Find that this partial differentiation is equal to 0

[Istiak]

Homework Statement

find a partial differentiation is equal to 0

Relevant Equations

Partial Differentiation

$$\sum_i (\frac{\partial}{\partial q_i}(\frac{\partial T}{\partial q_j}\dot{q}_i)+\frac{\partial}{\partial q_i}(\frac{\partial T}{\partial q_j})\ddot{q}_i)+\frac{\partial}{\partial t}(\frac{\partial T}{\partial \dot{q}_j})$$

They wrote that above equation is equal to $$\frac{d}{dt}(\frac{\partial T}{\partial \dot{q}_j})$$ hiwhc means differentiation of other functions are $0$. But, I was thinking if my explanation was correct. I thought that while differentiating respect to $q_i$ in the function $\sum_i \frac{\partial}{\partial q_i}(\frac{\partial T}{\partial q_j}\dot{q}_i)$ there's no variable $q$ so, while differentiating respect to that it will become $0$. I was thinking the same thing had happened to second one also.

But, actually $T=T(q_i,\dot{q}_i,t)$. So, T has q variable but although why it will be $0$? Is there something else I am missing?

 

[ergospherical]

$$\sum_i (\frac{\partial}{\partial q_i}(\frac{\partial T}{\partial q_j}\dot{q}_i)+\frac{\partial}{\partial q_i}(\frac{\partial T}{\partial q_j})\ddot{q}_i)+\frac{\partial}{\partial t}(\frac{\partial T}{\partial \dot{q}_j})$$They wrote that above equation is equal to $\dfrac{d}{dt}(\dfrac{\partial T}{\partial \dot{q}_j})$.

Did you make a typo? Ought to be $\dfrac{d}{dt} \dfrac{\partial T}{\partial \dot{q}_j} = \dfrac{\partial^2 T}{\partial q_i \partial \dot{q}_j} \dot{q}_i + \dfrac{\partial^2 T}{\partial \dot{q}_i \partial \dot{q}_j} \ddot{q}_i + \dfrac{\partial^2 T}{ \partial t\partial \dot{q}_j}$.

 

[Istiak]

that's what author wrote. I don't think I made typo.

 

[ergospherical]

You did make a few typos, but the picture in #3 is right. It's the chain rule for $\dfrac{\partial T}{\partial \dot q_j}(q,\dot{q}, t)$.

 

[Istiak]

Since, my another thread was deleted mod was saying it's dup of that so adding info here.

In the last second line they wrote that,

$$\frac{d}{dt}(\frac{\partial T}{\partial \dot{q}_j})-\frac{\partial T}{\partial q_j}=\frac{\partial \dot{T}}{\partial \dot{q}_j}-\frac{\partial T}{\partial q_j}-\frac{\partial T}{\partial q_j}$$

which isn't looking correct to me. Cause, we know that $\frac{d}{dt}(\frac{\partial T}{\partial \dot{q}_j})=\frac{\partial \dot{T}}{\partial \dot{q}_j}$

 

[Istiak]

It's the chain rule for $\dfrac{\partial T}{\partial \dot q_j}(q,\dot{q}, t)$.

$$\frac{\partial T}{\partial \dot{q}_j}=\frac{\partial T}{\partial q_j}\frac{\partial q_j}{\partial \dot{q}_j}+\frac{\partial T}{\partial \dot{q}_j}+\frac{\partial T}{\partial t}\frac{\partial t}{\partial \dot{q}_j}$$ I found it. I think something is wrong. Cause, I still can't make them $0$ (after putting that in the main equation).