The distances \(u\) and \(v\) of an object and its image respectively from a lens of focal length \(f\) are connected by the formula \[\begin{equation*} \frac{1}{u} + \frac{1}{v} = \frac{1}{f}. \end{equation*}\]

Determine the speed of the image of an object when the object is \(\quantity{3}{m}\) from a lens of focal length \(\quantity{\dfrac{1}{2}}{m}\) and is moving away from the lens at \(\quantity{5}{m\,s^{-1}}\).

We want to find the speed of the image, \(\dfrac{dv}{dt}\), at the moment when \(u=3\) and \(\dfrac{du}{dt}=5\). So let’s try implicitly differentiating the formula with respect to \(t\). \[\begin{align*} u^{-1} + v^{-1} &= 2 \\ \implies\quad -u^{-2}\frac{du}{dt} -v^{-2}\frac{dv}{dt} &= 0 \\ \iff\quad \frac{1}{u^2}\frac{du}{dt} &= -\frac{1}{v^2}\frac{dv}{dt} \end{align*}\] Substituting in the given values, we have \[\begin{equation} \frac{dv}{dt} = -\frac{5}{9}v^2. \label{eq:dvdt} \end{equation}\] Substituting values into the original formula, we have \[\begin{align*} \frac{1}{3} + \frac{1}{v} &= 2 \\ \implies\quad v^2 &= \frac{9}{25} \end{align*}\] and combining with \(\eqref{eq:dvdt}\) we have \[\begin{equation*} \frac{dv}{dt} = -\frac{5}{9}\frac{9}{25} = -\frac{1}{5}. \end{equation*}\]

So the speed of the image is \(\quantity{0.2}{m\,s^{-1}}\).

Alternatively, we could have rearranged the given formula as \[ v=\frac{u}{2u-1} \] and differentiated using the quotient rule to achieve the same result.