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.

- anonymous

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- anonymous

|dw:1433048694364:dw|

- 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

Speed = 700 x 2π x 10

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## More answers

- alekos

This will be in metres/min

- anonymous

hi

- anonymous

|dw:1433080009777:dw|

- anonymous

(/pi)

- zepdrix

hmm this one is tricky :(
Feels like it might be too physics'y for me...

- zepdrix

|dw:1433130117150:dw|

- 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

I'm not sure what variable we're solving for

- anonymous

Is it this?
given dz/dt= 10times/min,
find dy/dt=? when z=700
???

- zepdrix

hmm i have no idea :(
@freckles @amistre64

- anonymous

@freckles Please help

- freckles

the light from the beacon would be the diagonal right?

- anonymous

I think so..

- freckles

I think I want to go ahead and convert your rot/min to rad/min

- freckles

\[\frac{10 rot}{1 \min} \cdot \frac{2\pi rad}{1 rot}=20 \pi \frac{rad}{\min} \\ \]

- freckles

himmm...
so I guess we will use a trig ratio

- freckles

|dw:1433133523545:dw|
\[\cos(b)=\frac{z}{300}\]

- freckles

we know db/dt and we are looking for dz/dt

- freckles

but in your pic you say we are looking for dy/dt?

- anonymous

I was guessing...

- freckles

so I guess that means the becan isn't constantly 300m from the runaway

- freckles

not sure how to interpret the least distance part

- 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

and lol I have my trig ratio thing upside down

- 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

seems like a really big number

- unknownunknown

35km/minute?

- unknownunknown

|dw:1433135659930:dw|

- ganeshie8

|dw:1433136033005:dw|

- anonymous

So z=100sqrt58
and we're trying to find dx/dt?

- ganeshie8

Yes...

- ganeshie8

lets draw a bit more neat picture

- ganeshie8

|dw:1433136958532:dw|

- 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

next suppose you're located at (x, 0) on x axis :
|dw:1433137645938:dw|

- ganeshie8

\[\tan \theta = \dfrac{x}{300}\]
differentiate both sides implicitly with respect to \(t\), what do you get ?

- anonymous

\[\sec^2\theta(\frac{ d \theta }{ dt })=\frac{ 1 }{ 300 }(\frac{ dx }{ dt })\]
Is this correct?

- ganeshie8

looks good! i hope you can mange the rest

- anonymous

I acutally not sure how to go further. hint?

- ganeshie8

plugin the value of \(\dfrac{d\theta}{dt}\)

- anonymous

(58/9)(20pi)=(1/300)(dx/dt) ??

- anonymous

then I got dx/dt= 348000/9 rad/min ??

- ganeshie8

careful about the units
x is not angle, x is the horizontal distance, unit should be "meter"

- ganeshie8

it should be :
dx/dt= 348000/9 pi meters/min

- anonymous

Oops! Thank you for correcting me! :) and thank you for everything!

- ganeshie8

yw!

- anonymous

And also thank you @freckles! :) Thank you for helping me and trying to solve this question. :)

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