## MoonlitFate one year ago Find the exact length of the curve x= 9 cos t - cos 9t , y = 9 sin t - sin 9t , 0<t<pi

1. moonlitfate

This is what I have so far: $L = \int\limits_{0}^{\pi}\sqrt{(9 \cos t -9 \cos 9t)^2 + (-9 \sin t + 9 \sin9t)^2}dt$

2. moonlitfate

Then you expand that out to be $\int\limits_{0}^{\pi}\sqrt{81 \cos^2t-162 \cos t \cos 9t + 81 \cos^2 9t + 81 \sin^2 t - 162 \sin t \sin 9t + 81 \sin^2 9t} dt$

3. IrishBoy123

take the 9 outside the root!

4. moonlitfate

The issue that I'm having is just simplifying everything.

5. IrishBoy123

1/ combine the cos^2 + sin^2 pairs

6. moonlitfate

Would that also include the sin^2(9t) and cos^2(9t) as well. I apologize if that seems like a silly question. I just wasn't sure.

7. IrishBoy123

yes! but please get those 81's and 162'2 out and in fornt of the $$\int$$!!! it's trivial but it doesn't help

8. moonlitfate

Okay! Just give me a moment.:) Working this out as I go, and I thank you for your help and patience so far.

9. IrishBoy123

take your time, there is no rush

10. moonlitfate

It's easier for me to group these. Way to easy to get a bit thrown off by all the stuff going on under the radical. $9 \sqrt{9 (1) + 9(1) -9 [2 (\cos t \cos 9t + 9 \sin t \sin 9t)}$

11. moonlitfate

Hopefully, I didn't make a mistake somewhere.

12. IrishBoy123

the 9 should go out completely. you know you can copy your latex and re-do it, instead or re-typing it all?

13. moonlitfate

I definitely forget that I can do that.

14. IrishBoy123

double tap or right click on the equation and one of the options will be to show the latex code this will save you so much time in the long run if you can do that i use chrome then you can paste it back inside $$or $and$$ or$

15. IrishBoy123

https://gyazo.com/f991fb8f9a929c36365c5c8843d3543d take show TEX then Ctrl A and Ctrl C!!

16. moonlitfate

Let me see if I actually have the correct simplification of this radical. So far. $9 \sqrt{(1 + 1 ) -2 (\cos t \cos 9t + \sin t \sin 9t)}$ $9 \sqrt{(2 ) -2 (\cos t \cos 9t + \sin t \sin 9t)}$

17. IrishBoy123

$3 \sqrt{2} \sqrt{1 - (\cos t \cos 9t + \sin t \sin 9t)}$

18. moonlitfate

Always that algebra messing me up, but that's okay. At least I know I'm on the right track.

19. IrishBoy123

do you agree with the algebra? need my own work checked, you see :-)

20. moonlitfate

It looks fine me.

21. moonlitfate

So, the part that confuses me is that trig piece.

22. IrishBoy123

nah, it's: $9 \sqrt{2} \sqrt{1 - (\cos t \cos 9t + \sin t \sin 9t)}$

23. IrishBoy123

we can simplify: $(\cos t \cos 9t + \sin t \sin 9t)$

24. IrishBoy123

hint double angle formulae?? for cosine??

25. moonlitfate

I have to look that up real quick. ;/ Something I should know off the top of my head at this point.

26. IrishBoy123

no rush!

27. moonlitfate

Double angle for cosine: $\cos (2u) = \cos^2 u - \sin ^2 u = 2 \cos^2 u - 1= 1 - 2 \sin^2 u$

28. IrishBoy123

yes last one looks good ;-)

29. moonlitfate

Trying to get that fit into the double angle is just what's confusing me. I'm sorry.

30. IrishBoy123

sorry, my bad, i was looking a step ahead let's start with $$cos (A- B) = cosA cosB + sinA sinB$$ can you work that and $$(\cos t \cos 9t + \sin t \sin 9t)$$

31. moonlitfate

$\cos (t- 9t) = costcos9t + sintsin9t$

32. IrishBoy123

brill and cos (-x) = cos x so next step is?

33. moonlitfate

Would that give you $\cos (-8t) = - \cos(8t)$

34. IrishBoy123

no + cos 8t

35. moonlitfate

Oh, mixed up my signs. I got it.

36. IrishBoy123

cos(-x) = cos x [going to relaunch this thread as my copy/ paste thing is being funny "not" going away - and this one has a really interesting ending

37. moonlitfate

So, would I have to plug that in to the last part of the double angle formula for cosine and get: $\cos(8t) = 1- 2 \sin^2(8t)$

38. IrishBoy123

yes post it!

39. moonlitfate

So, I would plug that back into the integral with the crazy trig: $9\sqrt{2} \int\limits_{0}^{\pi}\sqrt{1-(1-2 \sin^2(8t)} dt$

40. IrishBoy123

soz again $\cos(8t) = 1- 2 \sin^2(4t)$ there was a typo there - because we half the angle yes :p

41. IrishBoy123

take it from here $9\sqrt{2} \int\limits_{0}^{\pi}\sqrt{1-(1-2 \sin^2(4t)} ) \ dt$

42. IrishBoy123

that is what we are measuring! believe it or not. the folks at desmos have asked me if it can be used as some kind of staff favourite. feels like a practical joke!!

43. IrishBoy123

|dw:1441652367386:dw|

44. moonlitfate

Sorry simplifying, still. $9\sqrt{2} \int\limits\limits_{0}^{\pi}\sqrt{1-(1-2 \sin^2(4t)} ) \ dt$ $9\sqrt{2} \int\limits\limits_{0}^{\pi}\sqrt{(2 \sin^2(4t)} ) \ dt$

45. moonlitfate

$18\int\limits\limits\limits_{0}^{\pi}( \sin(4t) ) \ dt$

46. IrishBoy123

what do you get ?!?!

47. phi

strictly speaking (and here we have to be) $18 \int\limits\limits_{0}^{\pi}\sqrt{\sin^2(4t)} \ dt \\ =18 \int\limits\limits_{0}^{\pi}|\sin(4t)| \ dt$

48. IrishBoy123

@phi yes

49. moonlitfate

Why is it coming out to 0. Nooo.

50. phi

and fyi, your plot shows 1/4 of the curve (t=0 to pi/2). we get 1/2 of the curve for t=0 to pi.

51. IrishBoy123

cool!!! that's the right conclusion we need to think about what we have done

52. IrishBoy123

we can start with this idea $9\sqrt{2} \int\limits\limits_{0}^{\pi}\sqrt{(2 \sin^2(4t)} ) \ dt$ $= 18\int\limits\limits\limits_{0}^{\pi} \pm \sin(4t) \ dt$

53. IrishBoy123

|dw:1441653100623:dw|

54. IrishBoy123

i am very happy to let @phi take over from here

55. moonlitfate

Is it legal to do this: $18\int\limits\limits\limits\limits_{0}^{\pi} \pm \sin(4t) \ dt$ $18 * 4 \int\limits\limits\limits\limits_{0}^{\pi} \pm \sin(t) \ dt$

56. IrishBoy123

no! what we need are 4 of these: |dw:1441653185317:dw|

57. moonlitfate

That never happened, you saw nothing...

58. IrishBoy123

|dw:1441653233509:dw| because these cancel out

59. IrishBoy123

this is a pig of a question the sting in the tail but do you see that we need to change the integration limit?

60. moonlitfate

u- substitution... would u = 4t in this case? :O

61. IrishBoy123

nooooooo :-))) we just need to draw in the integration interval we have taken something complicated and transformed it into a much simpler integral. look back up the thread! but because the simpler sin(4t) oscillates up and down, we know that we have to be careful. as th sin4t integral gives us zero so we take the first +ve integral and times that by 4 if @phi can teach this $18 \int\limits\limits_{0}^{\pi}\sqrt{\sin^2(4t)} \ dt \\ =18 \int\limits\limits_{0}^{\pi}|\sin(4t)| \ dt$ in a simpler way, i am all ears

62. IrishBoy123

|dw:1441653788624:dw|

63. IrishBoy123

so it 's $4 \times 18 \int\limits\limits\limits_{0}^{\pi/4}( \sin(4t) ) \ dt$ n'est ce pas??

64. moonlitfate

Working on getting to the point. :) Super close.

65. IrishBoy123

take your time i hope @phi comes back as there might be a better way to do this, and if so @phi would know it.

66. phi

I would do what you are doing...we want to integrate over a "1/4 cycle" of sin(4t) (that is for 4t= 0 to pi, because we know sin will be positive over that interval). that means t ranges from 0/4 to pi/4 or 0 to pi/4 once we have that value, we know that there are 4 such intervals in the range from t=0 to pi, so multiply by 4 to get the final length.

67. phi

to integrate sin(4t) dt let u = 4t du = 4 dt (so dt = 1/4 du) lower limit t=0 means u=0 upper limit t= pi/4 means u = pi thus do $\frac{1}{4} \int_0^\pi \sin(u) \ du$

68. IrishBoy123

you see @MoonlitFate !! your sub was inspired despite my feedback.......

69. phi

length is $4 \cdot 18\cdot \frac{1}{4} \int_0^\pi \sin x \ dx \\ =18 \int_0^\pi \sin x \ dx \\$

70. phi

FYI, for real numbers, here is how some people define the absolute value function. See eq 1 in https://en.wikipedia.org/wiki/Absolute_value#Real_numbers It does not particularly help us here, except it alerts us to be careful about how we integrate regions where sin(4t) is negative (we must negate these areas to get a positive value). the easiest approach is to exploit symmetry i.e. multiply by 4 the arc length found for t=0 to pi/4

71. IrishBoy123

i really like the "| |" idea as an explanatory tool......so we have $18 \int\limits\limits_{0}^{\pi}|\sin(4t)| \ dt = 18 \left( \int\limits\limits_{0}^{\pi/4} \sin(4t) \ dt + (-1)\int\limits\limits_{\pi/4}^{\pi/2} \sin(4t) \ dt \\ + \int\limits\limits_{\pi/2}^{3\pi/4} \sin(4t) \ dt + (-1)\int\limits\limits_{3\pi/4}^{\pi} \sin(4t) \ dt\right)$

72. IrishBoy123

@MoonlitFate hope this is ending well for you :p

73. moonlitfate

@IrishBoy123 @phi The confusion is still pretty real, but I'll come back to this problem later to see if anything makes a bit more sense. :) Thank you both for helping! I will definitely be posting more problems. Parametrics is just really confusing me.

74. IrishBoy123

@MoonlitFate it must be quite confusing to go from this |dw:1441657483924:dw| to this |dw:1441657602178:dw| if you haven't really seen it before good luck :p