Engineering Calculus Notes 214

Engineering Calculus Notes 214 - 202 CHAPTER 2. CURVES The...

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Unformatted text preview: 202 CHAPTER 2. CURVES The arclength of the parabola y = x2 between (0, 0) and ( 1 , 1 ) can be calculated using x as a parameter 24 → − (x) = (x, x2 ) 0 ≤ t ≤ 1 p with 1 + (2x)2 dx ds = 1 + 4x2 dx. = Thus s (C ) = 1 2 1 + 4x2 dx 0 which is best done using the trigonometric substitution 2x = tan θ 2 dx = sec2 θ dθ 1 dx = sec2 θ dθ 2 2 = sec θ 1 + 4x x=0↔θ=0 π 1 x= ↔θ= 2 4 and 2 π /4 1 + 4x2 dx = 0 0 π /4 = 0 1 (sec θ )( sec2 θ dθ ) 2 1 sec3 θ dθ 2 integration by parts (or cheating and looking it up in a table) yields π /4 0 11 1 1 sec3 θ dθ = sec θ tan θ + ln |sec θ + tan θ | 2 22 2 √ √ 1 = ( 2 + ln(1 + 2). 4 π /4 0 ...
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