MoonlitFate
  • MoonlitFate
Find the exact length of the curve x= 9 cos t - cos 9t , y = 9 sin t - sin 9t , 0
Mathematics
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katieb
  • katieb
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MoonlitFate
  • MoonlitFate
This is what I have so far: \[L = \int\limits_{0}^{\pi}\sqrt{(9 \cos t -9 \cos 9t)^2 + (-9 \sin t + 9 \sin9t)^2}dt\]
MoonlitFate
  • MoonlitFate
Then you expand that out to be \[\int\limits_{0}^{\pi}\sqrt{81 \cos^2t-162 \cos t \cos 9t + 81 \cos^2 9t + 81 \sin^2 t - 162 \sin t \sin 9t + 81 \sin^2 9t} dt\]
IrishBoy123
  • IrishBoy123
take the 9 outside the root!

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MoonlitFate
  • MoonlitFate
The issue that I'm having is just simplifying everything.
IrishBoy123
  • IrishBoy123
1/ combine the cos^2 + sin^2 pairs
MoonlitFate
  • 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.
IrishBoy123
  • 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
MoonlitFate
  • MoonlitFate
Okay! Just give me a moment.:) Working this out as I go, and I thank you for your help and patience so far.
IrishBoy123
  • IrishBoy123
take your time, there is no rush
MoonlitFate
  • 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. \[9 \sqrt{9 (1) + 9(1) -9 [2 (\cos t \cos 9t + 9 \sin t \sin 9t)}\]
MoonlitFate
  • MoonlitFate
Hopefully, I didn't make a mistake somewhere.
IrishBoy123
  • IrishBoy123
the 9 should go out completely. you know you can copy your latex and re-do it, instead or re-typing it all?
MoonlitFate
  • MoonlitFate
I definitely forget that I can do that.
IrishBoy123
  • 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 \]
IrishBoy123
  • IrishBoy123
https://gyazo.com/f991fb8f9a929c36365c5c8843d3543d take show TEX then Ctrl A and Ctrl C!!
MoonlitFate
  • MoonlitFate
Let me see if I actually have the correct simplification of this radical. So far. \[9 \sqrt{(1 + 1 ) -2 (\cos t \cos 9t + \sin t \sin 9t)}\] \[9 \sqrt{(2 ) -2 (\cos t \cos 9t + \sin t \sin 9t)}\]
IrishBoy123
  • IrishBoy123
\[3 \sqrt{2} \sqrt{1 - (\cos t \cos 9t + \sin t \sin 9t)}\]
MoonlitFate
  • MoonlitFate
Always that algebra messing me up, but that's okay. At least I know I'm on the right track.
IrishBoy123
  • IrishBoy123
do you agree with the algebra? need my own work checked, you see :-)
MoonlitFate
  • MoonlitFate
It looks fine me.
MoonlitFate
  • MoonlitFate
So, the part that confuses me is that trig piece.
IrishBoy123
  • IrishBoy123
nah, it's: \[9 \sqrt{2} \sqrt{1 - (\cos t \cos 9t + \sin t \sin 9t)}\]
IrishBoy123
  • IrishBoy123
we can simplify: \[(\cos t \cos 9t + \sin t \sin 9t)\]
IrishBoy123
  • IrishBoy123
hint double angle formulae?? for cosine??
MoonlitFate
  • MoonlitFate
I have to look that up real quick. ;/ Something I should know off the top of my head at this point.
IrishBoy123
  • IrishBoy123
no rush!
MoonlitFate
  • MoonlitFate
Double angle for cosine: \[\cos (2u) = \cos^2 u - \sin ^2 u = 2 \cos^2 u - 1= 1 - 2 \sin^2 u \]
IrishBoy123
  • IrishBoy123
yes last one looks good ;-)
MoonlitFate
  • MoonlitFate
Trying to get that fit into the double angle is just what's confusing me. I'm sorry.
IrishBoy123
  • 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 \((\cos t \cos 9t + \sin t \sin 9t)\)
MoonlitFate
  • MoonlitFate
\[\cos (t- 9t) = costcos9t + sintsin9t\]
IrishBoy123
  • IrishBoy123
brill and cos (-x) = cos x so next step is?
MoonlitFate
  • MoonlitFate
Would that give you \[\cos (-8t) = - \cos(8t) \]
IrishBoy123
  • IrishBoy123
no + cos 8t
MoonlitFate
  • MoonlitFate
Oh, mixed up my signs. I got it.
IrishBoy123
  • 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
MoonlitFate
  • MoonlitFate
So, would I have to plug that in to the last part of the double angle formula for cosine and get: \[\cos(8t) = 1- 2 \sin^2(8t) \]
IrishBoy123
  • IrishBoy123
yes post it!
MoonlitFate
  • MoonlitFate
So, I would plug that back into the integral with the crazy trig: \[9\sqrt{2} \int\limits_{0}^{\pi}\sqrt{1-(1-2 \sin^2(8t)} dt\]
IrishBoy123
  • IrishBoy123
soz again \[\cos(8t) = 1- 2 \sin^2(4t)\] there was a typo there - because we half the angle yes :p
IrishBoy123
  • IrishBoy123
take it from here \[9\sqrt{2} \int\limits_{0}^{\pi}\sqrt{1-(1-2 \sin^2(4t)} ) \ dt\]
IrishBoy123
  • 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!!
IrishBoy123
  • IrishBoy123
|dw:1441652367386:dw|
MoonlitFate
  • MoonlitFate
Sorry simplifying, still. \[9\sqrt{2} \int\limits\limits_{0}^{\pi}\sqrt{1-(1-2 \sin^2(4t)} ) \ dt\] \[9\sqrt{2} \int\limits\limits_{0}^{\pi}\sqrt{(2 \sin^2(4t)} ) \ dt\]
MoonlitFate
  • MoonlitFate
\[18\int\limits\limits\limits_{0}^{\pi}( \sin(4t) ) \ dt\]
IrishBoy123
  • IrishBoy123
what do you get ?!?!
phi
  • phi
strictly speaking (and here we have to be) \[18 \int\limits\limits_{0}^{\pi}\sqrt{\sin^2(4t)} \ dt \\ =18 \int\limits\limits_{0}^{\pi}|\sin(4t)| \ dt \]
IrishBoy123
  • IrishBoy123
@phi yes
MoonlitFate
  • MoonlitFate
Why is it coming out to 0. Nooo.
phi
  • 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.
IrishBoy123
  • IrishBoy123
cool!!! that's the right conclusion we need to think about what we have done
IrishBoy123
  • IrishBoy123
we can start with this idea \[9\sqrt{2} \int\limits\limits_{0}^{\pi}\sqrt{(2 \sin^2(4t)} ) \ dt\] \[= 18\int\limits\limits\limits_{0}^{\pi} \pm \sin(4t) \ dt\]
IrishBoy123
  • IrishBoy123
|dw:1441653100623:dw|
IrishBoy123
  • IrishBoy123
i am very happy to let @phi take over from here
MoonlitFate
  • MoonlitFate
Is it legal to do this: \[18\int\limits\limits\limits\limits_{0}^{\pi} \pm \sin(4t) \ dt\] \[ 18 * 4 \int\limits\limits\limits\limits_{0}^{\pi} \pm \sin(t) \ dt\]
IrishBoy123
  • IrishBoy123
no! what we need are 4 of these: |dw:1441653185317:dw|
MoonlitFate
  • MoonlitFate
That never happened, you saw nothing...
IrishBoy123
  • IrishBoy123
|dw:1441653233509:dw| because these cancel out
IrishBoy123
  • 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?
MoonlitFate
  • MoonlitFate
u- substitution... would u = 4t in this case? :O
IrishBoy123
  • 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
IrishBoy123
  • IrishBoy123
|dw:1441653788624:dw|
IrishBoy123
  • IrishBoy123
so it 's \[4 \times 18 \int\limits\limits\limits_{0}^{\pi/4}( \sin(4t) ) \ dt\] n'est ce pas??
MoonlitFate
  • MoonlitFate
Working on getting to the point. :) Super close.
IrishBoy123
  • 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.
phi
  • 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.
phi
  • 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 \[\frac{1}{4} \int_0^\pi \sin(u) \ du \]
IrishBoy123
  • IrishBoy123
you see @MoonlitFate !! your sub was inspired despite my feedback.......
phi
  • phi
length is \[ 4 \cdot 18\cdot \frac{1}{4} \int_0^\pi \sin x \ dx \\ =18 \int_0^\pi \sin x \ dx \\ \]
phi
  • 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
IrishBoy123
  • IrishBoy123
i really like the "| |" idea as an explanatory tool......so we have \[18 \int\limits\limits_{0}^{\pi}|\sin(4t)| \ dt = 18 \left( \int\limits\limits_{0}^{\pi/4} \sin(4t) \ dt + (-1)\int\limits\limits_{\pi/4}^{\pi/2} \sin(4t) \ dt \\ + \int\limits\limits_{\pi/2}^{3\pi/4} \sin(4t) \ dt + (-1)\int\limits\limits_{3\pi/4}^{\pi} \sin(4t) \ dt\right)\]
IrishBoy123
  • IrishBoy123
@MoonlitFate hope this is ending well for you :p
MoonlitFate
  • 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.
IrishBoy123
  • 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

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