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\bf{Question}

(i) Calculate the arc length of the curve $\gamma$ in Qn.1(iv)
from $(0,2)$ to $(2\pi,2)$.

(ii) Calculate the arc length of the curve $\gamma$ in Qn.1(v)
from $(1,0)$ to $(e^{2\pi k},0)$.

(iii) Calculate the path length of the curve $y^2=x^3$ from
$(1,1)$ to $(-1,1)$.



\bf{Answer}

\begin{description}
\item{(i)}

\begin{eqnarray*}
\gamma(t) & = & (t - \sin t, 1 + \cos t)\\ \| \gamma'(t) \| & = &
((1-c)^2 + s^2)^{\frac{1}{2}}\\ & = & (2(1-c))^{\frac{1}{2}}\\ & =
& 2\sin \frac{t}{2}, \ \ \ (\textrm{note }: \geq 0 \ \textrm{for }
0 \geq y \geq 2\pi)\\ \Rightarrow \textrm{length} & = &
\int_0^{2\pi} 2 \sin \frac{t}{2} \,dt\\ & = & \left [ -4\cos
\frac{t}{2} \right ]_0^{2\pi}\\ & = & 4 - (-4) = 8, \ \ \
(\textrm{Check }: 2\pi < 8<2\pi + 4)
\end{eqnarray*}

\item{(ii)}
\begin{eqnarray*}
\gamma(t) & = & (e^{kt} \cos t , e^{kt} \sin t)\\ \| \gamma'(t) \|
& = & e^{kt}((kc - s)^2 + (ks+c)^2)^{\frac{1}{2}}\\ & = &
e^{kt}.(k^2+1)^{\frac{1}{2}}\\ \Rightarrow  \textrm{length} & = &
\int_0^{2\pi} e^{kt} (k^2+1)^{\frac{1}{2}} \,dt\\ & = &
\frac{1}{k} (k^2+1)^{\frac{1}{2}} (e^{2\pi k} -1).
\end{eqnarray*}

\item{(iii)}

\begin{eqnarray*}
\gamma(t) & = & (t^2, t^3), \ : \ -1 \geq t \geq 1\\
\textrm{length} & = & \int_{-1}^1 \| (2t, 3t^2) \| \,dt\\ & = &
\int_{-1}^1 (4t^2+9t^4)^{\frac{1}{2}} \,dt\\ & = & \int_{-1}^1 |t|
(4+9t^2)^{\frac{1}{2}} \,dt\\ & = & 2 \times \left [ \frac{1}{27}
(4 + 9t^2)^{\frac{3}{2}} \right ]_0^1\\ & = & \frac{2}{27}
(13^{\frac{3}{2}} - 4^{\frac{3}{2}}).
\end{eqnarray*}
\end{description}


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