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

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

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

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

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

• 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. Feb 16, 2021 at 1:08
• @Shrey I did this but I don't know if it's okay Feb 16, 2021 at 1:20
• @Shrey $\frac{dF}{dt}=\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt}+\frac{\partial F}{\partial t}\frac{dt}{dt}$ $\frac{dF}{dt}=(\frac{\partial F}{\partial x}\hat{i}+\frac{\partial F}{\partial y}\hat{j}+\frac{\partial F}{\partial z}\hat{k})\cdot(\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}+\frac{dz}{dt}\hat{k})+\frac{\partial F}{\partial t}$ $\frac{dF}{dt}=\nabla F\cdot \frac{d\vec{r}}{dt}+\frac{\partial F}{\partial t}$ Feb 16, 2021 at 1:20
• @Shrey I modified my question and put my solution Can you see if it is correct? Feb 16, 2021 at 1:22
• Looks fine to me. Feb 16, 2021 at 1:25

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

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.

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