BBy Bot
Nov 03'24
Exercise
[math]
\newcommand{\ex}[1]{\item }
\newcommand{\sx}{\item}
\newcommand{\x}{\sx}
\newcommand{\sxlab}[1]{}
\newcommand{\xlab}{\sxlab}
\newcommand{\prov}[1] {\quad #1}
\newcommand{\provx}[1] {\quad \mbox{#1}}
\newcommand{\intext}[1]{\quad \mbox{#1} \quad}
\newcommand{\R}{\mathrm{\bf R}}
\newcommand{\Q}{\mathrm{\bf Q}}
\newcommand{\Z}{\mathrm{\bf Z}}
\newcommand{\C}{\mathrm{\bf C}}
\newcommand{\dt}{\textbf}
\newcommand{\goesto}{\rightarrow}
\newcommand{\ddxof}[1]{\frac{d #1}{d x}}
\newcommand{\ddx}{\frac{d}{dx}}
\newcommand{\ddt}{\frac{d}{dt}}
\newcommand{\dydx}{\ddxof y}
\newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}}
\newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}}
\newcommand{\dist}{\mathrm{distance}}
\newcommand{\arccot}{\mathrm{arccot\:}}
\newcommand{\arccsc}{\mathrm{arccsc\:}}
\newcommand{\arcsec}{\mathrm{arcsec\:}}
\newcommand{\arctanh}{\mathrm{arctanh\:}}
\newcommand{\arcsinh}{\mathrm{arcsinh\:}}
\newcommand{\arccosh}{\mathrm{arccosh\:}}
\newcommand{\sech}{\mathrm{sech\:}}
\newcommand{\csch}{\mathrm{csch\:}}
\newcommand{\conj}[1]{\overline{#1}}
\newcommand{\mathds}{\mathbb}
[/math]
The same curve can be defined by more than one parametrization:
- lab{10.2.7a}
Draw the curve defined parametrically by
[[math]] \dilemma{x(t) = t,}{y(t) = t, & 0 \leq t \leq 1.} [[/math]]
- lab{10.2.7b}
Draw the curve defined parametrically by
[[math]] \dilemma{x(t) = \sin\pi t,}{y(t) = \sin\pi t, & 0 \leq t \leq 1.} [[/math]]
- Compute the arc lengths from [math]t=0[/math] to [math]t=1[/math] for the parametrizations in \ref{ex10.2.7a} and \ref{ex10.2.7b}.
- Give a geometric interpretation which explains the difference between the arc lengths obtained for the two parametrizations.