Arc length of a curve (trigonometric identity)

  • Thread starter t_n_p
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  • #1
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Homework Statement



find arc length of the segment of the 2space curbe that is defined by the parametric equations

x(t) = t-sin(t)
y(t) = 1+cos(t)
0 ≤ t ≤ 4π

The Attempt at a Solution


I've found dx/dt and dy/dt respectively and put them into the arc length equation, i.e. sqrt[(dx/dt)²+(dy/dt)²]

dx/dt = 1-cost
dy/dt = -sint

therefore arc length L = sqrt[(1-cost)²+(-sint)²]
this leads to L = sqrt[1-2cost+cos²t+sin²t]

I am then told to use the double angle formula 2sin²t = 1-cos2t to simplify the integrand. I cannot see how this applies though!
If someone could point me in the right direction o:)
 

Answers and Replies

  • #2
danago
Gold Member
1,122
4
If cos 2t = 1-2sin²t, then cos t = 1-2sin²(t/2).

Sub that into the integrand and it should make it easier, and use the pythagorean identity to reduce sin²t+cos²t to 1.
 
  • #3
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cool!
so I used that and trig identity and got sqrt[4sin²(t/2)]

I am given a second hint that says sqrt(sin²t) = |sint|,
my question now is when I integrate |sin²(t/2)|, does anything change or will I perform integration as normal?
 
  • #4
danago
Gold Member
1,122
4
I would think that you'd need to integrate over different parts of the interval separately.

For 0 < t < 2π, sin(t/2) is positive, so that |sin(t/2)|=sin(t/2)

However for 2π < t < 4π, sin(t/2) is negative so that |sin(t/2)|=-sin(t/2)
 
  • #5
595
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ah yes, thank you! I went on to find an arc length of 16.
 
Last edited:

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