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anonymous

  • one year ago

The beacon at the airport is located 300m from a straight runway ay its least distance. The beacon rotates 10times per minute, how fast is the beam of light moving along the runway when the beam is 700 from the closest point to the beacon on the runway.

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  1. anonymous
    • one year ago
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    |dw:1433048694364:dw|

  2. anonymous
    • one year ago
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    If your drawing is correct, you should just use the algorithm of Pythagoras to solve it: \[a^2+b^2=c^2\]|dw:1433054251458:dw|

  3. alekos
    • one year ago
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    Speed = 700 x 2π x 10

  4. alekos
    • one year ago
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    This will be in metres/min

  5. anonymous
    • one year ago
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    hi

  6. anonymous
    • one year ago
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    |dw:1433080009777:dw|

  7. anonymous
    • one year ago
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    (/pi)

  8. zepdrix
    • one year ago
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    hmm this one is tricky :( Feels like it might be too physics'y for me...

  9. zepdrix
    • one year ago
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    |dw:1433130117150:dw|

  10. zepdrix
    • one year ago
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    it's like a rotational speed that they're asking about, yes? they want to know instantaneous rate of change of some type of ... rotational speed? ugh

  11. anonymous
    • one year ago
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    I'm not sure what variable we're solving for

  12. anonymous
    • one year ago
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    Is it this? given dz/dt= 10times/min, find dy/dt=? when z=700 ???

  13. zepdrix
    • one year ago
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    hmm i have no idea :( @freckles @amistre64

  14. anonymous
    • one year ago
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    @freckles Please help

  15. freckles
    • one year ago
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    the light from the beacon would be the diagonal right?

  16. anonymous
    • one year ago
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    I think so..

  17. freckles
    • one year ago
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    I think I want to go ahead and convert your rot/min to rad/min

  18. freckles
    • one year ago
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    \[\frac{10 rot}{1 \min} \cdot \frac{2\pi rad}{1 rot}=20 \pi \frac{rad}{\min} \\ \]

  19. freckles
    • one year ago
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    himmm... so I guess we will use a trig ratio

  20. freckles
    • one year ago
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    |dw:1433133523545:dw| \[\cos(b)=\frac{z}{300}\]

  21. freckles
    • one year ago
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    we know db/dt and we are looking for dz/dt

  22. freckles
    • one year ago
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    but in your pic you say we are looking for dy/dt?

  23. anonymous
    • one year ago
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    I was guessing...

  24. freckles
    • one year ago
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    so I guess that means the becan isn't constantly 300m from the runaway

  25. freckles
    • one year ago
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    not sure how to interpret the least distance part

  26. freckles
    • one year ago
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    just because I know @ganeshie8 is awesome I will call him he has great math eyes but I will still try to read this thingy a few more times

  27. freckles
    • one year ago
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    and lol I have my trig ratio thing upside down

  28. freckles
    • one year ago
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    |dw:1433135234352:dw| I still not entirely sure what to do with the least distance part but I think it is just solving dz/dt from \[\cos(b)=\frac{300}{z}\] I could be wrong though you will have to find sin(b)=opp/hyp so find the opp measurements given those other two measurements haven't you guys been looking at related rates forever now?

  29. freckles
    • one year ago
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    seems like a really big number

  30. unknownunknown
    • one year ago
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    35km/minute?

  31. unknownunknown
    • one year ago
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    |dw:1433135659930:dw|

  32. ganeshie8
    • one year ago
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    |dw:1433136033005:dw|

  33. anonymous
    • one year ago
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    So z=100sqrt58 and we're trying to find dx/dt?

  34. ganeshie8
    • one year ago
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    Yes...

  35. ganeshie8
    • one year ago
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    lets draw a bit more neat picture

  36. ganeshie8
    • one year ago
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    |dw:1433136958532:dw|

  37. ganeshie8
    • one year ago
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    we're given frequency, \(f=10 ~\text{revolutions per minute}\) So, \[\dfrac{d\theta}{d t}=\omega =2\pi f = 2\pi (10) = 20\pi~\text{radians per minute} \]

  38. ganeshie8
    • one year ago
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    next suppose you're located at (x, 0) on x axis : |dw:1433137645938:dw|

  39. ganeshie8
    • one year ago
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    \[\tan \theta = \dfrac{x}{300}\] differentiate both sides implicitly with respect to \(t\), what do you get ?

  40. anonymous
    • one year ago
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    \[\sec^2\theta(\frac{ d \theta }{ dt })=\frac{ 1 }{ 300 }(\frac{ dx }{ dt })\] Is this correct?

  41. ganeshie8
    • one year ago
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    looks good! i hope you can mange the rest

  42. anonymous
    • one year ago
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    I acutally not sure how to go further. hint?

  43. ganeshie8
    • one year ago
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    plugin the value of \(\dfrac{d\theta}{dt}\)

  44. anonymous
    • one year ago
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    (58/9)(20pi)=(1/300)(dx/dt) ??

  45. anonymous
    • one year ago
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    then I got dx/dt= 348000/9 rad/min ??

  46. ganeshie8
    • one year ago
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    careful about the units x is not angle, x is the horizontal distance, unit should be "meter"

  47. ganeshie8
    • one year ago
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    it should be : dx/dt= 348000/9 pi meters/min

  48. anonymous
    • one year ago
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    Oops! Thank you for correcting me! :) and thank you for everything!

  49. ganeshie8
    • one year ago
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    yw!

  50. anonymous
    • one year ago
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    And also thank you @freckles! :) Thank you for helping me and trying to solve this question. :)

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