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anonymous

  • 5 years ago

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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  1. anonymous
    • 5 years ago
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    is this a calculus problem?

  2. anonymous
    • 5 years ago
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    yea

  3. anonymous
    • 5 years ago
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    what topic is it in ?

  4. anonymous
    • 5 years ago
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    thats what i was thinking .....because its right off the final review

  5. anonymous
    • 5 years ago
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    damn you are typing a lot rob O.o

  6. anonymous
    • 5 years ago
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    is this calc 1 mmonish91 ?

  7. anonymous
    • 5 years ago
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    yes

  8. anonymous
    • 5 years ago
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    sorry... i can't think of the topic this question involves

  9. anonymous
    • 5 years ago
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    thanks for your help though

  10. anonymous
    • 5 years ago
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    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.

  11. anonymous
    • 5 years ago
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    hmm.. interesting

  12. watchmath
    • 5 years ago
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    \(\cos \theta = y/800\). 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.

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

  14. anonymous
    • 5 years ago
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    @watchmath-\[\cos 4\pi = 800/y\cosine is adjacent over hyp

  15. watchmath
    • 5 years ago
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    That is correct. The y is the hypothenuse, the 800 is the adjacent.

  16. anonymous
    • 5 years ago
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    yes, so the one that i gave is correct?

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

  18. anonymous
    • 5 years ago
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    but, we are talking here with the cosine.

  19. watchmath
    • 5 years ago
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    what are you trying to say about cosine function?

  20. anonymous
    • 5 years ago
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    cosine is adjacent over hypothenuse

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

  22. watchmath
    • 5 years ago
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    I agree with you, what I didn't agree is that you plug in \(4\pi\) for the angle.

  23. anonymous
    • 5 years ago
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    yeah..i don't have problem with that.

  24. anonymous
    • 5 years ago
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    i'm sorry, it must be no 4.

  25. watchmath
    • 5 years ago
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    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.

  26. anonymous
    • 5 years ago
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    lol, you are welcome:)

  27. watchmath
    • 5 years ago
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    (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?)

  28. anonymous
    • 5 years ago
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    you are welcome.

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