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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

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  1. moonlitfate
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    take the 9 outside the root!

  4. moonlitfate
    • one year ago
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    The issue that I'm having is just simplifying everything.

  5. IrishBoy123
    • one year ago
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    1/ combine the cos^2 + sin^2 pairs

  6. moonlitfate
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    take your time, there is no rush

  10. moonlitfate
    • one year ago
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    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
    • one year ago
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    Hopefully, I didn't make a mistake somewhere.

  12. IrishBoy123
    • one year ago
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    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
    • one year ago
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    I definitely forget that I can do that.

  14. IrishBoy123
    • one year ago
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    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
    • one year ago
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    https://gyazo.com/f991fb8f9a929c36365c5c8843d3543d take show TEX then Ctrl A and Ctrl C!!

  16. moonlitfate
    • one year ago
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    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
    • one year ago
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    \[3 \sqrt{2} \sqrt{1 - (\cos t \cos 9t + \sin t \sin 9t)}\]

  18. moonlitfate
    • one year ago
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    Always that algebra messing me up, but that's okay. At least I know I'm on the right track.

  19. IrishBoy123
    • one year ago
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    do you agree with the algebra? need my own work checked, you see :-)

  20. moonlitfate
    • one year ago
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    It looks fine me.

  21. moonlitfate
    • one year ago
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    So, the part that confuses me is that trig piece.

  22. IrishBoy123
    • one year ago
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    nah, it's: \[9 \sqrt{2} \sqrt{1 - (\cos t \cos 9t + \sin t \sin 9t)}\]

  23. IrishBoy123
    • one year ago
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    we can simplify: \[(\cos t \cos 9t + \sin t \sin 9t)\]

  24. IrishBoy123
    • one year ago
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    hint double angle formulae?? for cosine??

  25. moonlitfate
    • one year ago
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    I have to look that up real quick. ;/ Something I should know off the top of my head at this point.

  26. IrishBoy123
    • one year ago
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    no rush!

  27. moonlitfate
    • one year ago
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    Double angle for cosine: \[\cos (2u) = \cos^2 u - \sin ^2 u = 2 \cos^2 u - 1= 1 - 2 \sin^2 u \]

  28. IrishBoy123
    • one year ago
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    yes last one looks good ;-)

  29. moonlitfate
    • one year ago
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    Trying to get that fit into the double angle is just what's confusing me. I'm sorry.

  30. IrishBoy123
    • one year ago
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    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
    • one year ago
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    \[\cos (t- 9t) = costcos9t + sintsin9t\]

  32. IrishBoy123
    • one year ago
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    brill and cos (-x) = cos x so next step is?

  33. moonlitfate
    • one year ago
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    Would that give you \[\cos (-8t) = - \cos(8t) \]

  34. IrishBoy123
    • one year ago
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    no + cos 8t

  35. moonlitfate
    • one year ago
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    Oh, mixed up my signs. I got it.

  36. IrishBoy123
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    yes post it!

  39. moonlitfate
    • one year ago
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    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
    • one year ago
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    soz again \[\cos(8t) = 1- 2 \sin^2(4t)\] there was a typo there - because we half the angle yes :p

  41. IrishBoy123
    • one year ago
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    take it from here \[9\sqrt{2} \int\limits_{0}^{\pi}\sqrt{1-(1-2 \sin^2(4t)} ) \ dt\]

  42. IrishBoy123
    • one year ago
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    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
    • one year ago
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    |dw:1441652367386:dw|

  44. moonlitfate
    • one year ago
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    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
    • one year ago
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    \[18\int\limits\limits\limits_{0}^{\pi}( \sin(4t) ) \ dt\]

  46. IrishBoy123
    • one year ago
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    what do you get ?!?!

  47. phi
    • one year ago
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    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
    • one year ago
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    @phi yes

  49. moonlitfate
    • one year ago
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    Why is it coming out to 0. Nooo.

  50. phi
    • one year ago
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    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
    • one year ago
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    cool!!! that's the right conclusion we need to think about what we have done

  52. IrishBoy123
    • one year ago
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    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
    • one year ago
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    |dw:1441653100623:dw|

  54. IrishBoy123
    • one year ago
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    i am very happy to let @phi take over from here

  55. moonlitfate
    • one year ago
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    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
    • one year ago
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    no! what we need are 4 of these: |dw:1441653185317:dw|

  57. moonlitfate
    • one year ago
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    That never happened, you saw nothing...

  58. IrishBoy123
    • one year ago
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    |dw:1441653233509:dw| because these cancel out

  59. IrishBoy123
    • one year ago
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    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
    • one year ago
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    u- substitution... would u = 4t in this case? :O

  61. IrishBoy123
    • one year ago
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    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
    • one year ago
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    |dw:1441653788624:dw|

  63. IrishBoy123
    • one year ago
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    so it 's \[4 \times 18 \int\limits\limits\limits_{0}^{\pi/4}( \sin(4t) ) \ dt\] n'est ce pas??

  64. moonlitfate
    • one year ago
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    Working on getting to the point. :) Super close.

  65. IrishBoy123
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    you see @MoonlitFate !! your sub was inspired despite my feedback.......

  69. phi
    • one year ago
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    length is \[ 4 \cdot 18\cdot \frac{1}{4} \int_0^\pi \sin x \ dx \\ =18 \int_0^\pi \sin x \ dx \\ \]

  70. phi
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    @MoonlitFate hope this is ending well for you :p

  73. moonlitfate
    • one year ago
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    @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
    • one year ago
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    @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

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