Find Arc Length of Curve over [0, 2π]

[quietrain]

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]

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

 

[quietrain]

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]

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 )

 

[quietrain]

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 ! that's 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]

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

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

 

[tiny-tim]

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

 

[quietrain]

haha... i see thanks everyone!

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