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 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 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

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

No comment

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 You mean k(A.B) = (kA).B is not correct ,k being a real. . No where in my proof i did that mistake .

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 .

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??

. 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??

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

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 To that statement i agree fully. A lot of Ph. Doctors lack even traces of common sense

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.

[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

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]

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]??

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

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]