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- Thread starter joy
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- Thread starter
- #1

- Mar 1, 2012

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$\dfrac{d\theta}{dt}$ = 5 rpm = $\dfrac{10\pi}{60 \, sec} = \dfrac{\pi}{6}$ rad/sec

consider the right triangle formed by the light beam, the shoreline, and the perpendicular distance from the light house to the shoreline (recommend you make a sketch)

let $\theta$ be the angle between the light beam and the perpendicular distance, and $x$ be the distance from where the light beam intersects the shoreline to where the perpendicular distance segment intersects the shoreline. You are given the fixed perpendicular distance.

Using the aforementioned right triangle, write a trig equation that relates $x$, $\theta$, and the 16 km distance.

Take the time derivative of the equation and determine $\dfrac{dx}{dt}$ when $x=3$ km.

Mind your units.

consider the right triangle formed by the light beam, the shoreline, and the perpendicular distance from the light house to the shoreline (recommend you make a sketch)

let $\theta$ be the angle between the light beam and the perpendicular distance, and $x$ be the distance from where the light beam intersects the shoreline to where the perpendicular distance segment intersects the shoreline. You are given the fixed perpendicular distance.

Using the aforementioned right triangle, write a trig equation that relates $x$, $\theta$, and the 16 km distance.

Take the time derivative of the equation and determine $\dfrac{dx}{dt}$ when $x=3$ km.

Mind your units.

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