## anonymous 4 years ago A kite 50ft above the ground moves horizontally at a speed of 8ft/s. At what rate is the angle between the string and the horizontal decreasing when 300ft of string has been let out?

1. anonymous

|dw:1351143411523:dw|

2. anonymous

tan(theta) = 50/x differentiate both sides with respect to time find x when sqrt(x^2+50^2) =300 (*spoiler*) find theta when x= sqrt(300^2 -50^2) plug x, theta and dx/dt into the result of the differentiation to find d(theta)/dt

3. anonymous

can you explain the diferrentiatie both sides with respect to time part?

4. anonymous

@Algebraic!

5. anonymous

I guess, it's just implicit differentiation really...or the chain rule... or whatever you are comfortable thinking of it as... d/dt ( tan ( f(t) ) =

6. anonymous

derivative of the 'outside' is...?

7. anonymous

sec^2

8. anonymous

:)

9. anonymous

so it's x = 50/sec^2(theta)?

10. anonymous

yep, so sec^2 (f(t) ) * f '(t)

11. anonymous

f(t) is theta yeah, I wrote it that way because theta depends on time and I wanted you to see the chain rule

12. anonymous

so LHS is sec^2(theta) * d(theta) /dt RHS still needs to be differentiated...

13. anonymous

d/dt ( 1/(g(t)) ) =...?

14. anonymous

-1/ ..........

15. anonymous

what is g(t) in this case?

16. anonymous

well, just because I used f(t) for theta... now we're talking about x, which is a different function of time...

17. anonymous

ah okay

18. anonymous

-1/x^2*dx/dt

19. anonymous

great:)

20. anonymous

so:$\frac{ d \theta }{dt } = \frac{ -1 }{x^2 \sec^2\theta }\frac{ d x}{ dt }$

21. anonymous

dx/dt is given.. x and theta are easy to find...

22. anonymous

dx/dt = 8

23. anonymous

d(theta)/dt = -8/x^2*sec(theta)

24. anonymous

x= sqrt(300^2 -50^2) theta = arctan (50/x)

25. anonymous

wouldn't x = sqrt(300^2 + 50^2)?

26. anonymous

x = 304.138

27. anonymous

28. anonymous

@Algebraic!

29. anonymous

so x is longer than the hypotenuse?

30. anonymous

x = 295.803

31. anonymous

k:)

32. anonymous

so d(theta)/dt = -8/((295.803)^2*(sec^2(.16744 rads)) which is incorrect.

33. anonymous

-8.87E-5?

34. anonymous

well you said d(theta)/dt = -1/(x^2*sec^2(theta)) * dx/dt

35. anonymous

yes

36. anonymous

you didn't get -8.87E-5?

37. anonymous

yes I did

38. anonymous

and it says it's wrong...

39. anonymous

yep

40. anonymous

dunno let me look it all over. I don't see any glaring mistakes...

41. anonymous

alright

42. anonymous

-8.89E-5 rad.s/sec .. best I can do... let's see who else is on, might be able to see if I made a mistake...

43. anonymous

May I have the answer to the question?

44. anonymous

go for it.

45. anonymous

@eSpeX

46. anonymous

@RolyPoly id like the answer too!

47. anonymous

@callisto is checking it over...

48. anonymous

ty @Callisto

49. Callisto

I... am no good at maths... $tan \theta = \frac{50}{x}$ Differentiate both sides with respect to x. Probably you won't get: $\frac{ d \theta }{dt } = \frac{ -1 }{x^2 \sec^2\theta }\frac{ d x}{ dt }$

50. Callisto

My bad, I meant with respect to t

51. anonymous

you're saying the differentiation is wrong?

52. Callisto

$tan \theta = \frac{50}{x}$ Diff. both sides w.r.t. t $sec^2\theta \frac{d\theta}{dt} = \frac{-50}{x^2} \frac{dx}{dt}$$\frac{d\theta}{dt} = \frac{-50}{x^2sec^2\theta } \frac{dx}{dt}$

53. anonymous

arg

54. anonymous

forgot the 50 rfl

55. anonymous

you so smart cally

56. anonymous

:)

57. anonymous

58. anonymous

thank you @Callisto and @Algebraic!

59. anonymous

yes thanks @Callisto !

60. Callisto

$sec^2\theta=\frac{x^2+50}{x^2}$You can cancel the x^2 since $\frac{d\theta}{dt} = \frac{-50}{x^2sec^2\theta } \frac{dx}{dt}=\frac{-50}{x^2(\frac{x^2+50}{x^2}) } \frac{dx}{dt}$$=\frac{-50}{(x^2+50) } \frac{dx}{dt}$And x^2 is easy to find

61. anonymous

only took me 4hrs but i got my question worth 1pt right now to get some sleep

62. anonymous

heh

63. anonymous

you guys have a good night thanks again

64. Callisto

Good night!!~

65. anonymous

A revised Mathematica solution, posted on 25 October 2012, 22:15 California time, is attached.