# Arc length

## Homework Statement

find arc length of curve over the interval t(0,2pi)

r(t) = 10cos3t i + 10sin3t j

## The Attempt at a Solution

i apply the formula integral ||r'(t)|| over the interval 0,2pi

i get integrate sqrt((-30cos2tsint)2 + (30sin2tcost)2)
and then finally get 15sin2t |0 to 2pi

and i get 0

but the ans for the length is 60... :(

i believe it has got to do with me traversing the curve r(t) more than once when t goes from 0 to 2pi. but how do i know or show that that is the case? and how do i rectify this problem?

thanks!

tiny-tim
Homework Helper
hi quietrain!

(have a pi: π and a square-root: √ and an integral: ∫ )
… i apply the formula integral ||r'(t)|| over the interval 0,2pi

i get integrate sqrt((-30cos2tsint)2 + (30sin2tcost)2)
and then finally get 15sin2t |0 to 2pi

and i get 0

but the ans for the length is 60... :(

how did you get sin2t ?
i believe it has got to do with me traversing the curve r(t) more than once when t goes from 0 to 2pi. but how do i know or show that that is the case? and how do i rectify this problem?

when you take the square-root, you must be careful to make sure it's always positive

you may need to do the integral from 0 to π, and then double it

oh i simplify the expression to
∫30sintcost from 0 to 2π (is this wrong?)

so it becomes
15sin2t
or
-15cos2t

but integrating from pi to 0 or 2pi to 0 for either gives me 0 :(

tiny-tim
Homework Helper
hi quietrain!
oh i simplify the expression to
∫30sintcost from 0 to 2π (is this wrong?)

no, that integrand correct …

but you could simplify it to 15sin2t
so it becomes
15sin2t
or
-15cos2t

but integrating from pi to 0 or 2pi to 0 for either gives me 0 :(

you didn't do what i said

you need ∫30|sintcost| dt (or ∫|sin2t| dt )

so it will be 15∫|sin2t| dt from 0 to pi which is all positive and i double it so its to 2pi

so -30(1/2)cos2t from 0 to pi

so -15 ( 1 - 1 ) = 0 :(

or must i change my limits since it is now cos? so to get +ve cos i need to integrate from 0 to pi/2 and x4 to get 2pi?

so 15∫|sin2t| dt = -30cos2t (after x4) from 0 to pi/2
so = -30 ( -1 -1) = 60 ! thats the answer ...

but it seems trial and error style :( ???

weird.. so if i want ∫|sin2t| dt , the integral of modulus of sin2t , i am actually wanting the integrated form , cos, to be positive? so i have to change my integrating limits too?

SammyS
Staff Emeritus
Homework Helper
Gold Member
$$\sin(2t)\geq 0 \text{ for }0\leq t\leq \frac{\pi}{2}\ .$$

$$\sin(2t)\leq 0 \text{ for }\frac{\pi}{2}\leq t\leq\pi \ .$$

tiny-tim
Homework Helper
so it will be 15∫|sin2t| dt from 0 to pi which is all positive and i double it so its to 2pi

(what happened to that π i gave you?)

nooo

as SammyS indicates, you should consider going to π/2 and quadrupling that, instead of π and doubling

haha... i see thanks everyone!

(oh the pi looks like n :( i thought it was confusing)