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Determine the arc length of the curve given by x = t cos t , y = t sin t from t=0

We have a parametrization of the curve as x=t\cos t,\,y=t\sin t

{\dot  {x}}=\cos t-t\sin t,\,{\dot  {y}}=\sin t+t\cos t

({\dot  {x}})^{2}+({\dot  {y}})^{2}=(\cos t-t\sin t)^{2}+(\sin t+t\cos t)^{2}=(\cos ^{2}t-2t\cos t\sin t+t^{2}\sin ^{2}t)+(\sin ^{2}t+2t\sin t\cos t+t^{2}\cos ^{2}t)

=\cos ^{2}t+\sin ^{2}t+t^{2}\sin ^{2}t+t^{2}\cos ^{2}t=1+t^{2}\,

In order to integrate we now need a substitution: Let t=\cosh u\Rightarrow dt=\sinh u\,du,{\sqrt  {1+t^{2}}}=\sinh u

Hence L=\int _{{\operatorname {arcosh}0}}^{{\operatorname {arcosh}u}}\sinh ^{2}u\,du=\left[{\frac  {\sinh 2u-2u}{4}}\right]_{{\operatorname {arcosh}0}}^{{\operatorname {arcosh}u}}=\left[{\frac  {2t{\sqrt  {1+t^{2}}}-2\operatorname {arcosh}\,t}{4}}\right]_{0}^{t}={\frac  {1}{2}}t{\sqrt  {1+t^{2}}}-{\frac  {1}{2}}\operatorname {arcosh}\,t