Find arc length of r = 2 cos ⁡ θ in the range 0 ≤ θ ≤ π ?

Question

Find arc length of   r  =  2  cos  ⁡  θ in the range   0  ≤  θ  ≤  π ?

gozaderaradiox5 · Accepted Answer

Explanation:  L  =  ∫  d  s  =      ∫                  θ                  1                                    θ                  2                      r  d  θ  s  =  r  θ so   d  s  =  r  d  θ , thus  2      ∫          0              π        cos  ⁡  θ  =  2  (  sin  ⁡  θ  )            |              0              π        =  2  (  sin  ⁡  π  −  sin  ⁡  0  )  =  .  .  .Considering that this is for a semicircle perimeter of a full cycle..  4  (  sin  ]  t  h  e  t  a  )            |              0                      π        2              =  4  (  sin  ⁡  (      π    2    )  −  sin  ⁡  0  )  =  4  (  1  −  0  )  =  4

Audrina Jackson · Accepted Answer

It is easy to see that the curve is a circle of radius 1. It's length is obviously

2 π

A more analytic solution would go as follows

d s^{2} = d r^{2} + r^{2} d θ^{2}

So, for

r = 2 \cos θ

, we have

d r = - 2 \sin θ d θ

and hence

d s^{2} = (- 2 \sin θ d θ)^{2} + (2 \cos θ)^{2} d θ^{2} = 4 d θ^{2} \Rightarrow

d s = 2 d θ

Thus, the arc length is

L = \int_{θ = 0}^{θ = π} d s = 2 π

Find arc length of r =

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