suppose $x=f(t)$ with a constant acceleration.

Then does $\frac{\text d x}{\text d t} = \frac{\partial (x)}{\partial(t)}$ since the position in $x$ only changes with time?

Then the acceleration in the x-direction from chain rule can be

$$\frac{\text d u}{\text d t}= \frac{\partial(u)}{\partial(t)} + \frac{\partial(u)}{\partial(x)}\cdot\frac{\text d x}{\text d t}$$

But then I see $\frac{\text d u}{\text d t}= 2\frac{\text du(u)}{\text d u(t)}$. which does not satifisy $x=a+bt+ct^2$

If velocity constant then the acceleration is zero. Then, $\frac{\text d u}{\text d t}= \frac{\partial(u)}{\partial(t)}$

Hence I am confused.

  • 2
    $\begingroup$ Hi and welcome to physics SE. Please, read the guide on how to write formulas correctly. You have to enclose formulae in-between of dollar symbols, and use LaTex notation inside. For example: $\frac{dx}{dt}$ $\endgroup$
    – FGSUZ
    Nov 16 '18 at 0:17
  • $\begingroup$ I have performed the edits @FGSUZ mentions for you this time. Please make sure nothing was changed to be incorrect. $\endgroup$ Nov 16 '18 at 0:37
  • $\begingroup$ $\frac{\partial x}{\partial t}$=$u$ this is my first time here. I am sorry for my equations. I have edited them. $\endgroup$
    – mnk kanna
    Nov 16 '18 at 0:48

For functions of only one variable, the total derivative of the function with respect to that variable is the same as the partial derivative of the function with respect to that variable.

Therefore $$\frac{\text d x}{\text d t}=\frac{\partial x}{\partial t}$$

For your chain rule on $\frac{\text d u}{\text d t}$, you will always get $a$, but each term will be different depending on how you express $u$. For example, if $u=u_0+at$, then $$\frac{\text d u}{\text d t}=\frac{\partial u}{\partial t}+0=a$$ Or if you use $u^2=u_0^2+2ax$, then $$\frac{\text d u}{\text d t}=0+\frac{\partial u}{\partial x}\cdot\frac{\text d x}{\text d t}=a$$

If you used other relations it would always work out so that $\frac{\text d u}{\text d t}=a$ This is because this is the definition of acceleration. If you don't get this then something must have gone wrong with your work. The chain rule always works.

Also note that we can play this same game with position. For example, the following relation is valid: $$x=\frac{u^2-u_0^2}{2a}$$

And you can mathematically say $x=x(u)$. But our definitions still hold:

$$\frac{\text d x}{\text d t}=\frac{\text d x}{\text d u}\cdot\frac{\text d u}{\text d t}=\frac{2u}{2a}\cdot a=u$$

Essentially what we can say is that our derivatives and chain rules should always agree with our definitions in order for everything to be self-consistent.

  • $\begingroup$ I am confused with the chain rule actually, That is ${ d u}$ = $\frac{\partial (u)}{\partial(t)}$* ${d t}$ +$ \frac{\partial (u)}{\partial(x)}$*${d x}$ can we apply this here? $\endgroup$
    – mnk kanna
    Nov 16 '18 at 1:08
  • $\begingroup$ @mnkkanna I have added information about that $\endgroup$ Nov 16 '18 at 1:13
  • $\begingroup$ Hi, Thank you for the explanation. Can you kindly let me know? If ${u}$={u(x,t)} which of them will be invalid? $\endgroup$
    – mnk kanna
    Nov 16 '18 at 1:18
  • $\begingroup$ @mnkkanna I'm not sure I understand your question. $\endgroup$ Nov 16 '18 at 1:21
  • $\begingroup$ if velocity is a fucntion of both x and t, is the chain rule is valid? $\endgroup$
    – mnk kanna
    Nov 16 '18 at 1:29

$\frac{dx}{dt}\neq \frac{du(x)}{du(t)}$... What you needed to do is $\frac{d^2x(t)}{dt^2}=a$, integrate two times, and you'll get the answer that you are searching :)


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