Rate of change involving trig functions (1 Viewer)

AnandDNA · May 21, 2009

A rotating light L is situated at sea 180 metres from the nearest point P on a straight shoreline, The light rotates through one revolution every 10 seconds. Show that the rate at which a ray of light moves along the shore at a point 300 metres from P is 136(pi) m/s

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jet · May 21, 2009

Diagram Later
$\text{If you draw a diagram you will have a line segment, AB with a point P 180 m away from L with } PL \perp AB \text{If you let an arbitrary point Q sit on the segment AB, and the distance PQ be } r \text{we can work from there. Let the angle PLQ = } \theta \\ \text{Now, the period of the revolution is 10 s, hence the speed at which the light is rotating is } \frac{d\theta}{dt} = \frac{2\pi}{10} = \frac{\pi}{5} rad \, s^{-1} \\ \text{Now, } tan\theta = \frac{r}{180} \\ r = 180tan\theta \\ \therefore \frac{dr}{dt} = \frac{dr}{d\theta} \cdot \frac{d\theta}{dt} \\ = \frac{180}{cos^2 \theta} \cdot \frac{\pi}{5}\\ = \frac{36\pi}{cos^2 \theta} \\ \text{Now, when } r = 300, tan\theta = 5/3 \\ \theta = 1.0304....\\ \therefore \frac{dr}{dt} = 136\pi \text{Exactly!}$

jet · May 21, 2009

AnandDNA · May 21, 2009

thanks for that

Rate of change involving trig functions (1 Viewer)

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