lec31 - Lecture 31 18.01 Fall 2006 Lecture 31: Parametric...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Lecture 31 18.01 Fall 2006 Lecture 31: Parametric Equations, Arclength, Surface Area Arclength, continued Example 1. Consider this parametric equation: x = t 2 y = t 3 for ≤ t ≤ 1 x 3 = ( t 2 ) 3 = t 6 ; y 2 = ( t 3 ) 2 = t 6 = ⇒ x 3 = y 2 = ⇒ y = x 2 / 3 ≤ x ≤ 1 ds dy dx ds dy dx Figure 1: Infinitesimal Arclength. ( ds ) 2 = ( dx ) 2 + ( dy ) 2 ( ds ) 2 = (2 t dt ) 2 + (3 t 2 dt ) 2 = (4 t 2 + 9 t 4 )( dt ) 2 ( dx ) 2 ( dy ) 2 t =1 1 1 Length = ds = 4 t 2 + 9 t 4 dt = t 4 + 9 t 2 dt t =0 1 = (4 + 9 t 2 ) 3 / 2 = 1 (13 3 / 2- 4 3 / 2 ) 27 27 Even if you can’t evaluate the integral analytically, you can always use numerical methods. 1 Lecture 31 18.01 Fall 2006 Surface Area (surfaces of revolution) y ds a b y x Figure 2: Calculating surface area ds (the infinitesimal curve length in Figure 2) is revolved a distance 2 πy . The surface area of the thin strip of width ds is 2 πy ds ....
View Full Document

This note was uploaded on 02/27/2009 for the course MATH 155b taught by Professor Staff during the Fall '08 term at Vanderbilt.

Page1 / 4

lec31 - Lecture 31 18.01 Fall 2006 Lecture 31: Parametric...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online