MATH 152 Chapter 9.3

« previous | Wednesday, October 6, 2010 | next »

Arc Length

Length of smooth curve with parametric equations:
$L=\int _{a}^{b}{{\sqrt {{\bigg (}{\frac {dx}{dt}}{\bigg )}^{2}+{\bigg (}{\frac {dy}{dt}}{\bigg )}^{2}}}\,dt}$

What about $y=f(x)$? We know how to find length of straight line:
$s={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}$

${\begin{aligned}ds={\sqrt {dx^{2}+dy^{2}}}\quad \therefore \quad s=&\int {{\sqrt {dx^{2}+dy^{2}}}\,{\bigg (}{\frac {dx}{dx}}{\bigg )}}&=\int {{\sqrt {1+{\bigg (}{\frac {dy}{dx}}{\bigg )}^{2}}}\,dx}\\s=&\int {{\sqrt {dx^{2}+dy^{2}}}\,{\bigg (}{\frac {dy}{dy}}{\bigg )}}&=\int {{\sqrt {{\bigg (}{\frac {dx}{dy}}{\bigg )}^{2}+1}}\,dy}\end{aligned}}$

Example 1

$y=1+{\bigg (}{\frac {3x}{2}}{\bigg )}^{2/3}\quad x={\bigg [}{\frac {2}{3}},{\frac {16}{3}}{\bigg ]}\,\!$

Method 1

${\begin{aligned}s=&\int _{2/3}^{16/3}{{\sqrt {dx^{2}+dy^{2}}}\,{\frac {dx}{dx}}}\;=\;\int _{2/3}^{16/3}{{\sqrt {1+{\bigg (}{\frac {dy}{dx}}{\bigg )}^{2}}}\,dx}\\=&\int _{2/3}^{16/3}{{\sqrt {1+{\bigg (}{\frac {3x}{2}}{\bigg )}^{-2/3}}}\,dx}&({\text{Let }}u={\frac {3x}{2}}\quad \therefore \quad du={\frac {3}{2}}\,dx)\\=&{\frac {2}{3}}\int _{1}^{8}{{\sqrt {1+u^{-2/3}}}\,du}\\=&{\frac {2}{3}}\int _{1}^{8}{u^{-1/3}{\sqrt {u^{2/3}+1}}\,du}&({\text{Let }}w=u^{2/3}+1\quad \therefore \quad dw={\frac {2}{3}}\,u^{-1/3}\,du\\=&\int _{2}^{5}{{\sqrt {w}}\,dw}\end{aligned}}$

Method 2

Solve for x:
$x={\frac {2}{3}}\,(y-1)^{3/2}\quad \quad {\frac {dx}{dy}}=(y-1)^{1/2}\,\!$

${\begin{aligned}s&=\int _{2}^{5}{{\sqrt {{\bigg (}{\frac {dx}{dy}}{\bigg )}^{2}+1}}\,dy}\\&=\int _{2}^{5}{{\sqrt {y-1+1}}\,dy}\\&={\frac {2}{3}}\,y^{3/2}{\bigg |}_{2}^{5}\\&={\frac {2}{3}}{\big (}5^{3/2}-2^{3/2}\end{aligned}}$

Friday, October 8, 2010

Example 2

$y={\frac {1}{6}}\,x^{3}+{\frac {1}{2x}}\quad x\in [1,2]\,\!$

${\begin{aligned}s&=\int _{1}^{2}{{\sqrt {1+{\bigg (}{\frac {1}{2}}\,x^{2}-{\frac {1}{2x^{2}}}{\bigg )}^{2}}}\,dx}\\&=\int _{1}^{2}{{\sqrt {{\frac {1}{4}}\,x^{4}+{\frac {1}{2}}+{\frac {1}{4x^{4}}}}}\,dx}\\&=\int _{1}^{2}{{\sqrt {{\bigg (}{\frac {1}{2}}\,x^{2}+{\frac {1}{2x^{2}}}{\bigg )}^{2}}}\,dx}\\&=\int _{1}^{2}{{\frac {1}{2}}\,x^{2}+{\frac {1}{2x^{2}}}\,dx}\\&={\frac {1}{6}}\,x^{3}-{\frac {1}{2}}\,x^{-1}{\bigg |}_{1}^{2}\\&={\bigg (}{\frac {8}{6}}-{\frac {1}{4}}{\bigg )}-{\bigg (}{\frac {1}{6}}-{\frac {1}{2}}{\bigg )}\end{aligned}}$

Warm Up

$x=t\cos {t}\quad y=t\sin {t}\quad t\in [0,2\pi ]$

${\begin{aligned}s=&\int _{0}^{2\pi }{{\sqrt {{\bigg (}{\frac {dx}{dt}}{\bigg )}^{2}+{\bigg (}{\frac {dy}{dt}}{\bigg )}^{2}}}\,dt}\\=&\int _{0}^{2\pi }{{\sqrt {(\cos {t}-t\sin {t})^{2}+(\sin {t}+t\cos {t})^{2}}}\,dt}\\=&\int _{0}^{2\pi }{{\sqrt {\cos ^{2}{t}-2t\sin {t}\cos {t}+t^{2}\sin ^{2}{t}+\sin ^{2}{t}+2t\sin {t}\cos {t}+t^{2}\cos ^{2}{t}}}\,dt}\\=&\int _{0}^{2\pi }{{\sqrt {1+t^{2}}}\,dt}\\=&\int _{0}^{\tan ^{-1}{2\pi }}{sec^{3}{\theta }\,d\theta }&\int {\sec ^{3}{\theta }\,d\theta }\dots {\text{LOOK IT UP}}\end{aligned}}$