How Does Vector Dot Product Differentiation Relate to Scalar Functions?

[enricfemi]

Homework Statement

d( \vec {F}.\vec {F})/dt=d(F*F)/dt=2*F*dF/dt (1)
d( \vec {F}.\vec {F})/dt=2* \vec{F}.d \vec{F}/dt (2)
so F*dF/dt=\vec{F}.d \vec{F}/dt (3)
??

Homework Equations

The Attempt at a Solution

 

[dextercioby]

Why would something be wrong in what you've written ?

Daniel.

 

[enricfemi]

Why would something be wrong in what you've written ?

Daniel.

do you mean it is right?

 

[enricfemi]

\vec{F} and d\vec{F}/dt are in the same direction?

 

[dextercioby]

What relevance does that have for your equation ?

Daniel.

 

[enricfemi]

but it really seems relevant

 

[dextercioby]

Not to me. Why is it to you ?

Daniel.

 

[enricfemi]

so which one do you think the relerance is weak?
(1) (2) (3)

 

[dextercioby]

I don't know what you mean, I told you that for your derivation, it doesn't matter that the 2 vectors are in the same direction or not. It's simply "irrelevant".

Daniel.

 

[arildno]

\vec{F} and d\vec{F}/dt are in the same direction?

Why do you believe this??

 

[enricfemi]

[Why do you believe this??

Because of the the equation of (3)
it seems the angle between the two vector =0

 

[arildno]

Why do you believe the rate of change of the magnitude of the vector F equals the magnitude of the vector that is the rate of change of the vector F??

That's totally wrong!
Here's why:
\vec{F}=|F|\vec{i}_{r}, \vec{i}_{r}\cdot\vec{i}_{r}=1
Therefore,
\frac{d\vec{F}}{dt}=\frac{d|F|}{dt}\vec{i}_{r}+|F|\frac{d\vec{i}_{r}}{dt}
Thus, if the unit vector \vec{i}_{r} changes with time, then your result doesn't hold.

 

[enricfemi]

yeah,it must be wrong.
but i don'nt know in which equation,while they all obey the rules of vector.
and in (2):

d(\vec{F}.\vec{F})/dt=\vec{F}.d\vec{F}/dt+d\vec{F}/dt.\vec{F}
=2*\vec{F}.d\vec{F}/dt

tell me,please!

 

[dextercioby]

I already told you that all 3 equations are correct.

Daniel.

 

[tim_lou]

\left |\frac{d\vec{F}}{dt} \right | \neq \frac{d|\vec{F}|}{dt}

F*\frac{dF}{dt}=|\vec{F}| \frac{d|\vec{F}|}{dt}=\vec{F}\cdot\frac{d\vec{F}}{dt}=|\vec{F}|\left |\frac{d\vec{F}}{dt} \right | \cos{\theta}

where \theta is the angle between \dot{\vec{F}} and \vec{F}

 

[Marco_84]

Hello,
i Think also that the vectors F and dF/dt are perpendicular if the square of the modulus do not change in time, i.e. something that in differential geometry can be interpretated as a mobile 2-D basis among a given curve a=a(t). And with the introduction of a third vector, let's call him n, perpendicular to both of them we have the so famous "Triedro di Frenet".
sorry for my bad english.

since the scalar product (,):VxV--->R is a bilinear form defined on a vector space and has value on the Real field numbers.

if we develop the calculus we obtain from a side:

\frac{d\vec{F}\vec{F}}{dt}=2 \vec{F}\frac{d\vec{F}}{dt}

but from the other side:

\vec{F}\vec{F}=|F|^{2}

and d/dt of this quantity is zero by hypothesis.

we can recognize de def. of perpendicularity of the two vec. F and dF/dt.

N.B.
i did'n use the dot for the scalar product
bye bye
Marco

 

[Marco_84]

i don't know what's going on with the tex compiler but i think everybody understood the meaning of my opinion.

bye

maRCO

 

[D H]

The inner product of any constant-magnitude vector and its time derivative is identically zero. That's not an opinion, its an identity. (Note that this alone disproves the OP's misconception.) That \vect F \cdot \frac {d\vect F}{dt} = F \frac {dF}{dt} is also an identity. That \vect F is parallel to \frac{d\vect F}{dt} is just plain wrong.

 

[enricfemi]

Thank you!
I can understand this problem thoroughly now.
best wishes for the coming of Christmas