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key - Montana State Universityy Ey o y l E o y o or 7 1 2 If y o then X o so o o But g 0,0 0 So this...

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key Let p 2 1,0 Then Pfp f l 2 5 X 2 2 gti SZ Q Php h o means that a small charge in the positive x direction won't change the altitude h_y E o means that a small change in the positive y direction will decrease the attitude
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Page 1: key - Montana State Universityy Ey o y l E o y o or 7 1 2 If y o then X o so o o But g 0,0 0 So this is extraneous If 7 1 2 this y It x If y then tfX t IX 1 and so x z and y i and

key

Let p 2 1,0 Then Pfp f l 2 5

X 2 2 gti SZ

Q

Php

h o means that a smallcharge in thepositive x

direction

won't change the altitude

h_yE o means that a smallchange in the positive y direction

will decrease the attitude

Page 2: key - Montana State Universityy Ey o y l E o y o or 7 1 2 If y o then X o so o o But g 0,0 0 So this is extraneous If 7 1 2 this y It x If y then tfX t IX 1 and so x z and y i and

Let p no 2 E ThenXCpt 02Y fp ytzlp tt23gep xtz pTyCp12

2 P T 2p xty p

2 3 1,0 00504 0

ps.nop 12Fr p

o p EE

1 i i I

Page 3: key - Montana State Universityy Ey o y l E o y o or 7 1 2 If y o then X o so o o But g 0,0 0 So this is extraneous If 7 1 2 this y It x If y then tfX t IX 1 and so x z and y i and

to Approach along the line y mx wherem E IR

E.im i eimoDN The limit does notexist

because the value depends on M theSlope of the line e gIf 10,0 is approached on the y o then the answer is 1 but if

approachedalong 5 x then the value is Lz Thus differentpathsproducedifferent values i.e the limit is path dependent and therefore doesnot

The expression 4 t x yexist

is defined at3 t X 3gSo direct substitution in

fin It I 4xg in

Define F Z 3 5 y 2 3 Then 3 F

zwere F 3 22 35 and Fz X 3g Z Thus

3 2z 3y222

Page 4: key - Montana State Universityy Ey o y l E o y o or 7 1 2 If y o then X o so o o But g 0,0 0 So this is extraneous If 7 1 2 this y It x If y then tfX t IX 1 and so x z and y i and

f 2x 2g fy y 2x 3 f y 2 fyf 2 fyg 2g D 4g 4 4cg l

f and fy are continuous on 1122 so

ZX 2g omg 2 3 0

y x x 2 2 3 0

X 3 Xtc O x 3 or x I

3,3 or f l l

D 3,3 8 O and f o localminimum at 3,3

D 1 1 8 0 saddle at CI

Page 5: key - Montana State Universityy Ey o y l E o y o or 7 1 2 If y o then X o so o o But g 0,0 0 So this is extraneous If 7 1 2 this y It x If y then tfX t IX 1 and so x z and y i and

fly g Xy e objective function gcx.us fx2ttzy2 i oconstraintfunction

f 77gTy x 7T x y

y Fx x ay y f ayy Ey o

y l E o y o or 7 1 2

If y o then X o so o o But g 0,0 0 So this

is extraneous

If 7 1 2 this y It x

If y then tfX t IX 1 and so x z and y i

and therefore t2 Il

If y Ix then IX t x and so 2 andy I 1

and therefore t2 Il

f tz ti It 2 ti 2 Absolute maximum of 2 at Iz l

f t2 Il I 2 Ii 2 Absolute minimum of 2 at t2 Il

t

Page 6: key - Montana State Universityy Ey o y l E o y o or 7 1 2 If y o then X o so o o But g 0,0 0 So this is extraneous If 7 1 2 this y It x If y then tfX t IX 1 and so x z and y i and

f 4 3 32 9 23g

Pf f4X 2gOf 4 3 f 16 6

Let it f IsDef Tf I

file 6 of Is E6 4ft

Pf 4,3 f 16 6

f 16 6L

Page 7: key - Montana State Universityy Ey o y l E o y o or 7 1 2 If y o then X o so o o But g 0,0 0 So this is extraneous If 7 1 2 this y It x If y then tfX t IX 1 and so x z and y i and

The plane 2 Stax toy has normalvector T f 6,10 1

Define F x Y Z Z 2 y Then the surface 2 1 x2tyzis equivalent to the implicit surface 1 19,21 1 This

surface has normal Vector TF If the target place to

Flayz L is to beparallel to 2 St Cox toy then DF must

be parallel to rt that is they must be scalarmultiples of one

another Hence

FF IFZ It 3 C sf 2x 2g 1 7 6,10 1

Thus 2 1 and so3 and y 5 andtherefore the point is 3 5,35

2 It 7 x l tog

O9 It 25in o 2 I l t 7 09 1 26.2I 0.7 to 4

0.7g


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