A revolving search light, which is 800 yards from the shore, makes 2 revolutions (4
radians) per minute. How fast is the light traveling along the straight beach when it
is 1000 yards from the lighthouse?
HELP

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- anonymous

- katieb

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- anonymous

is this a calculus problem?

- anonymous

yea

- anonymous

what topic is it in ?

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## More answers

- anonymous

thats what i was thinking .....because its right off the final review

- anonymous

damn you are typing a lot rob O.o

- anonymous

is this calc 1 mmonish91 ?

- anonymous

yes

- anonymous

sorry... i can't think of the topic this question involves

- anonymous

thanks for your help though

- anonymous

Imagine a rigid rod 1000 yards long, attached to the searchlight rotating @ 2 revolutions per minute or 120 revolutions per hour.
The circumference of a circle of radius 1000 is 2000 Pi.
Multiply 2000 Pi by 120 revolutions per hour and by 3 to convert to feet.
2000 * Pi * 120 * 3 = 720000 Pi feet /hr
The tip of the rod travels (720000 Pi)/5280 or 428.4 miles per hour over the beach in a circular motion.
I hope I got this right.

- anonymous

hmm.. interesting

- watchmath

\(\cos \theta = y/800\). We want to find \(dy/dt\) when \(y=1000\).
Take the derivative implicitly we have
\(-\sin\theta \cdot d\theta/dt=1/800\cdot dy/dt\qquad(*)\)
When \(y=1000\) the opposite side of the angle theta is \(\sqrt{1000^2-800^2}=360\).
In that case \(\sin\theta = 360/1000=3.6\)
plug in \(\sin \theta =3.5\) and \(y=1000\) to \((*)\) we have
\(-3.6\cdot 4\pi=1/800\cdot dy/dt\)
\(dy/dt=-3.6\cdot 4\pi\cdot 800\) yards/minute.

- watchmath

Sorry the length of the opposite side should be \(600\) and \(\sin \theta =0.6\)
So the answer is
\(dy/dt=-0.6\cdot 4\pi\cdot 800\)

- anonymous

@watchmath-\[\cos 4\pi = 800/y\cosine is adjacent over hyp

- watchmath

That is correct. The y is the hypothenuse, the 800 is the adjacent.

- anonymous

yes, so the one that i gave is correct?

- watchmath

The theta is also changing with respect to t. So we can only say that
\(\cos(\theta)=y/800\)

- anonymous

but, we are talking here with the cosine.

- watchmath

what are you trying to say about cosine function?

- anonymous

cosine is adjacent over hypothenuse

- watchmath

The \(4\pi\) is the rate of how the theta changes, i.e., \(d\theta/dt=4\pi\) and not an actual angle itself.

- watchmath

I agree with you, what I didn't agree is that you plug in \(4\pi\) for the angle.

- anonymous

yeah..i don't have problem with that.

- anonymous

i'm sorry, it must be no 4.

- watchmath

Ah I see... what you meant now ....yes. I should write 800/y instead of y/800. I didn't draw the picture. I just did it in my head :D. Thanks for the correction.

- anonymous

lol, you are welcome:)

- watchmath

(Please check again pat18)
I think I have to fix this before I off go to bed
\(\cos \theta =800/y\)
\(-\sin\theta\cdot d\theta/dt=-800/y^2\cdot dy/dt\)
\(0.6\cdot 4\pi=800/10^6\cdot dy/dt\)
\(dy/dt=3000\pi\text{ yards/minute}\)
(I am not good at physics, is that make sense?)

- anonymous

you are welcome.

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