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anonymous
 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
anonymous
 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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is what I have so far: \[L = \int\limits_{0}^{\pi}\sqrt{(9 \cos t 9 \cos 9t)^2 + (9 \sin t + 9 \sin9t)^2}dt\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then you expand that out to be \[\int\limits_{0}^{\pi}\sqrt{81 \cos^2t162 \cos t \cos 9t + 81 \cos^2 9t + 81 \sin^2 t  162 \sin t \sin 9t + 81 \sin^2 9t} dt\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2take the 9 outside the root!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The issue that I'm having is just simplifying everything.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.21/ combine the cos^2 + sin^2 pairs

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would 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
 one year ago
Best ResponseYou've already chosen the best response.2yes! but please get those 81's and 162'2 out and in fornt of the \(\int\)!!! it's trivial but it doesn't help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay! Just give me a moment.:) Working this out as I go, and I thank you for your help and patience so far.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2take your time, there is no rush

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It'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. \[9 \sqrt{9 (1) + 9(1) 9 [2 (\cos t \cos 9t + 9 \sin t \sin 9t)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hopefully, I didn't make a mistake somewhere.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2the 9 should go out completely. you know you can copy your latex and redo it, instead or retyping it all?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I definitely forget that I can do that.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2double 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
 one year ago
Best ResponseYou've already chosen the best response.2https://gyazo.com/f991fb8f9a929c36365c5c8843d3543d take show TEX then Ctrl A and Ctrl C!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let 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
 one year ago
Best ResponseYou've already chosen the best response.2\[3 \sqrt{2} \sqrt{1  (\cos t \cos 9t + \sin t \sin 9t)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Always that algebra messing me up, but that's okay. At least I know I'm on the right track.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2do you agree with the algebra? need my own work checked, you see :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, the part that confuses me is that trig piece.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2nah, it's: \[9 \sqrt{2} \sqrt{1  (\cos t \cos 9t + \sin t \sin 9t)}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2we can simplify: \[(\cos t \cos 9t + \sin t \sin 9t)\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2hint double angle formulae?? for cosine??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have to look that up real quick. ;/ Something I should know off the top of my head at this point.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Double angle for cosine: \[\cos (2u) = \cos^2 u  \sin ^2 u = 2 \cos^2 u  1= 1  2 \sin^2 u \]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2yes last one looks good ;)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Trying to get that fit into the double angle is just what's confusing me. I'm sorry.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2sorry, my bad, i was looking a step ahead let's start with \(cos (A B) = cosA cosB + sinA sinB\) can you work that and \((\cos t \cos 9t + \sin t \sin 9t)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\cos (t 9t) = costcos9t + sintsin9t\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2brill and cos (x) = cos x so next step is?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would that give you \[\cos (8t) =  \cos(8t) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, mixed up my signs. I got it.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2cos(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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, would I have to plug that in to the last part of the double angle formula for cosine and get: \[\cos(8t) = 1 2 \sin^2(8t) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, I would plug that back into the integral with the crazy trig: \[9\sqrt{2} \int\limits_{0}^{\pi}\sqrt{1(12 \sin^2(8t)} dt\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2soz again \[\cos(8t) = 1 2 \sin^2(4t)\] there was a typo there  because we half the angle yes :p

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2take it from here \[9\sqrt{2} \int\limits_{0}^{\pi}\sqrt{1(12 \sin^2(4t)} ) \ dt\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2that 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441652367386:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry simplifying, still. \[9\sqrt{2} \int\limits\limits_{0}^{\pi}\sqrt{1(12 \sin^2(4t)} ) \ dt\] \[9\sqrt{2} \int\limits\limits_{0}^{\pi}\sqrt{(2 \sin^2(4t)} ) \ dt\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[18\int\limits\limits\limits_{0}^{\pi}( \sin(4t) ) \ dt\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2what do you get ?!?!

phi
 one year ago
Best ResponseYou've already chosen the best response.1strictly speaking (and here we have to be) \[18 \int\limits\limits_{0}^{\pi}\sqrt{\sin^2(4t)} \ dt \\ =18 \int\limits\limits_{0}^{\pi}\sin(4t) \ dt \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why is it coming out to 0. Nooo.

phi
 one year ago
Best ResponseYou've already chosen the best response.1and 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
 one year ago
Best ResponseYou've already chosen the best response.2cool!!! that's the right conclusion we need to think about what we have done

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2we 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441653100623:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2i am very happy to let @phi take over from here

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is 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
 one year ago
Best ResponseYou've already chosen the best response.2no! what we need are 4 of these: dw:1441653185317:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That never happened, you saw nothing...

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441653233509:dw because these cancel out

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2this is a pig of a question the sting in the tail but do you see that we need to change the integration limit?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0u substitution... would u = 4t in this case? :O

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2nooooooo :))) 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441653788624:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2so it 's \[4 \times 18 \int\limits\limits\limits_{0}^{\pi/4}( \sin(4t) ) \ dt\] n'est ce pas??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Working on getting to the point. :) Super close.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2take 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
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1to 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
 one year ago
Best ResponseYou've already chosen the best response.2you see @MoonlitFate !! your sub was inspired despite my feedback.......

phi
 one year ago
Best ResponseYou've already chosen the best response.1length is \[ 4 \cdot 18\cdot \frac{1}{4} \int_0^\pi \sin x \ dx \\ =18 \int_0^\pi \sin x \ dx \\ \]

phi
 one year ago
Best ResponseYou've already chosen the best response.1FYI, 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
 one year ago
Best ResponseYou've already chosen the best response.2i 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
 one year ago
Best ResponseYou've already chosen the best response.2@MoonlitFate hope this is ending well for you :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.2@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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