Question

Find the exact length of the curve

x= 9 cos t - cos 9t , y = 9 sin t - sin 9t , 0 10 years ago

Answer 1

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

Answer 2

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

Answer 3

take the 9 outside the root!

Answer 4

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

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

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

Answer 9

take your time, there is no rush

Answer 10

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.
\[9 \sqrt{9 (1) + 9(1) -9 [2 (\cos t \cos 9t + 9 \sin t \sin 9t)}\]

Answer 11

Hopefully, I didn't make a mistake somewhere.

Answer 12

the 9 should go out completely.

you know you can copy your latex and re-do it, instead or re-typing it all?

Answer 13

I definitely forget that I can do that.

Answer 14

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 \]

Answer 15

https://gyazo.com/f991fb8f9a929c36365c5c8843d3543d

take show TEX

then Ctrl A and Ctrl C!!

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

do you agree with the algebra?

need my own work checked, you see :-)

Answer 20

It looks fine me.

Answer 21

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

Answer 22

nah, it's:


\[9 \sqrt{2} \sqrt{1 - (\cos t \cos 9t + \sin t \sin 9t)}\]

Answer 23

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

Answer 24

hint

double angle formulae??

for cosine??

Answer 25

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

Answer 26

no rush!

Answer 27

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

Answer 28

yes

last one looks good ;-)

Answer 29

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

Answer 30

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 \((\cos t \cos 9t + \sin t \sin 9t)\)

Answer 31

\[\cos (t- 9t) = costcos9t + sintsin9t\]

Answer 32

brill

and cos (-x) = cos x

so next step is?

Answer 33

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

Answer 34

no

+ cos 8t

Answer 35

Oh, mixed up my signs. I got it.

Answer 36

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

Answer 37

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

Answer 38

yes

post it!

Answer 39

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

Answer 40

soz again

\[\cos(8t) = 1- 2 \sin^2(4t)\]

there was a typo there - because we half the angle

yes :p

Answer 41

take it from here

\[9\sqrt{2} \int\limits_{0}^{\pi}\sqrt{1-(1-2 \sin^2(4t)} ) \ dt\]

Answer 42

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!!