Question

The last one for the day please

Answer 1

Anybody

Answer 2

k

Answer 3

Coming one sec.

Answer 4

No 2 please

Answer 5

Are you able to download the attachment.

Answer 6

i have no clue how to do that

Answer 7

Kk.

Answer 8

i need help

Answer 9

i will msg u

Answer 10

With ?

Answer 11

k

Answer 12

which one are you doing?

Answer 13

@xapproachesinfinity no 2 sir.

Answer 14

ok so you want domain of g(x,y) corrrect

Answer 15

yes please.

Answer 16

we are given that g(0,0)=0 so 0 is part of the domain

Answer 17

OK.

Answer 18

now we focus our attention of \[g(x,y)=\frac{x^3-x^2y}{x^2+y^2}\]
this undefined for x^2+y^2=0

Answer 19

so we solve for that equation to find the domain

Answer 20

y= plus or -x

Answer 21

\[x^2+y^2=0\] but realize that x^2+y^2 is always positive no matter what x and y are

Answer 22

the only case is when x=0 and y=0

Answer 23

its a polynomial so its always continuous.

Answer 24

?

Answer 25

but here they told us g(x,y) is defined for (0,0)
so the domain must be all (x,y)'s with no exceptions

Answer 26

this the case of removable discontinuity

Answer 27

yes poly are always continues, but this is not just a polynomial it is a rational ply
you need to check the bottom all the time

Answer 28

OK.

Answer 29

in set notation Domain: {(x,y): x in R and y in R}

Answer 30

there is no restriction

Answer 31

you good?

Answer 32

Yeah c and d part to no 2 please

Answer 33

oh hold 2 didn't ask for Domain did it?

Answer 34

No

Answer 35

ok so
we need to find lim g(x,y) at (00)

Answer 36

sorry i read 1 i thought they were related

Answer 37

that is for which letter c or d

Answer 38

for c

Answer 39

OK yes

Answer 40

did you find the limit?

Answer 41

i did b and lim is 1

Answer 42

Not sure thou

Answer 43

\[\lim_{(x,y) \to (0.0)} \frac{x^3-x^2y}{x^2+y^2}\]
we faced with a limit of several variable approaching a point can take many directions
we choose two at least and see if the limit does not exist or not

Answer 44

I approach from different line am getting 0 always

Answer 45

could it be that the problem is approaching and then we can use squeeze theorem

Answer 46

to prove it?

Answer 47

if you check two paths you will get 0 for both, but this is not sufficient to say the limit is 0
so we convert to polar
\[\lim_{r\to 0^+}\frac{r^3\cos^3\theta -r^2\cos^2\theta r\sin\theta }{r^2\cos^2\theta +r^2\sin^2\theta }\]

Answer 48

no can't use squeeze theorem

Answer 49

now we a single variable limit

Answer 50

ok

Answer 51

cleaning that limit to get
\[\lim_{r\to 0^+}r(\cos^3\theta -\cos^2\theta \sin \theta)=0 \]

Answer 52

limit depends on r only so it does not matter what theta is in this case

Answer 53

so limit is 0
and we know that g(0,0)=0
so g is contineous function

Answer 54

yes thanks
How will I go about solving D please

Answer 55

we just concluded C

Answer 56

for d we need to find gx first

Answer 57

using quotient rule

Answer 58

?

Answer 59

\[\frac{\partial }{\partial x}g(x,y)=\]

Answer 60

yes we need quotient rule

Answer 61

{(3x^2-2xy)-(2x^4-2x^3y)}/(x^2+y^2)

Answer 62

gx (0,0) won't exist anyway
you will have (x^2+y^2 )^2 in denominator
so gx is not continuous

Answer 63

you just substitute zero

Answer 64

you don't need to just say gx is undefined for (0,0)

Answer 65

sorry how about the a part

Answer 66

just wanna be sure

Answer 67

you mean gx?

Answer 68

2a

Answer 69

for all xy = 0

Answer 70

you part b

Answer 71

gx(0,0)

Answer 72

No part a question 2a

Answer 73

Calculate gx(x, y) for all (x, y) NOT EQUAL TO (0,0) . Simplify your answer

Answer 74

2a is what you did just now gx(x,y)

Answer 75

they asked you before finding continuity because you need it

Answer 76

oh I thought we just did D

Answer 77

d asked for continuity to find continuity you need gx(0,0) and gx(x,y)
you did those for a and b

Answer 78

how did you do gx(0,0)

Answer 79

which number

Answer 80

letter

Answer 81

2b

Answer 82

I used definition

Answer 83

x=h and y=0 and my lim=1?

Answer 84

\[\lim_{\Delta x \to 0}\frac{g(x+\Delta x ,y)-g(x,y)}{\Delta x }\]

Answer 85

so for part a) g(xy) is not continuous. because we have x^2+Y^2 at the denominator so why did it say simplify your answer.

Answer 86

yes that's what i used and lim =1? is it

Answer 87

g(x,y) is conitnuous we proved it

Answer 88

Oh my I am all confused here.

Answer 89

gx(x,y) i said it was not continuous but i didn't look at gx(0,0)
you should first find that
then you look at limit of gx(x,y) at (0,0)

Answer 90

you sure it is 1 the gx(0,0)=1?

Answer 91

yes

Answer 92

1

Answer 93

if gx is coninuos then limit gx =1 too
check that if it is true

Answer 94

but let me see your work for gx(0,0)

Answer 95

lim f(0+h, 0) - f(00)/h

Answer 96

f(h,0)-f(0)/h

Answer 97

h-0/h