## 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.

1. anonymous

|dw:1433048694364:dw|

2. 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|

3. alekos

Speed = 700 x 2π x 10

4. alekos

This will be in metres/min

5. anonymous

hi

6. anonymous

|dw:1433080009777:dw|

7. anonymous

(/pi)

8. zepdrix

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

9. zepdrix

|dw:1433130117150:dw|

10. 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

11. anonymous

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

12. anonymous

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

13. zepdrix

hmm i have no idea :( @freckles @amistre64

14. anonymous

15. freckles

the light from the beacon would be the diagonal right?

16. anonymous

I think so..

17. freckles

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

18. freckles

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

19. freckles

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

20. freckles

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

21. freckles

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

22. freckles

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

23. anonymous

I was guessing...

24. freckles

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

25. freckles

not sure how to interpret the least distance part

26. 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

27. freckles

and lol I have my trig ratio thing upside down

28. 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?

29. freckles

seems like a really big number

30. unknownunknown

35km/minute?

31. unknownunknown

|dw:1433135659930:dw|

32. ganeshie8

|dw:1433136033005:dw|

33. anonymous

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

34. ganeshie8

Yes...

35. ganeshie8

lets draw a bit more neat picture

36. ganeshie8

|dw:1433136958532:dw|

37. ganeshie8

we're given frequency, $$f=10 ~\text{revolutions per minute}$$ So, $\dfrac{d\theta}{d t}=\omega =2\pi f = 2\pi (10) = 20\pi~\text{radians per minute}$

38. ganeshie8

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

39. ganeshie8

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

40. anonymous

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

41. ganeshie8

looks good! i hope you can mange the rest

42. anonymous

I acutally not sure how to go further. hint?

43. ganeshie8

plugin the value of $$\dfrac{d\theta}{dt}$$

44. anonymous

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

45. anonymous

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

46. ganeshie8

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

47. ganeshie8

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

48. anonymous

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

49. ganeshie8

yw!

50. anonymous

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