Asked by Jon
A rotating beacon is located 1 kilometer off a straight shoreline. If the beacon rotates at a rate of 3 revolutions per minute, how fast (in kilometers per hour) does the beam of light appear to be moving to a viewer who is 1/2 kilometer down the shoreline.
I need to show work, so formatting answers in this manner would be most appreciated. Thanks in advance! :) :)
I need to show work, so formatting answers in this manner would be most appreciated. Thanks in advance! :) :)
Answers
Answered by
Anonymous
Evaluate the limit as h -> 0 of:
[tan (pi/6 + h) - tan(pi/6)]/h
I thought the answer was √3/3, or tan(pi/6, but apparently that is wrong, any tips here?
[tan (pi/6 + h) - tan(pi/6)]/h
I thought the answer was √3/3, or tan(pi/6, but apparently that is wrong, any tips here?
Answered by
Anonymous
Sorry, I mean to make a new question, disregard above 'answer'.
Answered by
Jon
Gee, now you're going to make people who open this topic think the answer has already been provided. Thanks a ton...
Answered by
SuperGirl
You're a jerk.
Answered by
H H Chau
tan(θ)=x/1
cos(θ)=1/sqrt(x^2+1)
sec^2(θ)=x^2+1
dθ/dt=3 rev/min=6π rad/min
x=tan(θ)
dx/dt=sec^2(θ) dθ/dt
At x=0.5 km
dx/dt=(0.5^2+1)*6π=23.6 km/min=1414 kph
cos(θ)=1/sqrt(x^2+1)
sec^2(θ)=x^2+1
dθ/dt=3 rev/min=6π rad/min
x=tan(θ)
dx/dt=sec^2(θ) dθ/dt
At x=0.5 km
dx/dt=(0.5^2+1)*6π=23.6 km/min=1414 kph
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