Revision as of 00:02, 3 November 2024 by Bot (Created page with "<div class="d-none"><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}...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
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]

Find the arc lengths of the following parametrized curves.

  • [math]\dilemma{x = t+1,} {y = t^{\frac32}, & \mbox{from [/math](2,1)[math] to [/math](5,8)[math].}}[/math]
  • [math]\dilemma{x = t^2,} {y = \frac23 (2t+1)^\frac32, & \mbox{from [/math]\left(x(0),y(0)\right) = (0, \frac23)[math] to to [/math]\left(x(4), y(4)\right) = (16,18)[math].}}[/math]
  • [math]P(t) = (t^2, t^3)[/math], \quad from [math]P(0)[/math] to [math]P(2)[/math].
  • [math]\dilemma{x(\theta) = a \cos^3\theta, & a \gt 0,} {y(\theta) = a \sin^3\theta, & \mbox{from [/math]\left(x(0), y(0)\right) = (a,0)[math] to [/math]\left(x(\frac{\pi}2), y(\frac{\pi}2)\right) = (0,a)[math].}}[/math]