 5 2 5 2 5 2 55 5 55 5

1 5 5 5

2 52

5

5 55

5

5

y

y

x y x

z y zλ

−

−

−

= ± = ± = ± = ±⋅

⎛ ⎞⎜ ⎟= ⋅ = ⋅ ± = ± = ± = ± ⋅⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟= − = − ⋅ ± = = = ⋅⎜ ⎟

⇒

⇒

⇒⎝ ⎠

∓ ∓ ∓

Finally, we evaluate the distance-squared function at the two points we found:

2 2 23 1 3 3 1 3 6 2 64 4 4 4 4 4 4 4 4

3 1 3 3 1 3 112 2 2 2 2 2 2

1 1 12 2 2

2 5 , 5 , 5 2 5 5 5 4 5 5 5

4 5 5 5 5 5 5 5 5

25 5 2 5 0.8944271909999158785636694674925

f− − − − − − − − −

− − − − − − −

− − −

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

= ⋅ + + = ⋅ + = +

= + = ⋅ =

∓ ∓

The closest points to the origin on the surface (it can’t be a farthest point, since there are points on this surface 2 1xy yz− = that are arbitrarily large distances from the origin) are the points

3 1 34 4 42 5 ,5 , 5

− − −⎛ ⎞⋅ −⎜ ⎟

⎝ ⎠

3 1 34 4 4 and 2 5 , 5 , 5

− − −⎛ ⎞− ⋅ −⎜ ⎟⎝ ⎠

, both with distance from the origin equal to 2 0.8944275≅ .

The picture to the right shows the origin and the line segment of

length 25

connecting it so the closest point on the surface,

3 1 34 4 42 5 ,5 , 5

− − −⎛ ⎞⋅ −⎜ ⎟

⎝ ⎠.

Extending that red line to the other branch of the surface intersects

that surface at the other closest point, 3 1 34 4 42 5 , 5 , 5

− − −⎛ ⎞− ⋅ −⎜ ⎟⎝ ⎠

.

27. You need to construct a tank consisting of a right circular cylinder with length h and radius r, capped with two hemispherical ends with radius r, as shown in the figure. If the material for the hemispherical ends cost $24/m2, and the material for the cylindrical side costs $12/m2, find the value of r and h that minimize the cost of the materials for this tank, assuming that the volume must be 1000π m3.

Then the ratio, ?2hr=

The two hemispheric ends have a combined area of is 24 rπ (the area of a sphere), so it will cost

( ) [ ]2 22

24 24 $ 9 $6r rmm

π π⎡ ⎤⎡ ⎤⎣ ⎦ ⎢⎛ ⎞

=⎜ ⎟⎝ ⎠⎥⎣ ⎦

. The area of the cylindrical side is 2 rhπ , so it will cost

( ) [ ]222 12 24$ $rh rhm

mπ π⎡ ⎤⎡ ⎤⎣ ⎦ ⎢

⎛ ⎞=⎜ ⎟

⎝ ⎠⎥⎣ ⎦. We seek to minimize the overall cost function,

( )( ) ( )

296 24 24 4

, 24 4 ,

C r rh r r h

C r h r r h

π π π

π

= + = +

= +

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

( ) ( )

3 2 3 2 3 2

3 2 2

4 41000 1000 3000 4 33 3

, 4 3 4 3 3000

V r r h r r h r r h

G r h r r h r r h

π π π ⇒ ⇒= = + = + = +

= + = + =

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

2

2

8 4 212 624 & 3

3r h r hr rh

C G rr rr

π+ +⎡ ⎤+⎡ ⎤ ⎡ ⎤

∇ = ∇ = =⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

( ) ( )

We may assume that 0.

2

* *

8 8 824

8 6 2 24 8 6 2

8 24 3

r

r h r r h h h

r r r

π π ππ λ π λλ λ λ

ππ λλ

≠

↓

↑ ↑

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

=

⎫⇒ ⎪

⎪⎪⎬⎪⎪⇒⎪⎭

=

64 1624 48

64 16 64 32 64 3

22 2

2

32 32

h h

h h h h h h

h h

π ππ πλ λ

π π π π π πλ λ λ λ λ λ

π πλ λ

⎛ ⎞ ⎛ ⎞+ = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞+ = + + = + − = −⎜ ⎟⎝ ⎠

⇒=⇒ =

⇒

⇒ ⇒ ⇒

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

3232 2

82 162

hr

ππ λλ

π λ πλ

⎛ ⎞⎜ ⎟⎝ ⎠= = =⎛ ⎞⎜ ⎟⎝ ⎠

The length of the cylinder is twice the diameter of the hemispherical ends. This takes advantage of the fact that the tubular side of the cylinder costs half as much per unit area as the hemisphere’s material. We are done, as far as the quiz problem is concerned. But we will finish this off anyway… We know then that 4h r= . So our constraint equation becomes:

( ) ( )2 2 3 3

3 3

30004 3 3000 4 3 4 3000 16 300016

375 3755.723571 4 4 22.8942852

2

r r h r r r r r

r h r h

+ = +⇒ ⇒ ⇒

⇒

⋅ = = =

= ≅ = =⇒ ⇒ ≅

These quantities are in units of meters. The minimum possible cost is thus:

( ) ( ) ( )

[ ] [ ]

2

2

2 3

, 24 4 24 4 192

375192 192$ $19,759.9

4

$ 22min

C r h r

C

hr r r rr

r

π π π

π π

= + = + =

⎛ ⎞= = ≅⎜ ⎟⎜ ⎟

⎝ ⎠

__________________________________________________________________________________ 28. Suppose that p is on the level set defined by ( )F x eπ= . Assuming that the function F, 10 → , is

infinitely differentiable, and that the vector T is tangent to its level set ( )F x eπ= at p . Then ( ) ?T

D F p =

( ) ( ) 0T

D F p F p T= ∇ • = , since the gradient ( )F 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. __________________________________________________________________________________