[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]

For each of the following parametrizations [math]P(t) = (x(t),y(t))[/math], find the derived vector [math]\vec dP(t)[/math] for an arbitrary value of [math]t[/math] in the domain. Draw the vectors [math]\vec dP(0)[/math], [math]\vec dP(1)[/math], and [math]\vec dP(2)[/math] in the [math]xy[/math]-plane.

• [math]\dilemma{x(t) = t^2-1,} {y(t) = t^3, \quad -1\leq t \leq3.}[/math]
• [math]\dilemma{x(t) = \frac12(e^t+e^{-t}),} {y(t) = \frac12(e^t-e^{-t}), \quad -\infty \lt t \lt \infty.}[/math]
• [math]\dilemma{x(t) = t^2,} {y(t) = \frac23(3t+1)^{\frac32}, \quad -\frac13 \leq t \leq 5.}[/math]
• [math]\dilemma{x(t) = t^2+t+1,} {y(t) = \frac{t^3}3 + t^2 - 1, \quad -\infty \lt t \lt \infty.}[/math]