A-lighthouse-L-is-located-on-a-small-island-2-km-from-the-nearest-point-A-on-a-long-the-straigh-shoreline-If-the-lighthouse-lamp-rotates-at-3-revolutions-per-minute-how-fast-is-the-illuminated-s

Question Number 132225 by bemath last updated on 12/Feb/21

$$ \\ $$$$\rightarrow\mathrm{A}\:\mathrm{lighthouse}\:\mathrm{L}\:\mathrm{is}\:\mathrm{located}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{small}\:\mathrm{island}\:\mathrm{2}\:\mathrm{km}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{nearest}\:\mathrm{point}\:\mathrm{A}\:\mathrm{on}\:\mathrm{a}\:\mathrm{long}\:\mathrm{the} \\ $$$$\mathrm{straigh}\:\mathrm{shoreline}\:.\:\mathrm{If}\:\mathrm{the} \\ $$$$\mathrm{lighthouse}\:\mathrm{lamp}\:\mathrm{rotates}\:\mathrm{at}\:\mathrm{3} \\ $$$$\mathrm{revolutions}\:\mathrm{per}\:\mathrm{minute}.\:\mathrm{how}\:\mathrm{fast}\: \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{illuminated}\:\mathrm{spot}\:\mathrm{P}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{shoreline}\:\mathrm{moving}\:\mathrm{along}\:\mathrm{the} \\ $$$$\mathrm{shoreline}\:\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{4}\:\mathrm{km}\:\mathrm{from}\:\mathrm{A}\: \\ $$

Answered by mr W last updated on 12/Feb/21

Commented by mr W last updated on 12/Feb/21

x=2 tan θ (dx/dt)=(dx/dθ)×(dθ/dt)=(2/(cos^2 θ))×(dθ/dt)=2(1+tan^2 θ)×(dθ/dt) =2[1+((x/2))^2 ]×(dθ/dt) at x=4: (dx/dt)=2[1+((4/2))^2 ]×((3×2π)/(60))=π=3.141 km/s

$${x}=\mathrm{2}\:\mathrm{tan}\:\theta \\ $$$$\frac{{dx}}{{dt}}=\frac{{dx}}{{d}\theta}×\frac{{d}\theta}{{dt}}=\frac{\mathrm{2}}{\mathrm{cos}^{\mathrm{2}} \:\theta}×\frac{{d}\theta}{{dt}}=\mathrm{2}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)×\frac{{d}\theta}{{dt}} \\ $$$$=\mathrm{2}\left[\mathrm{1}+\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} \right]×\frac{{d}\theta}{{dt}} \\ $$$${at}\:{x}=\mathrm{4}: \\ $$$$\frac{{dx}}{{dt}}=\mathrm{2}\left[\mathrm{1}+\left(\frac{\mathrm{4}}{\mathrm{2}}\right)^{\mathrm{2}} \right]×\frac{\mathrm{3}×\mathrm{2}\pi}{\mathrm{60}}=\pi=\mathrm{3}.\mathrm{141}\:{km}/{s} \\ $$

Commented by bemath last updated on 12/Feb/21

i got (dx/dt) ∣_(x=4 km) = 60π km/min sir. correct? 60π ≈ 188.495559 km/min

$$\mathrm{i}\:\mathrm{got}\:\frac{\mathrm{dx}}{\mathrm{dt}}\:\mid_{\mathrm{x}=\mathrm{4}\:\mathrm{km}} \:=\:\mathrm{60}\pi\:\mathrm{km}/\mathrm{min} \\ $$$$\mathrm{sir}.\:\mathrm{correct}? \\ $$$$\mathrm{60}\pi\:\approx\:\mathrm{188}.\mathrm{495559}\:\mathrm{km}/\mathrm{min} \\ $$$$ \\ $$

Commented by mr W last updated on 12/Feb/21