Question

arclength integration

Answer 1

\[x=sint-cost\]\[y=sint+cost\]\[\frac{ \pi }{ 4 }\le x \le \frac{ \pi }{ 2 }\]

Answer 2

alright when I plugged it in to the arclength integration equation for both I ended up with \[\int\limits_{\pi/4}^{\pi/2}\sqrt{2+\sin2t}dt\]

Answer 3

and Im not sure how to proceed from this point

Answer 4

correction the range is \[\frac{ \pi }{ 4 } \le t \le \frac{ \pi }{ 2 }\]

Answer 5

how did you get the 2+sin(2t)?

Answer 6

indeed, shouldn't it be just 2

Answer 7

yes

Answer 8

\[\int\limits \sqrt{1+ (f'(x))^{2}}\]\[x'=cost+sint\]\[\int\limits \sqrt{1+\cos^{2}t+\sin^{2}t+2sintcost}dt\]\[\sin^{2}t+\cos^{2}t=1\]\[2sintcost=\sin2t\]\[\int\limits \sqrt{2+\sin2t}dt\]

Answer 9

\[\int\limits_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\]

Answer 10

my bad thanks

Answer 11

personal opinion, *for what it's worth*: always always always do these [and all double/triple/ surface/vector integration] from first-ish principles, and then you don't need to worry about/ mess up formulae. it adds nothing to workload. so for arcs, start with:

∆s^2 = ∆x^2 + ∆y^2 + ∆z^2, and take it from there.
S = ∑∆s = ∫ds

every time.

Answer 12

@IrishBoy123 I got a bunch of question marks for your reply

Answer 13

hit F5 to refresh your computer

Answer 14

@ IrishBoy123 that's not how I was taught for the derivation of that equation so would \[S= \int\limits \sqrt{ (\frac{ dx }{ dt })^{2}+(\frac{ dy }{ dt })^{2}+(\frac{ dz }{ dt })^{2}}dt\] also be true?

Answer 15

indeed it would !