Question

Calculus 3: simple-ish simplification

Answer 1

uploading a picture, one sec.

Answer 2

I was trying to solve for this but the radical is confusing me.

Answer 3

so lets for example do d/dy [ z/sqrt(x^2+y^2+z^2) ]
this is :
z * (-0.5) * 2y * (x^2+y^2+z^2)^(-3/2)

Answer 4

Im just unsure how to take the partials. Clarifying my question, haha

Answer 5

-zy/(x^2+y^2+z^2)^(3/2)

Answer 6

Oh I see.

Answer 7

Then since all the bases are common, we can factor all of them out.

Answer 8

what do you mean ? you have to differentiate them

Answer 9

you mean factor out after the differentiation ?

Answer 10

the \((x^2+y^2+z^2)^{3/2}\)

Answer 11

yes ok

Answer 12

but you need to differentiate each term with it

Answer 13

Yes, i get that :P

Answer 14

you cant just take it out before the differentiation because there are variables in it

Answer 15

yup! but wait that is just denominator I think you will get
-(yz+xy+xz) in the numerator

Answer 16

I was trying to make it simpler for myself instead of including it all within the vector notation, to pull it out as a common factor

Answer 17

Ok, so i'll write up what i have, or try to.

Answer 18

you should get 0 eventually.

Answer 19

...I was trying to figure that out on my own, but thanks, lol

Answer 20

sorry

Answer 21

So if all the components derivitate like the one you showed me an example of, can i also factor out the negative from the (-1/2)? I know the 2 is eliminated and al that other stuff.

Answer 22

yes but remember there are different signs for each differentiation in the curl

Answer 23

But we're left with...\[-\frac{1}{(x^2+y^2+z^2)^{3/2}}\] right?

Answer 24

the stuff we've factored out i mean!

Answer 25

you should do all the differentiation first
then factor out cause i see that you are getting confused

Answer 26

\[\sf -\frac{1}{(x^2+y^2+z^2)^{3/2}}\left< yz - zy~,~-(xz-zx)~,~(xy-yx)\right>\]

Answer 27

is that right?

Answer 28

yes

Answer 29

Then a scalar \(\cdot\) < 0 vector> = \(\hat 0\) right?

Answer 30

Or rather, just a 0 vector, lol.

Answer 31

yes the result of the curl is vector

Answer 32

Ack, Im a little confused on how to find the divergence now :( ...

Answer 33

divergence is easier.
you have to take the derivative of the x component with respect to x
then add the derivative of the y component with respect to y
and then add the same for z

Answer 34

Yeah, I understand what you have to do... but it's doing it that's a little... weird I guess.

Answer 35

what is your first term ?

Answer 36

\[\frac{\partial }{\partial x} \left(\frac{x}{(x^2+y^+z^2)^{1/2}}\right)\] of this....

Answer 37

yes show me what you got

Answer 38

Question is... do i simply use the product rule?

Answer 39

yes and treat x as the only variable

Answer 40

\[\frac{\partial}{\partial x} \left[x(x^2+y^2+z^2)^{-1/2}\right]\]\[= 1\cdot (x^2+y^2+z^2)^{-1/2} -\frac{1}{2}(x^2+y^2+z^2)^{-3/2}(2x)\]\[=\frac{x^2+y^2+z^ -x}{(x^2+y^2+z^2)^{3/2}}\]

Answer 41

I understood halfway through writing it, heh.

Answer 42

the bottom line should be
y^2+z^2 divided by (x^2+y^2+z^2)^(3/2)

Answer 43

I am not sure why the negative etween the z and x seems to be missing, or seems like it, but yeah

Answer 44

Bottom line?

Answer 45

of the d/dx part

Answer 46

Oh yes, that's what i was missing, thank you

Answer 47

so now you know how the d/dy and d/dz will be

Answer 48

and you know the answer ..?

Answer 49

Well analyzing this format, I know that for each component, i will multiply the numerator of the first fraction by \((x^2+y^2+z^2)\), so theonly thing that will change is the variable i'm taking the derivative of.

Answer 50

therefore I don't really need to solve for each derivative to understand that I will have:

Answer 51

\[=\frac{x^2+y^2+z^2 -x}{(x^2+y^2+z^2)^{3/2}} +\frac{x^2+y^2+z^2 -y}{(x^2+y^2+z^2)^{3/2}}+\frac{x^2+y^2+z^2 -z}{(x^2+y^2+z^2)^{3/2}} \] right?

Answer 52

no,
it is
-x^2 instead of -x

-y^2 instead of -y

-z^2 instead of -z

Answer 53

why are they squared now? when I solved for the partial w.r.t x, i got \[=\frac{x^2+y^2+z^2 -x}{(x^2+y^2+z^2)^{3/2}}\]

Answer 54

nvm

Answer 55

but this is wrong
i told you you should left with
y^2+z^2 divided by (x^2+y^2+z^2)^(3/2)

Answer 56

I found my mistake

Answer 57

good

Answer 58

I didn't multiply in the x that was my first function

Answer 59

ok

Answer 60

so you can simplify your answer now

Answer 61

\[= 1\cdot (x^2+y^2+z^2)^{-1/2} -\frac{1}{2}(x^2+y^2+z^2)^{-3/2}(2x)(x)\] yeah.

Answer 62

And sorry if I'm a bit slow atm, it's 5:30 AM here, heh.

Answer 63

oh lol go get some sleep

Answer 64

so what do you get ?

Answer 65

\[=\frac{y^2+z^2 }{(x^2+y^2+z^2)^{3/2}} +\frac{x^2+z^2}{(x^2+y^2+z^2)^{3/2}}+\frac{x^2+y^2}{(x^2+y^2+z^2)^{3/2}}\]\[=\frac{2x^2+2y^2+2z^2}{(x^2+y^2+z^2)^{3/2}}\]

Answer 66

good which is..

Answer 67

it's 2 over the square root of the botton function.

Haha sorry, you're eager to go. I got it xD

Answer 68

Thank you though! You are quite helpful.

Answer 69

dont want to be rude but i have to leave lol

Answer 70

so it is correct

Answer 71

No i know what you mean, lol. Take care.

Answer 72

ty you too. have good time

Answer 73

still need help ?