Calculus on UCM

Calculus Level 3

A laser source, situated at a distance $d$ from a plane wall $P$ , rotates around an axis $E$ , with constant angular velocity $\omega$ , emitting a laser beam marks a red spot that moves in a straight line along the wall. The red spot has an instantaneous velocity $v$ . The picture represents a view from above. Find $v$ in terms of $\omega$ , $d$ and the angle $\theta$ .

v = \omega d \cos \theta

v = \dfrac {\omega d} {\cos \theta }

v = \dfrac {d} {\cos^{2} \theta }

v = \dfrac {\omega d} {\cos^{2}\theta }

1 solution

Vitor Santos
May 18, 2016

Let $D$ be the distance between the right angle and the Red Spot, is immediate that $\tan \theta = \dfrac {D}{d} \rightarrow d \tan \theta = D$ , but $D$ is the travelled space of the Red Spot, so by differenciating it we should be able to find the velocity. $\dfrac{\mathrm{d} (d \tan \theta)}{\mathrm{d} t} = \dfrac{\mathrm{d} D}{\mathrm{d} t} = v$ applying the chain rule(*), d is constant. $d \dfrac{\mathrm{d} (\tan \theta)}{\mathrm{d} t} =d \dfrac{\mathrm{d} (\tan \theta)}{\mathrm{d} \theta} \dfrac{\mathrm{d} \theta}{\mathrm{d} t}$ . Differenciating it $d \dfrac{\mathrm{d} (\tan \theta)}{\mathrm{d} \theta} \dfrac{\mathrm{d} \theta}{\mathrm{d} t}=\dfrac {\omega d} {\cos^{2}} = v$ .

(*): Here you could complain, "Why I have to use the chain rule and not just derivate just the tangent? " Well, as stated in the question the angle theta varies in the ratio of $\omega t$ , because is a uniform circular motion, so I can't just derivate the tangent, because I am doing that to respect to time and theta varies in time.

Calculus on UCM

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