simongagne53
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- May 21, 2010
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Hi everyone! Thanks in advance for your time
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How fast is this light beam moving?
- Thread starter simongagne53
- Start date
BigGlenntheHeavy
Senior Member
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- Mar 8, 2009
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\(\displaystyle Can't \ read \ the \ whole \ problem.\)
Hello, simongagne53!
The lighthouse is at \(\displaystyle A.\)
The nearest point on the shoreline is \(\displaystyle B.\)
The beacon is pointing at point \(\displaystyle C.\)
\(\displaystyle x = BC,\;u = \angle A.\)
\(\displaystyle \text{In right triangle }ABC:\;\tan u = \frac{x}{7200} \quad\Rightarrow\quad \boxed{x \:=\:7200\tan u}\) .[1]
\(\displaystyle \text{The beacon is turning at: }\:\frac{du}{dt} \;=\; \frac{7\text{rev}}{\text{ min}} \;=\;\frac{7\text{rev}}{\text{ min}}\cdot\frac{2\pi\text{ rad}}{\text{ rev}}\)
. . \(\displaystyle \text{Hence: }\;\boxed{\frac{du}{dt}\;=\;14\pi \text{rad/min}}\) .[2]
Differentiate [1] with respect to time:
. . \(\displaystyle \frac{dx}{dt} \;=\;7200\sec^2\!u\,\frac{du}{dt}\)
\(\displaystyle \text{When }x = 5400,\;AC = 9000 \quad\Rightarrow\quad \sec u \:=\:\frac{9000}{7200} \:=\:\frac{5}{4}\)
\(\displaystyle \text{We have: }\;\frac{dx}{dt} \;=\;7200\left(\tfrac{5}{4}\right)^2(14\pi)\)
. . \(\displaystyle \text{Therefore: }\;\boxed{\frac{dx}{dt} \;=\;157,500\pi\text{ ft/min}}\) .[3]
Code:. . A . . * . . |\ . . |u\ 7200 | \ . . | \ . . | \ B * - - * C x
A lighthouse is situtated on an island 7200 feet from a straight shoreline.
If the beacon on the lighthouse makes 7 revolutions per minute,
how fast is the light beam moving along the shoreline
when it is pointing 5400 feet awar from the nearest point?
To solve the problem, we first need to express the distance \(\displaystyle A.\)
along the shoreline in terms of angle \(\displaystyle u.\)
The lighthouse is at \(\displaystyle A.\)
The nearest point on the shoreline is \(\displaystyle B.\)
The beacon is pointing at point \(\displaystyle C.\)
\(\displaystyle x = BC,\;u = \angle A.\)
\(\displaystyle \text{In right triangle }ABC:\;\tan u = \frac{x}{7200} \quad\Rightarrow\quad \boxed{x \:=\:7200\tan u}\) .[1]
\(\displaystyle \text{The beacon is turning at: }\:\frac{du}{dt} \;=\; \frac{7\text{rev}}{\text{ min}} \;=\;\frac{7\text{rev}}{\text{ min}}\cdot\frac{2\pi\text{ rad}}{\text{ rev}}\)
. . \(\displaystyle \text{Hence: }\;\boxed{\frac{du}{dt}\;=\;14\pi \text{rad/min}}\) .[2]
Differentiate [1] with respect to time:
. . \(\displaystyle \frac{dx}{dt} \;=\;7200\sec^2\!u\,\frac{du}{dt}\)
\(\displaystyle \text{When }x = 5400,\;AC = 9000 \quad\Rightarrow\quad \sec u \:=\:\frac{9000}{7200} \:=\:\frac{5}{4}\)
\(\displaystyle \text{We have: }\;\frac{dx}{dt} \;=\;7200\left(\tfrac{5}{4}\right)^2(14\pi)\)
. . \(\displaystyle \text{Therefore: }\;\boxed{\frac{dx}{dt} \;=\;157,500\pi\text{ ft/min}}\) .[3]