Easy multivariable derivative

[malawi_glenn]

I am repeting some multivariable calculus.

I want to know if have done right now:

\mathbf{r} = \mathbf{r}(q_1, q_2, q_3)

\dfrac{\partial \mathbf{r}}{\partial q_1} = \left(\dfrac{\partial r_1}{\partial q_1} , \dfrac{\partial r_2}{\partial q_1} , \dfrac{\partial r_3}{\partial q_1} \right)

let

\mathbf{r} = (q_1 + 2q_3, q_2 + 3q_1 - q_3, q_1 - q_3)

\dfrac{\partial \mathbf{r}}{\partial q_1} = (1,3,1)

[HallsofIvy]

If I understand what your problem is correctly- that r is a three dimensional vector function of 3 variables, then, yes, your answer is correct.

[malawi_glenn]

thanx dude!

How about this one?

\dfrac{d}{d\theta}\left( \dfrac{d\theta}{dt}\right) = 0 \text{ ?}

[CompuChip]

Yes. What about it? What is \theta? And what is the question?

[malawi_glenn]

theta is a function, and I am wondering if

\dfrac{d}{d\theta}\left( \dfrac{d\theta}{dt}\right) = 0 \text{ ?}

Is correct in this case: i forgot to post the link.

http://en.wikipedia.org/wiki/Lagrangian_mechanics

I am trying to figure out what is happening in "Pendulum on a movable support"

[matness]

According to information presented in the link
you are right
\dfrac{d\theta}{dt} is only function of t so derivativee w.r.t. theta is 0

[malawi_glenn]

great thanx! Have not done calculus for a while, so I am repeting a bit before next semester=)

[HallsofIvy]

No, that is not true.

In general, if \theta is a function of t and f is any function of \theta then it is also a function of t and
\frac{df}{d\theta}= \frac{df}{dt}\frac{dt}{d\theta}
In particular, if f= d\theta /dt then
\frac{d}{dt}\frac{d\theta}{dt}= \frac{d^2\theta}{dt^2}\frac{dt}{d\theta}
which is not necessarily 0.

To take an easy example, if \theta= e^t then
\frac{d\theta}{dt}= e^t= \theta[/itex]
so that
[tex]\frac{d}{d\theta}\frac{d\theta}{dt}= \frac{d\theta}{d\theta}= 1

[matness]

Right but

In the link L is function of theta, theta- dot and t

So for L at least
\frac{\partial}{\partial\theta}\left( \dfrac{d\theta}{dt}\right) = 0
is true
I have considered the lagrangian and was not carefull:blushing:
thanks

[HallsofIvy]

Try being careful!