triclino

I do not understand ,why you give irrelevant hints ,if you already know the problem.

Sometimes an easy problem has deeper consequences and we need to investigate.

Also every problem has various levels of proofs and according to what kind of proof we are asked to produce a very easy problem can become extremely difficult

THis looks like homework, triclino. We don't do your homework for you. We do give hints.

This is not homework .This a problem i would like to investigate and i bring it to the attention of the forum and wait for any suggestions .

On the other hand you can baptise every problem as home work and give irrelevant hints ,since you will not pursue the problem any further because it is homework

This suggests a route by which you can prove this conjecture: Split the proof into two parts, one for[math]x\,\in(-1,0)[/math] and the other for [math]x>0[/math].

This suggests nothing ,because you do not say what must we do in those two parts

How do we prove that:

[math](1+x)^h<1 +xh [/math] ,where : 0<h<1 and [math]x\neq 0[/math] x>-1

3 bodies of equal mass specific heat capacities are at temperature 100k, and two at 300k. they are acted by a heat engine. what is the maximum temperature that one of the bodies can attain?(not mentioned which one?)

 

ans. 400k.

Can you be a little more clear .What do you mean "are acted by a heat engine"??

No comment

=D H;556451]Your mistake is treating vectors as numbers. While numbers do form a vector space ([math]\mathbb R^1[/math]), vector spaces have different rules than numbers.

The rules i just quoted are those of vector spaces and some of those rules coincide with those of a field .

For example those concerning the predicate of equality.

Commutativity.

Existence of identity element.....e.t.c........e.t.c

In particular, you treated the inner product and scalar multiplication as interchangeable operations.

You mean k(A.B) = (kA).B is not correct ,k being a real.

They are not. Given vectors [math]\vec a[/math] and [math]\vec b[/math], [math](\vec a \cdot \vec b)\vec b \ne \vec a (\vec b \cdot \vec b)[/math] in general. That is exactly what you did, and that is invalid.

No where in my proof i did that mistake

.

Where did you get this last equation? It isn't right, so let's drop it.

 

Next, regarding the first three equations: let's get the nomenclature right. The first is better written as [math]F_x[/math] rather than [math]F(x)[/math]. The former indicates the x-component of the force vector; the latter implies that F is a function of x -- and x only. Secondly, you are missing a minus sign. The convention is [math]\vec F = - \nabla \phi[/math].

 

That said, given that [math]F_x = 2xy+z^3[/math] suggests that the potential is of the form

 

[math]\phi(x,y,z) = -x(xy + z^3) + f(y,z)[/math].

 

Continuing with the y-component of the force vector,

 

[math]-\,\frac{\partial \phi(x,y,z)}{\partial y} = x^2 - \frac{\partial f(y,z)}{\partial y}[/math]

 

Since [math]F_y = x^2[/math], this says that f(y,z)[/math] has no dependence on y. So let's call it f(z). Doing the same with the z-component, we find that that f(z) can have no dependence on z, either. All that is left is a constant, so

 

[math]\phi(x,y,z) = -x(xy + z^3) + C[/math]

 

If you can construct a potential function you know that the force is a conservative force. The contrapositive is not necessarily true. Failure to construct a potential function might just mean that you couldn't construct a potential function. You would have to prove that no such function could possibly exist to prove that a force is not conservative.

 

One way to prove this is to find a pair of states [math](x_0, y_0, z_0)[/math] and [math](x_1, y_1, z_1)[/math] and two distinct paths between these points. If the change in mechanical energy depends on the path taken the force is necessarily not conservative.

Thank you all .

D.H ,how would you calculate the Φ(x,y,z) of the force field:

F = (x+2y+4z) i + (2x-3y-z) j + (4x-y +2z) k.

You do not have to use the minus sign .

I just want to see the procedure .

i just mean A.B.C

If B and C are on the same currier ,or parallel to A ,then :

A.B.C = A.(B.C) =(A.B).C = C.(B.A) =(C.B).A = B.(A.C) ........e.t.c e.t.c

Otherwise : A.(B.C)[math]\neq (A.B).C[/math]

Since F = gradΦ(x,y,z) =>

F(x) = [math]\frac{\partial\Phi(x,y,z)}{\partial x}[/math]

F(y) = [math]\frac{\partial\Phi(x,y,z)}{\partial y}[/math]

F(z) = [math]\frac{\partial\Phi(x,y,z)}{\partial z}[/math] and

[math]\Phi(x,y,z) = \int(3xz^3)dz + C[/math]

AND now what??

You missed one possibility: That you misinterpreted/misapplied them, and that is what happened. What you calculated was the orthogonal projection of the acceleration vector a onto the line defined by the radial vector r. There is nothing wrong with that; in fact that is a frequently used operation. Your mistake was assuming that this orthogonal projection is equal to the acceleration vector. That is not the case.

 

As a side exercise, I suggest you investigate [math]d/dt\left(\vec r \times \vec v\right)[/math] , and also the vector triple product, [math]\hat r\times(\vec a \times \hat r)[/math], in the case of non-uniform circular motion.

 

You might also want to re-reread post #20.

The rules i used are the following:

1) if A and B are vectors then :

[math]\frac{d(A.B)}{dt}=\frac{dA}{dt}.B + A.\frac{dB}{dt}[/math]

2) A.A = [math]A^2[/math]

3) k(A.B) = (kA).B ,where k is a real No

4) If we add the same No to both sides of an equation the equation does change

5) if we multiply with the same No both sides of the equation the equation does not change .

6) the dot product of unit vectors is 1.

Now where is the misapplication??

http://math.scienceforums.net/Vector%20calculus/Conservative%20fields

 

Quite a few, as I wrote in the above link.

Thanks ,how do we find the function Φ(x,y,z) ,such that :

F = gradΦ(x,y,z)

In how many different ways can we show that the following force field is conservative??

F= [math](2xy+z^3[/math]) i + [math]x^2 [/math]j + [math]3xz^2[/math] k

Sometimes a formal proof goes against common sense.

Apart from the irrelevant example that you mentioned can you write a small formal proof of any mathematical problem and show us in which way it goes against common sense

See for example the Monty Hall problem. Even having a PhD does not mean that your common sense matches the formal proof.

To that statement i agree fully.

A lot of Ph. Doctors lack even traces of common sense

What you did correctly, triclino, was to calculate the component of the acceleration vector in the direction of the radial vector. What you did wrong was to assume that this is equal to the acceleration vector.

 

In particular, you calculated [math](\hat r \cdot \vec a)\hat r[/math], and this is equal to the acceleration vector only in the case that the acceleration vector is parallel to the radial vector. The transverse component of the acceleration vector is not zero in the case of non-uniform circular motion.

I assumed nothing. All i did is certain mathematical operations involving vector Analysis.

The only assumption in the whole procedure is that :r andv

are perpendicular.

The final result was dictated by the rules of vector analysis.

Unless the rules i used are not correct.

Let's use that logic on the velocity vector.

 

[math]\hat r \cdot \vec v = 0[/math]

 

Multiply by [math]\hat r[/math] on both sides and:

 

[math]\vec v = 0[/math]

The velocity vector is identically zero!

 

Which of course is wrong, and it is wrong for exactly the same reason your derivation was wrong.

[math]\hat{\mathbf{r}}[/math] and v are rerpendicular to each other that's why you get that result .

But r and a are not perpendicular to each other.

An analogical example is not enough you have to pinpoint the mistake if there is one

That derivation shows that there is only radial acceleration.

If there was another component the derivation should not be correct

Do you consider it incorrect if it is incomplete? You neither proved nor assumed [y>|x| and [math]|x|\geq 0[/math]] => y>0 nor referred to a previous proof for it. It's common sense, yes, but common sense is not formal proof.

I simply wanted to show that by writing :|x|<y this automatically implies y>0

Some books even mention that y should be >0.

Common sense is common logic .Do you think that a formal proof has not common sense on it?? That would be the end of everything.

Look for the following derivation where the circular motion is not uniform.

WE have r.v=0

Differentiate with respect to time and:

[math]\frac{d({\mathbf{r.v}})}{dt}= \frac{d{\mathbf{r}}}{dt}.{\mathbf{v}} + {\mathbf{r.a}}[/math]=

= v.v +r.a=0 => r.a= [math]-v^2[/math] =>

=> [math]r\hat{\mathbf{r}}[/math].a=[math]-v^2[/math] => [math]\hat{\mathbf{r}}[/math].a = [math]\frac{-v^2}{r}[/math]

multiply by :[math]\hat{\mathbf{r}}[/math] both sides of the equation and:

a= -[math]\hat{\mathbf{r}}\frac{v^2}{r}[/math]

First and foremost, the hypothesis must be well-formed. This is not a well-formed hypothesis:

 

That statement is obviously not true in general (consider the case y=0). Try again, this time stating your hypothesis correctly.

You can easily see that there was o typo in my problem :

instead of prove : |x|<y <=> -y<x<y I wrote : prove : |x|<y which has no meaning:

But from the proof that followed one could easily concluded that i ment :

prove : |x|<y <=> -y<x<y. This a high school theorem and it can be found in any high school books.

Since y>|x| and [math]|x|\geq 0[/math] => y>0.

Go for the real thing ,is that proof correct or not??

i mean is still :

a = [math]\frac{v^2}{r}[/math]??

.

 

Well, no. I mean, obviously there were lots of premises and conditions and whatnot that you left out, but still, ugh. And the line of thought was backwards - your argument started with your conclusion and went on to justify which just, ugh, you know better than that.

A mathematical proof is not an abstract painting where everybody can say whatever he/she likes .

A mathematical proof consists of :

1) Axioms,theorems,definitions.

2) The laws of logic .

So if you say that the particular proof is wrong ,you have to pinpoint ,the particular axiom or theorem or definition or the application 0f the particular law of logic that is not correct.

Otherwise your remarks and comments are of no value at all .

But to say for certain whether a proof is correct or not ,one must get into the depths of the proof .

This a simple high school problem .One can imagine what actually is happening with other more complicated Analysis proofs e.t.c ......e.t.c

The question now is what is going to be, a if v is not constant

Yes indeed i did not notice

I proved it here:

 

[math]\dot{\mathbf{r}} = \mathbf{v} = r \dot{\theta}(-sin\theta\hat{\mathbf{x}} + cos\theta\hat{\mathbf{y}}) [/math]

 

and then you can take a dot product of v and r to see they are perpendicular.

Correct:

What is the use now of bringing in ω ?? we simply have now:

v=[math]r\dot{\theta}[/math] and

[math]

\mathbf{a} = \frac{d\mathbf{v}}{dt} = v (-\dot{\theta} cos\theta\hat{\mathbf{x}} - \dot{\theta}sin\theta\hat{\mathbf{y}})

[/math]

And |a|=a =[math]r\dot{\theta}^2 = \frac{v^2}{r}[/math]