Question

Help.....calculus III

Answer 1

1) Show that the function F (X, Y) = ln (sqrt (x ^ 2 + y ^ 2)) satisfies the Laplace equation in two-dimensional

  ((& PartialD,) ^ (2) F) / (& PartialD, x ^ (2)) + ((& PartialD,) ^ (2) F) / (& PartialD; ^ y (2)) = 0

Answer 2

@ganeshie8

Answer 3

@paki

Answer 4

@UnkleRhaukus

Answer 5

@nincompoop

Answer 6

@matricked

Answer 7

@fabiomartins in which grade u are?

Answer 8

@hartnn

Answer 9

@phi

Answer 10

where are you stuck ?

first find
\(\Large \dfrac{\partial }{\partial x}(\ln \sqrt{x^2+y^2})\)

means differentiate that treating y as constant
what do u get ?

Answer 11

once done, differentiate the answer once more w.r.t x treating y as constant
and you will get the first term
" ((& PartialD,) ^ (2) F) / (& PartialD, x ^ (2))"

Answer 12

x/(x^2+y^2) ??????????????????

Answer 13

thats the first partial derivative, and its correct! :)
go for one more derivative

Answer 14

yes, that is the first derivative. do it again

Answer 15

THEN THAT MY ANSWER IS CORRECT? WHAT DO I DO NOW?

Answer 16

differentiate
\(\dfrac{x}{x^2+y^2}\)
w.r.t x, treating y as constant,
to get
\(\dfrac{\partial^2 F}{\partial x^2}\)

Answer 17

using quotient rule

Answer 18

our, now I can not :(

Answer 19

where are you getting stuck ?
you know quotient rule right ?

Answer 20

\(\Large \dfrac{(x^2+y^2)(x)'-x(x^2+y^2)'}{(x^2+y^2 )^2} = ... \)

Answer 21

you could also use the product rule after writing it as
x*(x^2+y^2)^(-1)

Answer 22

(x(x)^2+y(x)^2-x(x^2+y^2))/(x^2+y^2)^2

@hartnn

Answer 23

\((x^2+y^2)'\)
means partial derivative of x^2+y^2
which will just be 2x

Answer 24

x'=1

\(\Large \dfrac{(x^2+y^2)(x)'-x(x^2+y^2)'}{(x^2+y^2 )^2} =\dfrac{(x^2+y^2)(1)-x(2x)}{(x^2+y^2 )^2}=...\)

Answer 25

=...?

Answer 26

(x(1)^2+y(1)^2-x(2*x))/(x^2+y^2)^2

Answer 27

?

Answer 28

i don't understand what you have done...

Answer 29

you never used quotient rule ?
if not try phi's method which uses product rule.

Answer 30

\((x^2+y^2)(1)=x^2+y^2\)

and not \(x(1)^2+y(1)^2\)

Answer 31

\(\Large \dfrac{\partial ^2 F}{\partial x^2} =\dfrac{(x^2+y^2)(x)'-x(x^2+y^2)'}{(x^2+y^2 )^2} \\ \Large =\dfrac{(x^2+y^2)(1)-x(2x)}{(x^2+y^2 )^2}=\dfrac{y^2-x^2}{(x^2+y^2)^2}\)

now similarly find
\(\dfrac{\partial ^2 F}{\partial y^2}\)

Answer 32

and then add them, and you will be done

Answer 33

so now what do I do more? which equation should I solve? or over?

Answer 34

@hartnn

Answer 35

could you find
\(\dfrac{\partial ^2 F}{\partial y^2}\)
using same steps ??
what did you get ?

Answer 36

I can not make that step :(

Answer 37

why not ?
which step are you stuck ?

Answer 38

please give us inputs on what you have tried or why you are not able to try ?
just saying 'I cannot do' , then we won't be able to help u

Answer 39

can make the next step? hence why I try again ..

Answer 40

First, what did you get for
d^2 F/dx^2 ?

Answer 41

d^2*F/dx^2

Answer 42

yes. I know you found
\[ \frac{\partial F}{\partial x} = x (x^2+y^2)^{-1}\]
what did you get for
\[ \frac{\partial^2 F}{\partial x^2} \]

Answer 43

???

Answer 44

this looks *almost* correct
(x(1)^2+y(1)^2-x(2*x))/(x^2+y^2)^2
if you write it as
(x^2+y^2-x(2*x))/(x^2+y^2)^2
that is correct

simplify to
(x^2+y^2-2x^2))/(x^2+y^2)^2

(-x^2 +y^2)/(x^2+y^2)^2

Answer 45

dF/dy is the same problem (with x and y swapped)

Answer 46

and d^2 F / dy^2 is also the same problem

Answer 47

but now I'll have to do more for to finalize this issue?

Answer 48

yes. so far you have
d^2 F / dx^2 =(-x^2 +y^2)/(x^2+y^2)^2
you need to find
d^2 F / dy^2 (which is *very* similar to finding d^2 F / dx^2 , which you know how to do)

Answer 49

I'm kinda confused, is the first step.

Answer 50

@phi

Answer 51

You are showing
\[ \frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2} = 0 \]

the partial derivative of F with respect to x means find the derivative of F but treat y as a constant.

the partial derivative of F with respect to y means find the derivative of F but treat x as a constant.

Answer 52

you need to find the first derivative of F with respect to y (while treating x as a constant)

can you find a derivative?

Answer 53

I'm kinda confused, you can finish this question for me? and then explain to me step by step? but does it all unless I get a little lost.

Answer 54

@phi

Answer 55

If you don't know how to take a derivative, you should not be studying Calculus III
Here is a video on partial derivatives.
http://www.khanacademy.org/math/multivariable-calculus/partial_derivatives_topic/partial_derivatives/v/partial-derivatives