Question

Find the length of the curve r=1+cos(theta)

Answer 1

i got it to
2*int from 0 to pi sqrt(2+2cos(theta)) d(theta). I do not understand how to simplify it to
2*int from 0 to pi sqrt((4(1+cos(2theta))/2) d(theta)

I can get the rest

Answer 2

r^2=1+2costheta+cos^2theta and dr/dtheta = -sin(theta).. dr/dtheta^2 = sin^2(theta).. right?

Answer 3

thus L = int from 0 to 2pi sqrt(2+2cos(theta) dtheta

Answer 4

sorry, my equation and draw buttons do not work today...

Answer 5

r² + (r')²= (1 + cosθ)² = 1 + 2cosθ + cos²θ + (-sinθ)²
= 1 + 2cosθ

Answer 6

well, the thing is √(1 + 2cosθ) can not be integrated

Answer 7

that would = 2+2cos(theta) since cos^2theta +sin^2theta =1, right?

Answer 8

oops yes. But still, it can not be integrated

Answer 9

how do I get it from sqrt(2+2cos(theta) to (4(1+cos(2theta))/2 ?

Answer 10

sqrt(4(1+cos(2theta))/2 i mean.

Answer 11

then they pull the sqrt4 out front.. and are left with sqrt[(1+cos)theta)/(2)], which they change to cos (theta/2) and then integrate.

Answer 12

unfortunately my trig identity skills are poor.

Answer 13

are you sure the book did that? The integrand can not be integrated, so even if there were a trig identity, wolfram alpha would have done that. But it say that it can not be done

Answer 14

hmm.. thats what my sol manual .. and also Chegg says. Chegg did the simplification slightly different, but they both show the same answers.

Answer 15

ok, let me check for some trig identity

Answer 16

what did they have as answer?

Answer 17

8

Answer 18

what is the limit of integration??

Answer 19

0 to 2pi for the cardioid.. i did 2* 0 to pi.. since its symmetric

Answer 20

can you take a snapshot of their work? I don't have microsoft word to read the file

Answer 21

hmm what do u want me to paste teh screenshot to?

Answer 22

that works

Answer 23

btw, answer is indeed 8. But it was found by wolfram alpha

Answer 24

i don't have a way besides pasting the screenshot into a word document. how else can i get the screenshot to you?

Answer 25

its okay, I will ask my professor I guess. I am not good enough apparently to figure this out lol. hopefully this will not be an exam question haha

Answer 26

I think i might found the trig identity

Answer 27

hahah FOUND IT!! XDDD

Answer 28

awesome

Answer 29

ok, so √(2 + 2cosθ) = √(2) √(1 + cosθ) yes?

Answer 30

yes

Answer 31

cosθ = cos²(x/2) - sin²(x/2) <--- this bad retriceis the key XDD

Answer 32

they changed a.s.s to retrice O.O

Answer 33

heheh

Answer 34

anyway, under the square root, we have
1 + cos²(x/2) - sin²(x/2)

Answer 35

but, 1 - sin²(x/2) = cos²(x/2)

Answer 36

so, cos²(x/2) + cos²(x/2) = 2cos²(x/2)

Answer 37

oooooooh dang!!!!!!

Answer 38

so you have √(2) √[ 2 cos²(x/2) ] = 2 | cos(x/2) |

Answer 39

2 ∫ |cos(x/2)| dx, from 0 to 2pi = 8

XDDD

Answer 40

that's so awesome.

Answer 41

thank you so much!

Answer 42

you're welcome :DDD

Answer 43

that identity is not on my identity sheet ... lol guess i will have to add it.

Answer 44

good idea lol