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where F is a differentiable function of $x, y,z, t$ and $x, y, z$ differentiable functions of $t$, Show that $\frac{dF}{dt}=\frac{\partial F}{\partial t}+\nabla F\cdot \frac{d\vec{r}}{dt}$

We define $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ as the position vector

I would like you to give me some suggestions. It has been difficult for me to know where to start

I did this but I don't know if it's okay

$\frac{dF}{dt}=\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt}+\frac{\partial F}{\partial t}\frac{dt}{dt}$

$\frac{dF}{dt}=(\frac{\partial F}{\partial x}\hat{i}+\frac{\partial F}{\partial y}\hat{j}+\frac{\partial F}{\partial z}\hat{k})\cdot(\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}+\frac{dz}{dt}\hat{k})+\frac{\partial F}{\partial t}$

$\frac{dF}{dt}=\nabla F\cdot \frac{d\vec{r}}{dt}+\frac{\partial F}{\partial t}$

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  • $\begingroup$ Hint: Consider the multivariable chain rule to find an expression for the total time derivative of F (involving time derivatives of x, y and z). You can show this is equivalent to your expression by expanding the gradient of F and the velocity into Cartesian components. $\endgroup$
    – Shrey
    Feb 16, 2021 at 1:08
  • $\begingroup$ @Shrey I did this but I don't know if it's okay $\endgroup$
    – Kale_1729
    Feb 16, 2021 at 1:20
  • $\begingroup$ @Shrey $\frac{dF}{dt}=\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt}+\frac{\partial F}{\partial t}\frac{dt}{dt}$ $\frac{dF}{dt}=(\frac{\partial F}{\partial x}\hat{i}+\frac{\partial F}{\partial y}\hat{j}+\frac{\partial F}{\partial z}\hat{k})\cdot(\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}+\frac{dz}{dt}\hat{k})+\frac{\partial F}{\partial t}$ $\frac{dF}{dt}=\nabla F\cdot \frac{d\vec{r}}{dt}+\frac{\partial F}{\partial t}$ $\endgroup$
    – Kale_1729
    Feb 16, 2021 at 1:20
  • $\begingroup$ @Shrey I modified my question and put my solution Can you see if it is correct? $\endgroup$
    – Kale_1729
    Feb 16, 2021 at 1:22
  • $\begingroup$ Looks fine to me. $\endgroup$
    – Shrey
    Feb 16, 2021 at 1:25

2 Answers 2

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Write the multivariate chain rule in the language of infinitesimals, viz.$$dF=\frac{\partial F}{\partial t}dt+\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+\frac{\partial F}{\partial z}dz.$$Note there is nothing different about $t$ here as compared with $x,\,y,\,z$. But if $x,\,y,\,z$ are differentiable functions of $t$ so an $F(t,\,x,\,y,\,z)$ differentiable with respect to each of its four arguments simplifies to a differentiable function of $t$, division by $dt$ gives$$\frac{dF}{dt}=\frac{\partial F}{\partial t}+\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt}.$$Finally, write the last three terms as a dot product:$$\frac{dF}{dt}=\frac{\partial F}{\partial t}+\nabla F\cdot\frac{d\vec{r}}{dt}.$$

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Use the chain rule to get, $$\dfrac{dF}{dt}=\dfrac{\partial F}{\partial t}+\dfrac{\partial F}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial F}{\partial y}\dfrac{dy}{dt}+\dfrac{\partial F}{\partial z}\dfrac{dz}{dt}$$

Should be clear from there.

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    $\begingroup$ $\frac{dF}{dt}=\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt}+\frac{\partial F}{\partial t}\frac{dt}{dt}$ $\frac{dF}{dt}=(\frac{\partial F}{\partial x}\hat{i}+\frac{\partial F}{\partial y}\hat{j}+\frac{\partial F}{\partial z}\hat{k})\cdot(\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}+\frac{dz}{dt}\hat{k})+\frac{\partial F}{\partial t}$ $\frac{dF}{dt}=\nabla F\cdot \frac{d\vec{r}}{dt}+\frac{\partial F}{\partial t}$ $\endgroup$
    – Kale_1729
    Feb 16, 2021 at 1:20
  • $\begingroup$ I did this but I don't know if it's okay $\endgroup$
    – Kale_1729
    Feb 16, 2021 at 1:20
  • $\begingroup$ Yes, that's correct $\endgroup$
    – user256872
    Feb 16, 2021 at 1:22
  • $\begingroup$ Great thank you very much. $\endgroup$
    – Kale_1729
    Feb 16, 2021 at 1:26

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