I would like to know how differentiate this equation it should be fairly simple but its been a long night and I cant seem to figure it out.

$r =\frac{f}{R . \hat z} R$

with respect to time.

  • f is some constant (focal length)
  • R is a position vector
  • $R . \hat z $ is inner product

Any help will be appreciated

  • $\begingroup$ Did you mean $\frac{f}{R.z\hat R}$? You get this as $\frac{f}{R.z\hat R}$. $\endgroup$ – Martin Sleziak Oct 24 '12 at 18:10
  • $\begingroup$ Please edit your question to make it more readable. $\endgroup$ – Hagen von Eitzen Oct 24 '12 at 18:10
  • $\begingroup$ I'm guessing it should be $R (\hat z \cdot R)$. $\endgroup$ – Javier Oct 24 '12 at 18:12
  • $\begingroup$ sorry about that, it has now been correctly formatted $\endgroup$ – Alex Oct 24 '12 at 18:13
  • 1
    $\begingroup$ Does $\hat{z}$ depends on time? $\endgroup$ – Pragabhava Oct 24 '12 at 18:15

Assuming the direction of $\hat z$ does not change with time, $$ \begin{align} \dot r &= \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{fR}{\langle R, \hat z\rangle}\right)\\ & = \frac{f}{\langle R, \hat z\rangle^2}\left[\langle R, \hat z\rangle \dot R- R\frac{\mathrm{d}}{\mathrm{d}t}\left(\langle R, \hat z\rangle\right)\right]\\ & = \frac{f}{\langle R, \hat z\rangle^2}\left[\langle R, \hat z\rangle \dot R- \langle \dot R, \hat z\rangle R\right] \end{align} $$


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