Question

Someone please help me figure this out

Answer 1

\[Y^{-2}_{2}(\theta,\psi)=\sqrt{\frac{ 15 }{ 32\pi }}e^{-2i \psi}(\sin)^{2}\]

Answer 2

is a spherical harmonic and describes a d-orbital
find an expression for:

Answer 3

\[(\frac{ dY^{-2}_{2} }{ d \psi})_{\theta}\]
and
\[(\frac{ dY^{-2}_{2} }{ d \sin(\theta)})_{\psi}\]

Answer 4

\[(\frac{ dY^{-2}_{2} }{ d \psi})_{\theta}=-2i\sqrt{\frac{ 15 }{ 32\pi }}e^{-2i \psi}(\sin)^{2}\]

Answer 5

treating sin(theta)=x
\[(\frac{ dY^{-2}_{2} }{ d \sin(\theta)})_{\psi}=\sqrt{\frac{ 15 }{ 32\pi }}e^{-2i \psi}2(x)\]
\[\rightarrow (\frac{ dY^{-2}_{2} }{ d \sin(\theta)})_{\psi}=\sqrt{\frac{ 15 }{ 32\pi }}e^{-2i \psi}2(sin(\theta))\]

Answer 6

Sorry I can't help, but what math is this?

Answer 7

it is called "mathematics for physical chemistry"

Answer 8

@zepdrix @satellite73 hey do you guys know how to partial derivatives I'm still learning and could use the help

Answer 9

ideas?

Answer 10

I don't understand this notation,\[\large\rm \left(\frac{ dY^{-2}_{2} }{ d \psi}\right)_{\theta}\]Are you taking a full derivative with respect to phi?
and a partial with respect to theta?

Answer 11

it means the derivative with respect to psi while keeping theta constant.

Answer 12

You just described a partial derivative.
So you meant to take partials of Y with respect to both?
Like this?\[\large\rm \left(\frac{ \partial Y^{-2}_{2} }{ \partial \psi}\right)_{\theta}\]Why use mixed notation then?
I'm still really confused...

Answer 13

yeah its a partial derivative. i was told to just make theta a constant and "carry it along" I'm just learning this so I'm not one to provide much clarity =(

Answer 14

\[\large\rm Y^{-2}_{2}=\sqrt{\frac{ 15 }{ 32\pi }}e^{-2i \psi}\sin^{2}\theta\]Ugh this is making no sense :(

If you were trying to do this,\[\large\rm \left(Y^{-2}_{2}\right)_{\phi}=-2i\sqrt{\frac{ 15 }{ 32\pi }}e^{-2i \psi}\sin^{2}\theta\]Then yes, you did that properly.
You treated the entire square root, and the sine stuff as constant, and just applied your exponent derivative rule.

Answer 15

But I still don't understand what that derivative notation is doing on the inside...

Answer 16

i dont understand it either it looks like a mess to me. what about the second one?

Answer 17

Oh I wrote the wrong greek letter :D haha sorry

Answer 18

\[\large\rm \left(Y^{-2}_{2}\right)_{\psi}=-2i\sqrt{\frac{ 15 }{ 32\pi }}e^{-2i \psi}\sin^{2}\theta\]

Answer 19

i didn't notice lol

Answer 20

i dropped the sign and i in mine... humm

Answer 21

Oh :x

Answer 22

so is that how you take a derivative in respect to sintheta? it just looks so wrong to me

Answer 23

Let's back up a second...
this \(\large\rm \left(\frac{ dY^{-2}_{2} }{ d \sin\theta}\right)_{\psi}\) is not the same as this \(\large\rm \left(Y^{-2}_{2}\right)_{\psi}\)

So if they want us to first take a full derivative, with respect to sin(theta), then let's do that.

Answer 24

\[\large\rm \frac{d Y^{-2}_{2}}{d \sin \theta}=\color{royalblue}{\left(\sqrt{\frac{ 15 }{ 32\pi }}e^{-2i \psi}\right)'}\sin^{2}\theta+\sqrt{\frac{ 15 }{ 32\pi }}e^{-2i \psi}\color{royalblue}{\left(\sin^{2}\theta\right)'}\]Product rule, ya?

Answer 25

i follow

Answer 26

\[\large\rm \frac{d Y^{-2}_{2}}{d \sin \theta}=\color{orangered}{\left(\sqrt{\frac{ 15 }{ 32\pi }}e^{-2i \psi}(-2i)\frac{d \psi}{d \sin \theta}\right)}\sin^{2}\theta+\sqrt{\frac{ 15 }{ 32\pi }}e^{-2i \psi}\color{royalblue}{\left(\sin^{2}\theta\right)'}\]This is going to give us a little bit of weird garbage though :D
So I dunno if this is the route they intended lol.
We end up with this psi' term, ya?

I've only differentiated the first term so far, hence the color change.

Answer 27

I see what you did but i'm not 100% on why you did it. I'm getting confused because this isn't how I was taught =(. can I show you the example I was given on how to do this?

Answer 28

Sure, that might help.

Answer 29

example:
the 2p orbital is described by the wave function:
\[\psi_(r,\theta,\phi)=\frac{1}{\sqrt{32a^{5}_{B}}}re^{\frac{-r}{2a_{B}}}sin(\theta)cos(\phi)\]
calculate:
\[(\frac{\partial\psi}{\partial\theta})_{r,\phi}\]
and
\[(\frac{\partial\psi}{\partial r})_{\theta,\phi}\]

Answer 30

answer:
\[\psi_(r,\theta,\phi)=Cre^{\frac{-r}{2a_{B}}}sin(\theta)cos(\phi)\]
\[(\frac{\partial\psi}{\partial\theta})_{r,\phi}=Cre^{-\frac{r}{2_{B}}}cos(\phi)cos(\theta)\]
\[(\frac{\partial\psi}{\partial r})_{\theta,\phi}=C(sin(\theta))(cos(\phi))[e^{\frac{-r}{2_{B}}}+r(\frac{-1}{2}a_{B})e^{\frac{-r}{2_{B} }}]\]

Answer 31

Woah, why did the stuff in the exponent change?

Answer 32

Where did the a go?

Answer 33

=( its supposed to be there still i dont know what happened

Answer 34

Oh ok

Answer 35

Why is there a C in the final answers?
That doesn't make any sense.
What is C?

Answer 36

Oh you renamed the ugly fraction at the start to be C.
Ok fair enough

Answer 37

yes lol and here:
\[(\frac{\partial\psi}{\partial r})_{\theta,\phi}=C(sin(\theta))(cos(\phi))[e^{\frac{-r}{2a_{B}}}+r(\frac{-1}{2}a_{B})e^{\frac{-r}{2a_{B} }}]\]

Answer 38

This isn't making sense to me :( Grrr
Like... here, the first one...\[\large\rm \left(\frac{\partial\psi}{\partial\color{orangered}{\theta}}\right)_{r,\phi}=Cre^{-\frac{r}{2_{B}}}\cos\phi\color{orangered}{\cos\theta}\]Ok sin(theta) turned into cos(theta), good.\[\large\rm \left(\frac{\partial\psi}{\partial\theta}\right)_{\color{orangered}{r},\phi}=C\color{orangered}{re^{-\frac{r}{2_{B}}}}\cos\phi\cos\theta\]Partial with respect to r gives us the same thing back for the exponential with -1/2a_b coming down in front.
Presumably this was absorbed into the C maybe, but where in the world is this r coming from?\[\large\rm \left(\frac{\partial\psi}{\partial\theta}\right)_{r,\color{orangered}{\phi}}=Cre^{-\frac{r}{2_{B}}}\color{orangered}{\cos\phi}\cos\theta\]And then the partial with respect to phi,
why aren't we ending up with -sin(phi) ??

This makes no seeeeeeeense :((

Answer 39

Thats where we are doing things different i was told to treat r an phi as constants. I'm not supposed to take the partial with respect to them

Answer 40

Oh oh, ok so that's what this weird notation means maybe.
But that still doesn't explain the r in front. Hmm.

Answer 41

Generally speaking, at least in math classes,
\(\large\rm \dfrac{\partial y}{\partial x}\) means the same thing as \(\large\rm y_x\)

So maybe you can see why this new notation is really bonkers.

Answer 42

you look like you are doing mixed partial derivatives which would look like this:
\[\frac{\partial\psi}{\partial\theta\partial r\partial\phi}\]

Answer 43

Oh the r was there the entire time :(
I wrote the problem down wrong woops t.t

Answer 44

its completely strange to me too. i learned this friday

Answer 45

Ok this is making a lot more sense then :)

Answer 46

\[\large\rm Y^{-2}_{2}(\theta,\psi)=\sqrt{\frac{15}{32\pi}}e^{-2i\psi}\sin^{2}\theta\]

Answer 47

yes so i was told to split that into two partials

Answer 48

When you're going to take the partial with respect to psi,
you can think of the function like this sort of,\[\large\rm Y^{-2}_{2}(\theta,\psi)=Ce^{-2i\psi}\]The exponential is the only thing you're differentiating.\[\large\rm \frac{\partial Y}{\partial \psi}=-2i Ce^{-2i\psi}=-2i\sqrt{\frac{15}{32\pi}}e^{-2i\psi}\sin^{2}\theta\]Treating the sin^2theta as constant in the process.

Answer 49

This is a tough problem to refresh on partials lol.
You should have covered them a little bit in Calc 3 :D

Answer 50

never did calc 3 D= i'm in physical chemistry math

Answer 51

Oh dear :o

Answer 52

yupp i got the same for the first one

Answer 53

Your other one looks good also.\[\large\rm \frac{\partial Y}{\partial \sin \theta}=\sqrt{\frac{15}{32\pi}}e^{-2i\psi}2\sin\theta\]Power rule on your sine,
and no chain rule because the the argument of our differentiation matches the next chain, ya?

You remember simple problems like this right?\[\large\rm \frac{d}{dx}\left(x\right)^2=(x)^1\frac{d}{dx}x\]You COULD chain like this ^
But notice the inner function matches the argument of differentiation.
So this dx/dx that we end up with is meaningless.
We always skip that step.

Same thing here, you end up with a d sin(theta)/d sin(theta) when you chain.

Answer 54

So all of that arguing and nonsense just to say, "yes you're right"
lol XD

Answer 55

well sweet lol

Answer 56

thank you

Answer 57

np :U

go eat a butterfinger, you've earned it.
Or snap into a slim jim.. or something

Answer 58

^I love food. Good job zep.