Multivariable Derivative Practice

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Homework Help Overview

The discussion revolves around multivariable calculus, specifically focusing on derivatives of vector functions and the implications of differentiating with respect to variables that are themselves functions of time.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the correctness of derivatives of vector functions and question the implications of differentiating with respect to a variable that is a function of time. There are attempts to clarify the context of the variables involved, particularly in relation to Lagrangian mechanics.

Discussion Status

The discussion includes varying interpretations of the derivative expressions, with some participants providing guidance on the mathematical relationships involved. There is acknowledgment of different perspectives on the implications of the derivatives discussed.

Contextual Notes

Participants reference specific functions and contexts, such as the Lagrangian mechanics of a pendulum, which introduces additional complexity to the discussion. There is mention of the need for careful consideration in applying calculus concepts.

malawi_glenn
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I am repeting some multivariable calculus.

I want to know if have done right now:

[tex]\mathbf{r} = \mathbf{r}(q_1, q_2, q_3)[/tex]

[tex]\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)[/tex]

let

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

[tex]\dfrac{\partial \mathbf{r}}{\partial q_1} = (1,3,1)[/tex]
 
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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.
 
thanx dude!

How about this one?

[tex]\dfrac{d}{d\theta}\left( \dfrac{d\theta}{dt}\right) = 0 \text{ ?}[/tex]
 
Yes. What about it? What is [itex]\theta[/itex]? And what is the question?
 
theta is a function, and I am wondering if

[tex]\dfrac{d}{d\theta}\left( \dfrac{d\theta}{dt}\right) = 0 \text{ ?}[/tex]

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"
 
According to information presented in the link
you are right
[tex]\dfrac{d\theta}{dt}[/tex] is only function of t so derivativee w.r.t. theta is 0
 
Last edited:
great thanx! Have not done calculus for a while, so I am repeting a bit before next semester=)
 
No, that is not true.

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

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

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

So for L at least
[tex]\frac{\partial}{\partial\theta}\left( \dfrac{d\theta}{dt}\right) = 0[/tex]
is true
I have considered the lagrangian and was not carefull:blushing:
thanks
 
  • #10
Try being careful!
 

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