Ummm... help?[Physics and Vector calculus involved]

[heartslaught]

I need someone to help me (this is NOT HOMEWORK, it's stuff on my own time) with this problem (4.18b out of Taylor's classical mechanics), deriving the surface normal formula:

(b) The direction of del(f) (f being a scalar function) at any point r (r being a position vector) is the direction in which f increases fastest as we move away from r. [Choose a small displacement dr = epsilon*u where u is a unit vector and epsilon is fixed and small. Find the direction of u for which the corressponding df is maximum, bearing in mind a dot b = mag(a)*mag(b)*cos(theta).]

I know the normal should be around del(f)/mag(del(f)) I just don't know how to formalize my intuition.

Thanks in advance!

[Woozie]

I need someone to help me (this is NOT HOMEWORK, it's stuff on my own time) with this problem (4.18b out of Taylor's classical mechanics), deriving the surface normal formula:

(b) The direction of del(f) (f being a scalar function) at any point r (r being a position vector) is the direction in which f increases fastest as we move away from r. [Choose a small displacement dr = epsilon*u where u is a unit vector and epsilon is fixed and small. Find the direction of u for which the corressponding df is maximum, bearing in mind a dot b = mag(a)*mag(b)*cos(theta).]

I know the normal should be around del(f)/mag(del(f)) I just don't know how to formalize my intuition.

Thanks in advance!

I'm not sure if I completely understand what you're asking. Are you asking "If we have a given surface, what is the formula for a vector normal to the surface a point r or are you asking something about the gradient vector?

Edit: Wait, I think I understand. Give me a minute.

[heartslaught]

Yeah, I had the same problem decrypting his verbosity, which he can have (the author) sometimes.

http://books.google.com/books?id=P1k...sult#PPA152,M1

That's the book via google books, the text of the problem is there in the preview, so you get a better idea Woozie.

[Woozie]

Okay, if you're asking about the formula for the normal vector at a given point of a surface.

Let the surface be described parametrically by:

x=f(s,t)
y=g(s,t)
z=h(s,t)

Then you take the vector who's three components are the partial derivatives with respect to the first variable, and find it's cross product with the vector who's components are the partial derivatives with respect to the second variable.

So the first vector would be

V1=<df/ds,dg/ds,dh/ds> [[the derivative of f with respect to s, the derivative of g with respect to s, then h with respect to s. Sorry, it's so hard to make notation look clear on these forums x_x)

V2 would be the same, but with the derivatives with respect to t instead of s.

The normal vector to the surface is then V1xV2.

[Woozie]

Oh okay, that helps, let me read the problem

[Woozie]

I see what they're asking now. They want you to prove that the gradient is the direction of maximum increase of a scalar. This was a VERY important theorem in calculus 3, and so I'm sure you've proved this before. But that doesn't help if it's been a while since you took calc three. Let me see if I can remember the proof.

[heartslaught]

Believe it or not, that proof was never proven in calc 3... not to my memory or notes anyhow.

[Woozie]

Believe it or not, that proof was never proven in calc 3... not to my memory or notes anyhow.

Wow, it's kind of important to be skipped. Slap your professor for me.

I'm going to assume you already know the formula for the directional derivative? If not, I'll derive it for you, but for now I'm going to assume you already know that the directional derivative is the rate of change of a scalar field in a given direction (i.e. the directional derivative at point f(x,y) in the direction of u is the instantaneous rate of change of f(x,y) if you were to travel in that direction.

I'm assuming you already know that the formula for the directional derivative in the direction of u=<a,b> is

Du=del(f)*u (where del(f) is the gradient of f and the * mean dot product).

So since that formula gives the rate of change in any given direction then we need to figure out which u gives the highest value for del(f)*u. Remember, u is just a unit vector who's direction is the direction we're finding the rate of change in. Del(f) is just the gradient vector.

So what value of u gives the highest value of the product? Well, the value of u is fixed because it's a unit vector. The only thing we change about u is the direction, not the magnitude. So what direction should we have it point?

Well, Du=del(f)*u=|delf|*|u|*cosθ = |delf|*cosθ [[since magnitude of u is 1]].

What value of cos gives the greatest value to the product? The highest possible value of cos is 1, and this occurs when the angle is 0. This means that the cosine of the angle between the gradient vector and u should be 0 degrees, or in other words, u is in the same direction as the gradient. So the directional derivative has it's maximum value in the direction of the gradient. Since the directional derivative gives the rate of increase, and it's maximum value is when the direction is that of the gradient, then the gradient's direction and the direction of maximum increase is always the same direction. In other words, the gradient always points in the direction of maximum increase.

[Kaylia]

I hate typing a long answer, press "preview post", and see that Woozie already typed a better and longer answer.

[Woozie]

My answers are too long and not always very good x_x I'm surprise people don't just ignore my walls of text and say "can someone give me a shorter answer"

[heartslaught]

Okay, so I got the angle part in a very messy fashion. I've adjusted my notes for the directional derivative, and because of that, I managed to also punch out the normal bit as well, so it works out (as 4.19 requires you have that normal bit to solve it).

And honestly, never learned the directional derivative that well... until I originally took this course (passed it with a D due to personal life kicking my ass at the time), it took two years until I saw it post calc3, in the mechanics course.

Now just to remember how to take the gravitational force on a point mas via a infinitely long rod, and I should be golden for this chapters exercises (yes, I'm doing every problem in every chapter, lol).

[Neosutra]

The gravitational force experienced by the point mass from the rod will be a finite force per unit length of the rod . Since the rod is infinitely long, it would be an infinite force.

Unless I am forgetting some derivation Q.Q.

[Woozie]

Most of the point masses on the rod are canceled out by point masses at some other point on the rod
^
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v


The rod is assumed to be infinitely long.

Since there's the same amount of mass above and below the mass, any force in the up or down direction cancel each other out, and the result is a finite force on the point pulling it to the left, towards the rod, with no up or down component. In undergrad level E&M courses, a common homework problem is to prove this using Gauss' law (different force but the exact same result).

Edit: Technically, even though the up/down components of force cancel each other out, there still is an infinite amount of forces pulling it to the left, but the inverse square law in gravity and E&M assures that the forces deteriorate fast enough such that the integral will converge when you're adding these forces, giving you a finite result. If the force variation with range was something different than inverse square, it's possible that leftward force could still turn out to be infinite.

[Neosutra]

Assuming the point is on the rod, and not at one end. But then again what kind of infinite rod has an end!

Been a while since I thought about some of these fun problems. I want to go back to school =/

[Woozie]

I once decided to see what would happen if I used a rod that did have an end (a half infinite rod. Well, I used a half infinite line, not rod). There was no symmetry, so gauss' law didn't simplify it at all, and no matter how I did it, I ended up with a huge freakin integral that I wasn't about to try to evaluate x_x

[Rhinox]

you know.. even before I clicked this thread I knew I would be over my head but I clicked it anyway and what'ya know!

[Blarg]

How much of a physicists daily time is spent talking about the length of a rod?

Oh and ummm express you answer in seconds assuming gravity is 9.81ms^2 and we are on the moon and upside down.

[Creeps]

Thank God some people are blessed with being able to understand this shit so I don't have to

I'm taking Calc I right now for some stupid fucking reason (I'm a history major), and I'm so lost

[Mizango]

That is what makes Newton my favorite Scientist! Neil Degrasse Tyson said it best "Newton is great! He basically invented Calculus on a DARE, in a matter of 3 months, simply to solve a problem that he had no answer to!"

[Kaylia]

How much of a physicists daily time is spent talking about the length of a rod?

Oh and ummm express you answer in seconds assuming gravity is 9.81ms^2 and we are on the moon and upside down.

Not that often, it's usually balls and various spherical object.

Thank God some people are blessed with being able to understand this shit so I don't have to

I'm taking Calc I right now for some stupid fucking reason (I'm a history major), and I'm so lost

Everything in maths is annoyingly hard... until you understand it, then it's really easy. It's like this for everyone (except Euler, that dude wasn't human).

Being lost is perfectly normal, especially when you do transition between different language (arithmetic -> algebra -> geometry -> calculus -> space/vector). Don't give up, and try to get information from different source if one isn't working for you. You don't have to stick to what your teacher give you if it's not working, there is the internet (forum, wiki..) and library to save you some time.


[edit] And before I get killed by a mathematician, I wasn't implying that there is different kind of mathematics, I'm simply saying the transition between those class usually required more works than usual.