Consider the function r = f ( θ ) in polar coordinates. The length of an arc of a circle is just S = θ rWhere r is the radius of the circle and θ is the angle that represents this arc. But since r = f ( θ ) and θ Should approach zero so that we can get the exact value of the arc, So d S = f ( θ ) d θIntegrating from θ 1 to θ 2 , we get : S = ∫ θ 1 θ 2 f ( θ ) d θBut the actual formula for the length of a curve in polar coordinates is ∫ θ 1 θ 2 f 2 ( θ ) + f ′ ( θ ) 2 d θ .I know that my approach isn’t rigorous enough, but it’s is still reasonable, so why it is different from the actual formula?

Question

Consider the function   r  =  f  (  θ  ) in polar coordinates. The length of an arc of a circle is just  S  =  θ  rWhere   r is the radius of the circle and   θ is the angle that represents this arc. But since   r  =  f  (  θ  ) and   θ Should approach zero so that we can get the exact value of the arc, So    d  S  =  f  (  θ  )    d  θIntegrating from       θ    1   to       θ    2  , we get :  S  =      ∫                  θ        1                            θ        2              f  (  θ  )    d  θBut the actual formula for the length of a curve in polar coordinates is      ∫                  θ        1                            θ        2                        f      2        (    θ    )    +          f      ′        (    θ          )      2          d  θ  .I know that my approach isn’t rigorous enough, but it’s is still reasonable, so why it is different from the actual formula?

Blaze Frank · Accepted Answer

Your equation for   d  S is incorrect. In polar coordinates the differential length would be   f  (  θ  )  d  θ only if   r is a constant. If you draw a little diagram you&#039;ll see that, in fact, the differential element of length is given by  d  S  =            f      2        +                  (                            d        f                    d        θ                                      )                    2           d  θ

Consider the function r = f ( θ ) in polar coordinates. The le

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