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

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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 0 t 1 x 3 = ( t 2 ) 3 = t 6 ; y 2 = ( t 3 ) 2 = t 6 = x 3 = y 2 = y = x 2 / 3 0 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 0 0 1 = (4 + 9 t 2 ) 3 / 2 = 1 (13 3 / 2 - 4 3 / 2 ) 27 0 27 Even if you can’t evaluate the integral analytically, you can always use numerical methods. 1
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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 π y . The surface area of the thin strip of width ds is 2 π y ds . Example 2. Revolve Example 1 ( x = t 2 , y = t 3 , 0 t 1 ) around the x-axis. Refer to Figure 3. y x Figure 3: Curved surface of a trumpet. 2
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Lecture 31 18.01 Fall 2006 1 3 1 2 π t t 4 + 9 t 2 dt 4 Area = 2 π y ds = 0 = 2 π t 4 + 9 t 2 dt y ds 0
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