# Easy multivariable derivative

1. Aug 8, 2007

### 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

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

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

2. Aug 8, 2007

### HallsofIvy

Staff Emeritus
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.

3. Aug 8, 2007

### malawi_glenn

thanx dude!

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

4. Aug 8, 2007

### CompuChip

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

5. Aug 8, 2007

### 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"

6. Aug 8, 2007

### 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

Last edited: Aug 8, 2007
7. Aug 8, 2007

### malawi_glenn

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

8. Aug 8, 2007

### HallsofIvy

Staff Emeritus
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
$$\frac{d\theta}{dt}= e^t= \theta[/itex] so that [tex]\frac{d}{d\theta}\frac{d\theta}{dt}= \frac{d\theta}{d\theta}= 1$$

9. Aug 8, 2007

### 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
thanks

10. Aug 8, 2007

### HallsofIvy

Staff Emeritus
Try being careful!