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Arc length

  • Thread starter quietrain
  • Start date
  • #1
654
2

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!
 

Answers and Replies

  • #2
tiny-tim
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hi quietrain! :smile:

(have a pi: π and a square-root: √ and an integral: ∫ :wink:)
… 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 ? :confused:
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 :wink:
 
  • #3
654
2
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 :(
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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hi quietrain! :smile:
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 :wink:
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 :redface:

you need ∫30|sintcost| dt (or ∫|sin2t| dt ) :wink:
 
  • #5
654
2
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?
 
  • #6
SammyS
Staff Emeritus
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[tex]\sin(2t)\geq 0 \text{ for }0\leq t\leq \frac{\pi}{2}\ .[/tex]

[tex]\sin(2t)\leq 0 \text{ for }\frac{\pi}{2}\leq t\leq\pi \ .[/tex]
 
  • #7
tiny-tim
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Homework Helper
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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 :redface:

as SammyS :smile: indicates, you should consider going to π/2 and quadrupling that, instead of π and doubling :wink:
 
  • #8
654
2
haha... i see thanks everyone!

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

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