Figure 5 Let o Pr = r and o Pr = r + r ∆ –1 x o Pr = and angle Pr o pr = θ ∆ θ 1 – 1 – Let N Pr be perpendicular to o Pr. 1 – Now N Pr = r and N Pr = r ∆ θ ∆ 1 – Hence (Pr Pr)2 = (N Pr)2 + (N Pr 1 1)2 – – = (∆ r)2 + (r ∆ θ)2 2 2 2 2 2  2  S 2 d  where and are the values of corresponding to at the end points of the curve. α β θ Example 6 = 0 to = 2π. Find the length of the equiangular spiral r = a ek from θ θ θ Solution The equation of the is given in polar form. The length of arc of a curve in polar form is given by 103  2  S 2 d  dae k 2 d = 2 k 0  d  a 2 2k = 0 =
Figure 5 Let o Pr –1 = r and o Pr = r + ∆ r x o Pr – 1 = θ and angle Pr – 1 o pr = ∆ θ Let N Pr – 1 be perpendicular to o Pr. Now N Pr – 1 = r ∆ θ and N Pr = ∆ r Hence (Pr – 1 Pr)2 = (N Pr)2 + (N Pr – 1)2 = (∆ r)2 + (r ∆ θ)2 2 2 2 2 2  S   2 d 2 where α and β are the values of θ corresponding to at the end points of the curve. Example 6 Find the length of the equiangular spiral r = a ek θ from θ = 0 to θ = 2π. Solution The equation of the is given in polar form. The length of arc of a curve in polar form is given by 103  S   2 d 2 = 0 k 2 dae k d 2 = 0 a 2 2k   d =
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