This is the equation of acardioid. The polar equation is expressed in terms of a periodic function, so we have to know its period to get the exact length (and not just a portion or some excess of it).
r=1+sinθ has period T=2π (the function can be obtained as a vertical translation of the sine function in the plane of coordinates (θ,r)).The length of a periodic polar curve can be computed by integrating the arc length on a complete period of the function, i.e. on an interval I of length T=2π:
L=∫Ids
where ds=r2+(dθdr)2dθ
So we have to compute the derivative:
dθdr=dθd(1+sinθ)=cosθ
and this implies
ds=(1+sinθ)2+cosθ2dθ=1+2sinθ+sin2θ+cos2θdθ
By the Pythagorean trigonometric identity cos
cos2θ+sin2θ=1
ds=2+2sinθdθ=2(1+sinθ)dθ
Integrating that square root is not easy... we'd better take advantage of trigonometric identities. We have 1+sinθ
(multiplied by a numerical coefficient) and we would like to get rid of that square root. The half angle formula for cosine states that
cos2(2α)=21+cosα.
We can use this with sine too, because if
α=θ−2π
then
cosα=cos(θ−2π)=sinθ.
So we substitute in the formula and get cos22θ−2π=21+cos(θ−2π)
and then
2cos2(4θ−4π)=1+sinθ
ds=4cos2(4θ−4π)dθ=2∣cos(4θ−4π)∣dθ
Now we can integrate this on an interval of length 2π. To do that, we would like to get rid of the absolute value.
We know that
cos(2x−4π)⪖0⟺−2π+2πk⪕2x−4π⪕2π+2πk
for k any integer value, so we get that −
−2ππ+4πk⪕x⪕23π+4πk
In the end, we can integrate the arc length on
I=[−2π;23π], since this is an interval of length 2π. We can remove the absolute value, since cos(2x−2π)⪖0 on I. ∫−2π23π2cos(2θ−4π)dθ=4sin2π−4sin2−π=4+4=8
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