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\bf{Question}

Sketch each of the plane curves $\gamma$ given by the following
parametrizations:

\medskip

\begin{tabular}{rlrl}
\hspace{1.5cm}(i)&\ $\gamma(t)=(t,e^t)$&\hspace{1cm}(iv)&\
$\gamma(t)=(t-\sin t,1+\cos t)$\\ (ii)&\
$\gamma(t)=(t^3,t^4)$&(v)&\ $\gamma(t)=(e^{kt}\cos t, e^{kt}\sin
t)\quad(k\ne0)$\\ (iii)&\ $\gamma(t)=(t-t^2,t+t^2)$&(vi)&\
$\gamma(t)=(t^2-1,t^3-t)$
\end{tabular}



\bf{Answer}

\begin{description}
\item{(i)}
\begin{eqnarray*}
K(t) & = & e^t(1+e^{2t})^{-\frac{3}{2}} >0\\ K'(t) & = &
e^t(1+2^t)^{-\frac{5}{2}}.(1-2e^{2t})
\end{eqnarray*}

\begin{center}
\epsfig{file=342-4A-1.eps, width=50mm}
\end{center}

So $K'=0$ where $e^{2t}=\frac{1}{2}$.

\item{(ii)}
 $K(t) = 12t^{-} (9+16t^2)^{-\frac{3}{2}}$, defined for all $t \ne
0$. (Note also $\gamma '(0)=0$).

$\left. \begin{array}{rllrl} |K| & \to \infty & \textrm{ as } & t
& \to 0\\ & \to 0 & \textrm{ as } & t & \to \infty
\end{array}
\right \}$

No vertices

\begin{center}
\epsfig{file=342-4A-2.eps, width=50mm}

At origin, deceptive@ $\gamma(t)$ slows to instantaneous halt.

$\frac{dy}{dx}$ exists here ($=0$), but $\frac{d^2y}{dx^2}$
doesn't.
\end{center}

\item{(iii)}
$x+y=2t$, $-x+y=2t^2$, so rotation by 45 degrees shows curve is a
parabola.

\begin{center}
\epsfig{file=342-4A-3.eps, width=50mm}
\end{center}

$K(t)=4(2+8t^2)^{-\frac{3}{2}}$, max when $t=0$.

\item{(iv)}
This is a cycloid: take standard cycloid (for circle of radius 1),
reflect in the $x$-axis and translate by $2$ in the $y$-direction.

\begin{center}
\epsfig{file=342-4A-4.eps, width=70mm}
\end{center}

$\gamma'(t) = (1-\cos t, -\sin t)$, which is zero when $t=2n\pi$
($n=0,\pm 1, \pm 2,\cdots)$.

\begin{eqnarray*}
K(t) & = & -\frac{(1-c).c +ss}{((1-c)^2+s^2)^{3/2}}\\ & = &
\frac{1-c}{(2-2c)^{\frac{3}{2}}}\\ & = & 2^{-\frac{3}{2}}
(1-c)^{-\frac{1}{2}}\\ & & \textrm{where } c = \cos t \ (\ne 1)\\
& & \textrm{and } s = \sin t\\ & = & 1/4\sin\frac{t}{2}
\end{eqnarray*}
defined except when $t=2n\pi$, ($n \in \textbf{Z}$).

So $k>0$, and as $t$ goes from $0$ to $2\pi$ we see $K$ increases
to $\frac{1}{4}$ (at $t=\pi$), then increases again.

\item{(v)}
$K(t) = e^{-kt}(1+K^2)^{-\frac{1}{2}}$.

$\begin{array}{lcl} \to 0 & \textrm{ as } & t \to + \infty\\ \to 0
\infty & \textrm{ as } & t \to - \infty, \ \textrm{i.e. as
}\gamma(t)\to (0,0)
\end{array}$

\begin{center}
\epsfig{file=342-4A-5.eps, width=50mm}

Spiral - radius increases exponentially.
\end{center}

\item{(vi)}

\begin{center}
$y(t)$ goes \ \ \ \epsfig{file=342-4A-6.eps, width=40mm}

While $x(t)$ goes \ \ \ \epsfig{file=342-4A-7.eps, width=40mm}
\end{center}

If $t \to -t$, then $x\to x$ and $y \to -y$.

\begin{center}
\epsfig{file=342-4A-8.eps, width=50mm}
\end{center}

$K(t)=(6t^2+2)(9t^4-2t^2+1)^{-\frac{3}{2}},>0$

defined for all $t \in \Re$ since $9u^2-2u+1$ has no real roots.

$K'(t)=-24t(9t^4+4t^2-1)(9t^4-2t^2+1)^{-\frac{5}{2}},=0$ when
$t=0$ or $t^2=\frac{1}{9}(-2 + \frac{13})$.
\end{description}


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