Question

Setup an integral that will give the arclength of the equation r = 1+2cos(theta) from 0 to pi. Do not evaluate, just the formula.

Answer 1

i guess you need to integrate something like this

Answer 2

So??

Answer 3

Where did you get "s"?

Answer 4

i thought the question is asking the arc length?! im not sure what the question needs you the do
the arc length s=r\(\theta \)

Answer 5

I thought arclength for polar coords was \[\int\limits_{0}^{1} \sqrt(\frac{ d(r) }{ d(\Theta) } + r(\Theta)^2)\]

Answer 6

I don't quite remember since i haven't touched this in awhile
is that ds

Answer 7

i mean that expression

Answer 8

since the question wants you to find arc length
you need to do \[\int\limits ds\]

Answer 9

I've never heard of doing "ds". Can you elaborate?

Answer 10

well since we are looking for the length of an arc we need to integrate all the smallest ds
to get the length. s being an infinitesimally small arc length

Answer 11

not sure about this stuff! @zepdrix

Answer 12

We usually denote an arc with \(\Large\rm s\).
So we break that arc into immeasurably small pieces of length \(\Large\rm ds\).

And then the total arc length is the sum of all of those little pieces.\[\Large\rm s=\int\limits ds\]

I remember the formula for arc length when you're dealing with parametric equations:\[\Large\rm \int\limits ds=\int\limits \sqrt{1+\left(\frac{dy}{dx}\right)^2}~~dx\]Looks like polar is somewhat similar.
I can't remember how you derive the equation but your formula you posted looks correct.\[\Large\rm \int\limits \sqrt{r^2+\left(\frac{d r}{d \theta}\right)^2}~~d \theta\]

Answer 13

We have our function:\[\Large\rm r(\theta)=1+2\cos \theta\]Taking a derivative is pretty straight forward, yes?

Answer 14

Hmm why are your bounds 0 and 1.
We're applying these bounds to theta, yes?
So our theta should range from 0 to pi I think?

Answer 15

yeah it should be 0 to pi

Answer 16

now you just need to integrate that stuff without evaluating the inegral

Answer 17

Yeah, the derivation is simple, and the range is what @xapproachesinfinity said, from 0 to pi, but it can't be THAT simple...can it? The prof left a decent amount of space, which leads me to assume it's a bit complex, but if that's it (of course, simplification aside), then I'm ok with that. Thank you both!

Answer 18

Yw