anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1433048694364:dw|
anonymous
  • anonymous
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|
alekos
  • alekos
Speed = 700 x 2π x 10

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alekos
  • alekos
This will be in metres/min
anonymous
  • anonymous
hi
anonymous
  • anonymous
|dw:1433080009777:dw|
anonymous
  • anonymous
(/pi)
zepdrix
  • zepdrix
hmm this one is tricky :( Feels like it might be too physics'y for me...
zepdrix
  • zepdrix
|dw:1433130117150:dw|
zepdrix
  • zepdrix
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
anonymous
  • anonymous
I'm not sure what variable we're solving for
anonymous
  • anonymous
Is it this? given dz/dt= 10times/min, find dy/dt=? when z=700 ???
zepdrix
  • zepdrix
hmm i have no idea :( @freckles @amistre64
anonymous
  • anonymous
@freckles Please help
freckles
  • freckles
the light from the beacon would be the diagonal right?
anonymous
  • anonymous
I think so..
freckles
  • freckles
I think I want to go ahead and convert your rot/min to rad/min
freckles
  • freckles
\[\frac{10 rot}{1 \min} \cdot \frac{2\pi rad}{1 rot}=20 \pi \frac{rad}{\min} \\ \]
freckles
  • freckles
himmm... so I guess we will use a trig ratio
freckles
  • freckles
|dw:1433133523545:dw| \[\cos(b)=\frac{z}{300}\]
freckles
  • freckles
we know db/dt and we are looking for dz/dt
freckles
  • freckles
but in your pic you say we are looking for dy/dt?
anonymous
  • anonymous
I was guessing...
freckles
  • freckles
so I guess that means the becan isn't constantly 300m from the runaway
freckles
  • freckles
not sure how to interpret the least distance part
freckles
  • freckles
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
freckles
  • freckles
and lol I have my trig ratio thing upside down
freckles
  • freckles
|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?
freckles
  • freckles
seems like a really big number
unknownunknown
  • unknownunknown
35km/minute?
unknownunknown
  • unknownunknown
|dw:1433135659930:dw|
ganeshie8
  • ganeshie8
|dw:1433136033005:dw|
anonymous
  • anonymous
So z=100sqrt58 and we're trying to find dx/dt?
ganeshie8
  • ganeshie8
Yes...
ganeshie8
  • ganeshie8
lets draw a bit more neat picture
ganeshie8
  • ganeshie8
|dw:1433136958532:dw|
ganeshie8
  • ganeshie8
we're given frequency, \(f=10 ~\text{revolutions per minute}\) So, \[\dfrac{d\theta}{d t}=\omega =2\pi f = 2\pi (10) = 20\pi~\text{radians per minute} \]
ganeshie8
  • ganeshie8
next suppose you're located at (x, 0) on x axis : |dw:1433137645938:dw|
ganeshie8
  • ganeshie8
\[\tan \theta = \dfrac{x}{300}\] differentiate both sides implicitly with respect to \(t\), what do you get ?
anonymous
  • anonymous
\[\sec^2\theta(\frac{ d \theta }{ dt })=\frac{ 1 }{ 300 }(\frac{ dx }{ dt })\] Is this correct?
ganeshie8
  • ganeshie8
looks good! i hope you can mange the rest
anonymous
  • anonymous
I acutally not sure how to go further. hint?
ganeshie8
  • ganeshie8
plugin the value of \(\dfrac{d\theta}{dt}\)
anonymous
  • anonymous
(58/9)(20pi)=(1/300)(dx/dt) ??
anonymous
  • anonymous
then I got dx/dt= 348000/9 rad/min ??
ganeshie8
  • ganeshie8
careful about the units x is not angle, x is the horizontal distance, unit should be "meter"
ganeshie8
  • ganeshie8
it should be : dx/dt= 348000/9 pi meters/min
anonymous
  • anonymous
Oops! Thank you for correcting me! :) and thank you for everything!
ganeshie8
  • ganeshie8
yw!
anonymous
  • anonymous
And also thank you @freckles! :) Thank you for helping me and trying to solve this question. :)

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