Question

Given w=f(x,y) x=rcos(theta) y=rsin(theta)
a) compute 2w/2x and 2w/2y
b)Prove that...

Answer 1

\[(\frac{ 2w }{ 2x })^2+(\frac{ 2w }{ 2y })^2=(\frac{ 2w }{ 2r })^2+(\frac{ 1 }{ r^2 })(\frac{ 2w }{ 2theta })^2\]

Answer 2

do u mean partials ?

Answer 3

yes

Answer 4

you may use \(\large \color{Red}{\text{\partial}}\) for showing \(\large \partial \)

Answer 5

I don't know how to start the problem though

Answer 6

start by finding each partial

Answer 7

\(\large \frac{\partial w}{\partial r} = ? \)


\(\large \frac{\partial w}{\partial \theta} = ? \)

Answer 8

\(\large \frac{\partial w}{\partial r} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial r} \)

\(\large \qquad= \frac{\partial w}{\partial x}\cos\theta + \frac{\partial w}{\partial y}\sin\theta \)

Answer 9

similarly find \(\large \frac{\partial w}{\partial \theta} \), and plug them in ur right hand side of the equation

Answer 10

and try to simplify it to get the left hand side

Answer 11

Is it \[\frac{ 2w }{ 2 \theta }=\frac{ 2w }{ 2x }\frac{ 2x }{ 2 \theta }+\frac{ 2w }{ 2y }\frac{ 2y }{ 2 \theta }\]?

Answer 12

thats right, and since \(\large x = r \cos \theta\) and \(\large y = r\sin \theta\)
just evaluate these partials and plug in above...

Answer 13

Is it
\[\frac{ 2w }{ 2 \theta }=\frac{ 2w }{ 2x }(-rsin \theta)+\frac{ 2w }{ 2y }(rcos \theta)\]?

Answer 14

looks perfect !

Answer 15

So then do I solve and see if I get the equation from the one I first posted?

Answer 16

To prove it?

Answer 17

just plug these on right hand side

Answer 18

Could you set it up, I'm still kind of confused

Answer 19

Right hand side of the equation you wanted to prove :


\(\large \left(\frac{\partial w}{\partial r} \right)^2 + \frac{1}{r^2}\left(\frac{\partial w}{\partial \theta} \right)^2 \)

Answer 20

ganesh=wrong... don't listen hehe

Answer 21

So plug the whole answers into 2w/2r and 2w/2theta to get the left hand side?

Answer 22

we have a tough guest here :/

Answer 23

yes !

Answer 24

ok ill give it a try

Answer 25

right after plugging in, most terms cancel out and take u to left hand side
should be easy...

Answer 26

good luck !

Answer 27

One question, how do I find 2w/2x and 2w/2y themselves? Or would 2w/2r and 2w/2theta be good enough?

Answer 28

leave them as they are.. you dont have them on ur right hand side, eh ?

Answer 29

Right hand side :

\(\large \left(\frac{\partial w}{\partial r} \right)^2 + \frac{1}{r^2}\left(\frac{\partial w}{\partial \theta} \right)^2 \)


\(\large \left(\frac{\partial w}{\partial x}\cos\theta + \frac{\partial w}{\partial y}\sin\theta \right)^2 + \frac{1}{r^2}\left(\frac{\partial w}{\partial x}(-r\sin\theta) + \frac{\partial w}{\partial y}(r\cos\theta)\right)^2 \)

Answer 30

That's what I got so far, I'm just squaring it right now, is it suppose to be long?

Answer 31

nope, next step is ur answer

Answer 32

Right hand side :

\(\large \left(\frac{\partial w}{\partial r} \right)^2 + \frac{1}{r^2}\left(\frac{\partial w}{\partial \theta} \right)^2 \)


\(\large \left(\frac{\partial w}{\partial x}\cos\theta + \frac{\partial w}{\partial y}\sin\theta \right)^2 + \frac{1}{r^2}\left(\frac{\partial w}{\partial x}(-r\sin\theta) + \frac{\partial w}{\partial y}(r\cos\theta)\right)^2 \)



\(\large \left(\frac{\partial w}{\partial x}\cos\theta + \frac{\partial w}{\partial y}\sin\theta \right)^2 + \frac{1}{r^2}\times r^2\left(\frac{\partial w}{\partial x}(-\sin\theta) + \frac{\partial w}{\partial y}(\cos\theta)\right)^2 \)

Answer 33

^^r^2 cancels out

Answer 34

where did you get the r^2 from?

Answer 35

Oh wait nvm you factored the r^2 out

Answer 36

Right hand side :

\(\large \left(\frac{\partial w}{\partial r} \right)^2 + \frac{1}{r^2}\left(\frac{\partial w}{\partial \theta} \right)^2 \)


\(\large \left(\frac{\partial w}{\partial x}\cos\theta + \frac{\partial w}{\partial y}\sin\theta \right)^2 + \frac{1}{r^2}\left(\frac{\partial w}{\partial x}(-r\sin\theta) + \frac{\partial w}{\partial y}(r\cos\theta)\right)^2 \)



\(\large \left(\frac{\partial w}{\partial x}\cos\theta + \frac{\partial w}{\partial y}\sin\theta \right)^2 + \frac{1}{r^2}\times r^2\left(\frac{\partial w}{\partial x}(-\sin\theta) + \frac{\partial w}{\partial y}(\cos\theta)\right)^2 \)


\(\large \left(\frac{\partial w}{\partial x}\cos\theta + \frac{\partial w}{\partial y}\sin\theta \right)^2 +\left(\frac{\partial w}{\partial x}(-\sin\theta) + \frac{\partial w}{\partial y}(\cos\theta)\right)^2 \)


\(\left(\frac{\partial w}{\partial x} \right)^2\cos^2\theta + \left(\frac{\partial w}{\partial y} \right)^2\sin^2\theta + \left(\frac{\partial w}{\partial x} \right)^2\sin^2\theta + \left(\frac{\partial w}{\partial y} \right)^2\cos^2\theta \)

Answer 37

2ab terms cancel away because of opposite signs.

Answer 38

you're left only with the square terms... see if you're okay so far

Answer 39

Yep I think I got it right so far, do I factor out the sin and cos?

Answer 40

factor the partials instead so that we can use sin^2 + cos^2 = 1

Answer 41

group 1st and 3rd terms

Answer 42

\[(\frac{ 2w }{ 2x })^2(\cos^2 \theta+\sin^2 \theta)+(\frac{ 2w }{ 2y })^2(\cos^2 \theta+\sin^2 \theta)\]Right?

Answer 43

yes !

Answer 44

then replace cos^2+sin^2 to 1 to get the final answer. Right?

Answer 45

Thanks!

Answer 46

u wlc :)