⧼exchistory⧽
13 exercise(s) shown, 0 hidden
BBy Bot
Nov 03'24
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Consider the spiral defined in polar coordinates by the equation [math]r=e^{2\theta}[/math]. Compute the arc length of this curve from [math]\theta=0[/math] to [math]\theta=\ln 10[/math].

BBy Bot
Nov 03'24
[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]
  • lab{10.7.12a} Using the integral formula for arc length in polar coordinates, compute the arc length of the polar graph of the equation [math]r=2\sec \theta[/math] from [math]\theta = -\frac\pi4[/math] to [math]\theta=\frac\pi4[/math].
  • Identify and draw the curve in part \ref{ex10.7.12a}, and verify from the geometry the value obtained for the arc length.
BBy Bot
Nov 03'24
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Consider the curve defined by the equation [math]r=2\cos^2 \frac\theta2[/math] in polar coordinates.

  • Find the arc length of this curve from [math]\theta=0[/math] to [math]\theta=\pi[/math].
  • Find the arc length of this curve from [math]\theta=0[/math] to [math]\theta=2\pi[/math].