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{\bf Exam Question

Topic: Arc Length}

Calculate the length of the curve given by $$x=t^2,y=2t,(0\le t\le
1).$$ Give your answer both in exact form and also as an
approximation rounded to four decimal places, using your
calculator. \vspace{0.5in}

{\bf Solution}


$$L=\int_0^1\sqrt{x'(t)^2+y'(t)^2}\ dt =\int_0^1\sqrt{4t^2+4}\ dt
= 2\int_0^1\sqrt{t^2+1}\ dt $$

Let $t=\sinh u;dt=\cosh u\ du.$

\begin{eqnarray*} L &amp; = &amp; \int_0^{\sinh^{-1}1}2\cosh^2u\ du \\ &amp; = &amp;
\int_0^{\sinh^{-1}1}(1+\cosh 2u\ du \\ &amp; = &amp; \left[u+\frac{\sinh
2u}{2}\right]_0^{\sinh^{-1}1}\\ &amp; = &amp; \left[u+\sinh u\cosh u
\right]_0^{\sinh^{-1}1} \\ &amp; = &amp;
\sinh^{-1}1+1.\cosh\left(\sinh^{-1}1\right) \\ &amp; = &amp;
\sinh^{-1}1+\sqrt{1+\sinh^2\left(\sinh^{-1}1\right)} \\ &amp; = &amp;
\sinh^{-1}1+\sqrt{2}=2.2956\ \mathrm{(4\ d.p.)}
\end{eqnarray*}








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