# Derivatives and Partial Derivatives

1. Sep 30, 2010

### Trenthan

ey guys

Generally i just do these without thinking, however i was checking some work today with a friend and he is adament i did my derivative wrong....

If i can double check with you

Well firstly
'c' is simply a constant
q1 and q2 are generalised coordinates
IZG1 is simply the moment of inertia
'T' being Kinetic energy

For those who dont notice what this is. Its apart of Lagrange's equation during the formation of the Equation of Motion

EDIT**, i should mention that q1 and q2 are both functions of time

$$T = \frac{1}{2}m_{1}c^{2}\dot{q_{1}}^2 + \frac{1}{2}I_{ZG1}\dot{q_{1}}^{2} + \frac{1}{2}m_{2}(\dot{q_{2}}^{2}+\dot{q_{1}}^{2}q_{2}^{2})+\frac{1}{2}I_{ZG2}\dot{q_{1}}^{2}$$

$$\\ \frac{\partial T}{\partial \dot{q_{1}}} = m_{1}c^{2}\dot{q_{1}} + I_{ZG1}\dot{q_{1}}+ m_{2}q_{2}^{2}\dot{q_{1}}+ I_{ZG2}\dot{q_{1}}$$

His version
$$\frac{d}{dt}\left ( \frac{\partial T}{\partial \dot{q_{1}}} \right ) = m_{1}c^{2}\ddot{q_{1}}+I_{ZG1}\ddot{q_{1}} + m_{2}\ddot{q_{1}}q_{2}^{2} + 2m_{2}q_{2}\dot{q_{1}} + I_{ZG2}\ddot{q_{1}}$$

Mine(below this), so unless im mistaken it should have
$$\dot{q_{2}}$$
in the third term????

$$\frac{d}{dt}\left ( \frac{\partial T}{\partial \dot{q_{1}}} \right ) = m_{1}c^{2}\ddot{q_{1}}+I_{ZG1}\ddot{q_{1}} + m_{2}\ddot{q_{1}}q_{2}^{2} + 2m_{2}q_{2}\dot{q_{2}}\dot{q_{1}} + I_{ZG2}\ddot{q_{1}}$$

Now if someone can clarify which version is correct big help. Ive gone back to my text book however i dont have anything similar. Everything is linear so no 'squared' stuff

Cheers Trent, and thanks in advance

Last edited: Sep 30, 2010
2. Sep 30, 2010

### Petr Mugver

Your version is the correct one.

3. Sep 30, 2010

### Trenthan

wohhh sorry i didnt make it clear

His version is the top

Mine is the lower one****

So you reffering to the top or bot

Soz lolz

4. Sep 30, 2010

### Petr Mugver

Bottom. I guess you could have guessed this answer!

5. Mar 23, 2011