the no-normie zone - 2017-10-09

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Anonymous

13:54

@BalarkaSen I need help with this vector calculus.Are you around?

Anonymous

I'm trying to prove $\nabla.\nabla^2\vec{F}=\nabla^2(\nabla. \vec{F})$ if $\nabla.\vec{F}=0$

Anonymous

Basically I wanted to prove curl(curl(curl(curl $\vec{F}$))) = $\nabla^4\vec{F}$

Balarka Sen

Ugh, I don't want to do this.

Anonymous

Is there any shortcut? :/

Balarka Sen

Wait, if you say $\nabla \cdot F = 0$, then the right hand side of what you wrote, $\nabla^2(\nabla \cdot F)$, is zero...?

Anonymous

13:59

Yes

Anonymous

If I can show the left side is 0 I'm done

Balarka Sen

I think the $\nabla \cdot (\nabla^2 F) = \nabla^2 (\nabla \cdot F)$ regardless of any gradient assumption on $F$ (it has to be $C^3$, of course).

Explicitly write it out.

Anonymous

Okay, trying

Anonymous

@BalarkaSen It's working using Clariut's. (Means it has to be $C^3$). Phew! Expanded and matched!

Balarka Sen

Excellent!

Anonymous

14:49

Q: Will $\phi_1=z-f(x,y)$ and $\phi_2=z-g(x,y)$ be orthogonal to each other at $(a,b,c)$?

If $z=f(x,y)$ and $z=g(x,y)$ are orthogonal at a point $(a,b)$ (and $z=c$ at that point) then will $\phi_1(x,y,z)=z-f(x,y)$ and $\phi_2(x,y,z)=z-g(x,y)$ also be orthogonal at the point $(a,b,c)$ ? Our professor used this result to solve a question but I'm not sure why the result is true.

Anonymous

@BalarkaSen Any ideas? ^

Balarka Sen

@Blue Uh, what does it mean for $\phi_1(x, y, z)$ and $\phi_2(x, y, z)$ to be orthogonal at $(a, b, c)$? Those are functions, not surfaces in $\Bbb R^3$.

Pretty sure you mean the level sets $\phi_1(x, y, z) = 0$ and $\phi_2(x, y, z) = 0$ are orthogonal, in which case the statement is tautologically true.

Anonymous

Oooooooooooooo. That clears my confusion. Got it!

Anonymous

16:37

@BalarkaSen Over here shouldn't it be $r^2drd(\theta)$ inside the integral instead of $rdrd(\theta)$ as we are converting to polar coordinates to parameterize the surface ? I added the timestamp (around 12 mins into the video)

Anonymous

Or am I missing something?

Balarka Sen

You are indeed misremembering. Do the change of variables calculation.

Anonymous

I mean when we convert from cartesian to polar don't we write $dxdy=rd\theta dr$ for the integrals?

Anonymous

$r$ is the value of the jacobian matrix determinant

Balarka Sen

We do.

Anonymous

16:42

Yes, but in that video they didn't add the $r$ for the jacobian

Anonymous

They just got $\sqrt{2}r$ from the cross product determinant

Balarka Sen

Oh, I didn't watch the video. Oh wow, yes, that is total crap

Anonymous

Ya, that's what I'm saying. Khan Academy usually doesn't make such errors. Strange

Balarka Sen

No. Wait a second.

They are integrating $\displaystyle \int z dS$

In polar coordinates, $z$ is $1 - r\cos(\theta)$

Anonymous

Yes, got it till there

Anonymous

16:45

Now dS is obtained from the cross product $(s,t)$ by $|\delta_s\times \delta_t|dsdt$ usually. But here they are changing to $(r,\theta)$

Balarka Sen

What is the parameterization they are doing? I think they're right, they're converting $dS$ to $rdrd\theta$ but the $\sqrt{2}$ comes from the speed of the parameterization

Actually let me see the video

Anonymous

@BalarkaSen $|\delta_r\times \delta_{\theta}|$ evaluates to $\sqrt{2}r$

Balarka Sen

So $S_3$ is a disk of radius, uh, $2$?

The top of the chopped off stump

Anonymous

The base has a radius of 1

Balarka Sen

Oh, it has diameter $\sqrt{2}$ I believe

Anonymous

16:52

So 45 degrees with that

Balarka Sen

No diameter, radius $\sqrt{2}$, right?

pls check my grade 7 geometry lmao

Anonymous

$\sqrt(2)\cos(45)=1$

Anonymous

Yes

Balarka Sen

@Blue Ok, now look.

I want to find out $\displaystyle \iint_{S_3} z dS$

Anonymous

Yup

Balarka Sen

16:56

Forget how $S_3$ is positioned in $\Bbb R^3$. Parameterize $S_3$ as the disk of radius $\sqrt{2}$; $(x, y) = (r\sin(\theta), r\cos(\theta))$ where $0 \leq \theta \leq 2\pi$ and $0 \leq r \leq \sqrt{2}$.

This is basically setting up polar coordinates on the plane that contains $S_3$.

Anonymous

@BalarkaSen Okay, so far so good

Balarka Sen

Use change of variables theorem. That means the integral is $\displaystyle \int_0^{2\pi} \int_0^{\sqrt{2}} (1 - r\cos(\theta)) r dr d\theta$.

Anonymous

Shouldn't $z$ be $0$ as we are setting up polar coordinates on the plane of $S_3$ itself?

Anonymous

I'm getting confused

Balarka Sen

Well, no, it's just height of each point in $S_3$.

Just what $z$ means

We're integrating over $S_3$, which is a slanted disk, not the disk on the xy-plane (which has height 0 everywhere)

Anonymous

17:04

@BalarkaSen I mean you parameterized the surface $S_3$ as $(r\sin(\theta),r\cos(\theta))$, right? So how can you take height from xy plane as $1-r\cos(\theta)$ ?

Anonymous

If you had parameterized the xy plane itself then you could have taken the height z to be $1-r\cos(\theta)$

Balarka Sen

Ah, you're right.

Thanks

So there's something to be taken care of. It's actually $(1 - r\cos(\theta)) \cos(\pi/4)$ I believe

No, no, no, I am confused.

We're integrating $f(x, y) = 1 - x$ over this weird ass $S_3$.

Anonymous

Let's think of it in a different way.Suppose we have a $r$ and a $\theta$ axis and the surface is $S(r,\theta)$

Balarka Sen

Actually I don't really want to think about it, I'm busy right now.

Anonymous

What does $dS/dr$ and $dS/d(\theta)$ give?

Balarka Sen

17:11

I don't know the original problem anyway

I feel like what they did was correct, you're misinterpreting

Like if do I change of variables on $\int z dS$

That'd give $\iint (1 - r\cos(\theta)) r dr d\theta$

The $\sqrt{2}$ factor comes from their limits being slightly different

Anonymous

I think they are already accounting for the increase in area when they do $|\delta_{r}\times \delta_{\theta}|$

Balarka Sen

No, the $|\delta_r \times \delta_\theta|$ is exactly the determinant of the Jacobian

that appears in change of variables

So it's definitely not $r^2$ like you say

But I am just confused where $\sqrt{2}$ comes from, but I have stopped thinking about it

Anonymous

@BalarkaSen Ah! Got it

Anonymous

Makes sense now!!

Anonymous

Thanks

Balarka Sen

17:30

I didn't help much so :P

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