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Given the unit vectors:

$\vec e_\rho=\bigl(\begin{smallmatrix} cos(\theta )\\ sin(\theta )\\0 \end{smallmatrix}\bigr); \vec e_\phi=\bigl(\begin{smallmatrix} -sin(\theta )\\ cos(\theta )\\0 \end{smallmatrix}\bigr); \vec e_z=\bigl(\begin{smallmatrix} 0\\ 0\\1 \end{smallmatrix}\bigr);$

and my position vector:

$\vec r=\rho\vec e_\rho+z\vec e_z$

If $,\rho, \phi,z$ are explicit functions of time ($\rho=\rho(t),\phi=\phi(t),z=z(t) $what is $\frac{d}{dt}\vec r $ and $\frac{d^2}{d^2t}\vec r$?

I am not entirely sure but don't I just differentiate "normally"?

$\vec r=\rho(t)\bigl(\begin{smallmatrix} cos(\theta )\\ sin(\theta )\\0 \end{smallmatrix}\bigr)+z(t)\bigl(\begin{smallmatrix} 0\\ 0\\1 \end{smallmatrix}\bigr)$

$\frac{d}{dt}\vec r=\rho'(t)\frac{d\theta}{dt}\bigl(\begin{smallmatrix} -sin(\theta )\\ cos(\theta )\\0 \end{smallmatrix}\bigr)+z'(t)\bigl(\begin{smallmatrix} 0\\ 0\\1 \end{smallmatrix}\bigr)$

Or am I just writing down nonsense?

I also wasn't sure how to differentiate the z-vector.

Maybe someone with a little more knowledge/experience can help me out here :) Thanks in advance

Edit:

$\frac{d}{dt}\vec r=\rho'(t)\bigl(\begin{smallmatrix} cos(\theta )\\ sin(\theta )\\0 \end{smallmatrix})+\rho(t)\frac{d\theta}{dt}\bigl(\begin{smallmatrix} -sin(\theta )\\ cos(\theta )\\0 \end{smallmatrix})+z'(t)(\begin{smallmatrix} 0\\ 0\\1 \end{smallmatrix}\bigr)$

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Suppose that $\theta$ depends on $t$ then $\frac{d}{dt}(\rho(t)\cos\theta)=\rho'(t)\cos\theta-\rho(t)\frac{\partial\theta}{\partial t}\sin\theta$.

It seems you missed out a term, if $\theta$ does not depend on $t$ then the extra term will be zero.

Also when you did the derivative it seems you tried to differentiate everything at once, but notice that the second term ($\rho(t)...$) is expressed as something multiplying $\vec e_\phi$.

Your deriavative of $z\vec e_z$ is fine.

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  • $\begingroup$ I edited my beginning post because somehow the formatting isn't working for me in the comments. I used the product rule now. I am not sure what you mean when you say "the second term (ρ(t)...) is expressed as something multiplying e⃗ ϕ" Do you just mean that the derivative of$\vec e_\roh=e_/phi$? $\endgroup$ – qmd Dec 10 '14 at 8:32
  • $\begingroup$ What you have is right now, all I meant was that the second vector is in terms of $\vec e_\phi$ $\endgroup$ – Ellya Dec 10 '14 at 8:38
  • $\begingroup$ Nice. Thank you very much. Have a nice day $\endgroup$ – qmd Dec 10 '14 at 8:40
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One thing to keep in mind is the unit vectors $\hat e_\rho$ and $\hat e_\theta$ are functions of $\theta$. So, to find the total differential $d\vec r$ we must consider the individual differentials $d\rho$, $d\theta$ and $dz$. It is easy to verify that $\partial\hat e_\rho/\partial\theta = \hat e_\theta$ and $\partial\hat e_\theta/\partial\theta = -\hat e_\rho$. This is reasonable, because $\hat e_\rho$ and $\hat e_\theta$ can change directions, but not magnitudes. The only way they can change is at right angles to themselves, not along their own directions. $\hat e_z$ always has the same direction and magnitude, so its differential is zero. The coefficients of $\hat e_\rho$, $\hat e_\theta$ and $\hat e_z$ may have non-zero differentials, or they may be constants, including zero.

To find the total differential of $\vec r(\rho,\theta,z)$, use the product rule. That is, take the differential of the coefficient times the unit vector, then add the coefficent times the differential of the unit vector. You should be able to write $$d\vec r = (d\rho) \hat e_\rho + \rho (d\theta \hspace{1 pt} \hat e_\theta) +(dz)\hat e_z $$ Divide both sides by $dt$, and you have the derivative. For practice, you should find the second derivative of $\vec r$.

To repeat: even though you don't see $\theta$ in the equation $\vec r=\rho \hat e_\rho+z \hat e_z$, it is there, hidden in the $\hat e_\rho$.

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