## anonymous one year ago A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out? i got -8/400 but it was wrong

1. freckles

|dw:1434325170769:dw| picture looks something like this

2. freckles

|dw:1434325227788:dw|

3. freckles

"A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out? i got -8/400 but it was wrong " --- Moves horizontally at a speed of 8ft/s this means x'=8ft/s we are looking for theta (in rad) when y=200ft

4. freckles

well what is theta when y=200ft

5. freckles

what trig function can you use here

6. freckles

|dw:1434325371019:dw|

7. freckles

hint you have opp and hyp measurements given

8. anonymous

ok i did sin of 100/200 and got .479

9. freckles

arcsin(1/2)? which is pi/6

10. anonymous

why arcsin?

11. freckles

to solve sin(theta)=1/2 you normally take arcsin( ) of both sides since arcsin(sin(x))=x where x is between -pi/2 and pi/2 |dw:1434325633246:dw| anyways we are going to use x (because we are given x's rate) and theta for sure and I will go ahead and use that constant side $\tan(\theta)=\frac{100}{x}$ differentiate both side

12. anonymous

sec^2x=0?

13. freckles

how did you get that equation?

14. freckles

sec is never 0

15. anonymous

im confused...

16. freckles

Like can you tell me what is confusing you please? Because I don't know how to help if I don't know where to start to help.

17. anonymous

the derivative of tan=100/x thats where i got confused

18. freckles

well the equation is actually $\tan(\theta)=\frac{100}{x} \text{ or you can write it as } \\ \tan(\theta)=100x^{-1}$ so you don't have to use quotient rule if you prefer

19. freckles

you will need to use chain rule on both sides though

20. freckles

do you know how to evluate either one of these: $\frac{d \tan(t)}{dt}=? \\ \frac{d(t^{-1})}{dt}=?$

21. freckles

for the first it is easy just to remember derivative of tan(t) w.r.t t is sec^2(t) and the second one I asked you about can be done by power rule have you every used power rule before? I hope so because this question is about application of chain rule

22. freckles

$\tan(\theta)=100x^{-1} \\ \text{ differentiate both sides } \\ \sec^2(\theta) \cdot \frac{d \theta}{d t}=100 \cdot \frac{d(x^{-1})}{dt}$ you still to differentiate the right hand side

23. anonymous

would it be 100/-x^2?

24. freckles

don't forget you need to write times dx/dt

25. freckles

$\sec^2(\theta) \cdot \frac{d \theta}{ d t}=-\frac{100}{x^2} \cdot \frac{dx }{dt}$

26. freckles

plug in all the values you found and are given and solve for dtheta/dt

27. anonymous

would it be sec^2(pi/6)*dtheta/dt=- 100/1ooroot3* 200?

28. freckles

well the x has a square on it and I thought dx/dt was given as 8ft/sec why you put 200 for dx/dt? dx/dt represents the rate and which the kite moves horizontally

29. freckles

$\sec^2(\frac{\pi}{6}) \cdot \frac{ d \theta}{ d t}=\frac{-100}{(100 \sqrt{3})^2} \cdot 8$

30. anonymous

i solved the right hand side but then how do i divide by the sec^2 im havinf trouble calculating that?

31. freckles

1/sec(k)=cos(k)

32. freckles

$\frac{d \theta}{dt} =\frac{1}{\sec^2(\frac{\pi}{6})} \frac{-100}{(100 \sqrt{3})^2} \cdot 8 \\ \frac{d \theta}{ dt}=\cos^2(\frac{\pi}{6}) \frac{-100}{(100\sqrt{3})^2} \cdot 8$

33. anonymous

okay i got -.o2?

34. freckles

ok and i have -1/50 those should be equal

35. freckles

don't forget to put your units

36. freckles

have fun

37. anonymous

wait

38. anonymous

my answer was -8/400 and i got it wrong?

39. anonymous

or is it not supposed to be negative?

40. freckles

you can try without did you put the units though ?

41. anonymous

42. freckles

k

43. freckles

have you got it yet?

44. anonymous

i only have one more try so im scared to type it and get it wrong

45. freckles

ok don't know what to tell you we have the answer above says it is decreasing at a rate of 1/50 rad/sec

46. anonymous

i know this thing is weird -_- but thank you for your help though!!

47. anonymous

it was supposed to be positive got it right!!!