Question

partial differentiation

Answer 1

\[z= \frac{-y}{x^2 + y^2}\\ z_y=?\]

Answer 2

substitute to polar

\[x=rcos \theta ~ y= rsin \theta \\ z=\frac{-\sin \theta}{r}\]

chain rule for partial derivatives:
\[\frac{\partial z}{\partial y}=\frac{\partial z}{\partial \theta} \times \frac{\partial \theta}{\partial y} + \frac{\partial z}{\partial r} \times \frac{\partial r}{\partial y}\]

\[\frac{\partial z}{\partial \theta}=\frac{-\cos \theta}{r} \\ \frac{\partial \theta}{\partial y}= (rcos \theta)^{-1} \\ \frac{\partial z}{\partial r}=\frac{\sin \theta}{r^2} \\ \frac{\partial r}{\partial y}= (\sin \theta)^{-1}\]

Answer 3

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

Answer 4

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

Answer 5

\[z_y= \frac{-1}{r^2}+\frac{1}{r^2}=0\]

Answer 6

@amity exactly! when i do it the regular way without polar, i get your answer. can you tell me where i'm going wrong?

Answer 7

Use the quotient rule and you'll get the answer amity gave.

Answer 8

yea.. i want to know what i've done wrong with polar @agent0smith

Answer 9

Is \[\frac{\partial r}{\partial y} = sin \theta\]

Answer 10

1/sin( )

Answer 11

Great you broke calculus

Answer 12

:o

Answer 13

Oh, I misread your above post. All your partial derivatives look correct. All your math looks right. I might have to actually get some paper and try it myself, I'm checking it all in my head.

Answer 14

\[\frac{\partial r}{\partial y} \sqrt{x^2+y^2} = \frac{1}{2} \frac{1}{\sqrt{x^2+y^2}} *2 y=\frac{rsin\theta}{r}\]

Answer 15

\[y=rsin \theta\\ \frac{\partial y}{\partial r}=\sin \theta \\ \frac{\partial r}{\partial y}= 1/\sin \theta\]

Answer 16

I wonder if there's some kind of dividing by zero thing going on or what lol

Answer 17

z is undefined at x=y=0
anything to do with that?

Answer 18

\[\frac{\partial y}{\partial r} = \frac{1}{\frac{\partial y}{\partial r}}\] but dr/dy is holding x constact not theta

Answer 19

∂y∂r=1∂y∂r
but dr/dy is holding x constant not theta. I think.

Answer 20

i'm considering y to be a function of only r and theta

i.e y=rsin( theta)

Answer 21

r is a function of x,y

Answer 22

Yeah, I'm thinking something funny happens somehow when you divide by r but I can't seem to pin it down. Just looking at the surface itself we know \(z_y \ne 0\)

Answer 23

ya, \[y_z \ne 0\]. and we all believe the differentiation not using polar.

Answer 24

@ganeshie8 @dan815 @ParthKohli @Astrophysics

Answer 25

is the math correct? as @Redcan points out, x needs to be held constant throughout

Answer 26

i get the feeling what i've calculated is
\[z=f(y, \theta) \\ z_y=0\]

Answer 27

i.e, y and theta are the independent variables,

Answer 28

Sounds believable... but I don't know how to think about verifying that thought.

Answer 29

@redcan ur right :)
i've calculated dr/dy holding theta constant and d(theta)/dy holding r constant

Answer 30

http://wwwf.imperial.ac.uk/~jdg/AECHAIN.PDF this might help, page 2.

Answer 31

the correct way would have been
\[z=\frac{-\sin \theta}{r} \\ r=(x^2+y^2)^{1/2}\\ \theta=\tan^{-1}\frac{y}{x}\]

Answer 32

There you go.

Answer 33

:D thanks a lot everyone!!

Answer 34

thanks @agent0smith , i messed up the chain rule :P