chap08-et-instructor-solutions

# chap08-et-instructor-solutions - FURT HER APPLICAT IONS 8...

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8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS 8.1 Arc Length and Surface Area Preliminary Questions 1. Which integral represents the length of the curve y = cos x between 0 and π ? Z 0 p 1 + cos 2 xdx , Z 0 q 1 + sin 2 SOLUTION Let y = cos x .Then y 0 =− sin x ,and1 + ( y 0 ) 2 = 1 + sin 2 x . Thus, the length of the curve y = cos x between 0 and is Z 0 q 1 + sin 2 . 2. How do the arc lengths of the curves y = f ( x ) and y = f ( x ) + C over an interval [ a , b ] differ ( C is a constant)? Explain geometrically and then justify using the arc length formula. The graph of y = f ( x ) + C is a vertical translation of the graph of y = f ( x ) ; hence, the two graphs should have the same arc length. We can explicitly establish this as follows: length of y = f ( x ) + C = Z b a s 1 + · d dx ( f ( x ) + C ) ¸ 2 = Z b a q 1 +[ f 0 ( x ) ] 2 = length of y = f ( x ). Exercises 1. Express the arc length of the curve y = x 4 between x = 2and x = 6 as an integral (but do not evaluate). Let y = x 4 y 0 = 4 x 3 and s = Z 6 2 q 1 + ( 4 x 3 ) 2 = Z 6 2 p 1 + 16 x 6 . 2. Express the arc length of the curve y = tan x for 0 x 4 as an integral (but do not evaluate). Let y = tan x y 0 = sec 2 x ,and s = Z / 4 0 q 1 + ( sec 2 x ) 2 = Z / 4 0 p 1 + sec 4 . 3. Find the arc length of y = 1 12 x 3 + x 1 for 1 x 2. Hint: Show that 1 + ( y 0 ) 2 = µ 1 4 x 2 + x 2 2 . Let y = 1 12 x 3 + x 1 y 0 = x 2 4 x 2 ( y 0 ) 2 + 1 = ± x 2 4 x 2 2 + 1 = x 4 16 1 2 + x 4 + 1 = x 4 16 + 1 2 + x 4 = ± x 2 4 + x 2 2 . Thus, s = Z 2 1 q 1 + ( y 0 ) 2 = Z 2 1 v ² ² t ± x 2 4 + 1 x 2 2 = Z 2 1 ¯ ¯ ¯ ¯ ¯ x 2 4 + 1 x 2 ¯ ¯ ¯ ¯ ¯

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SECTION 8.1 Arc Length and Surface Area 959 = Z 2 1 ± x 2 4 + 1 x 2 dx since x 2 4 + 1 x 2 > 0 = ± x 3 12 1 x ¯ ¯ ¯ ¯ 2 1 = 13 12 . 4. Find the arc length of y = ³ x 2 ´ 4 + 1 2 x 2 over [ 1 , 4 ] . Hint: Show that 1 + ( y 0 ) 2 is a perfect square. SOLUTION Let y = ³ x 2 ´ 4 + 1 2 x 2 .Then y 0 = 4 ³ x 2 ´ 3 µ 1 2 1 x 3 = x 3 4 1 x 3 and ( y 0 ) 2 + 1 = ± x 3 4 1 x 3 2 + 1 = x 6 16 1 2 + 1 x 6 + 1 = x 6 16 + 1 2 + 1 x 6 = ± x 3 4 + 1 x 3 2 . Hence, s = Z 4 1 q 1 + y 0 2 = Z 4 1 v ² ² t ± x 3 4 + 1 x 3 2 = Z 4 1 ¯ ¯ ¯ ¯ ¯ x 3 4 + 1 x 3 ¯ ¯ ¯ ¯ ¯ = Z 4 1 ± x 3 4 + 1 x 3 since x 3 4 + 1 x 3 > 0on [ 1 , 4 ] = ± x 4 16 + x 2 2 ¯ ¯ ¯ ¯ 4 1 = 525 32 . In Exercises 5–10, calculate the arc length over the given interval. 5. y = 3 x + 1, [ 0 , 3 ] Let y = 3 x + 1. Then y 0 = 3, and s = Z 3 0 1 + 9 = 3 10. 6. y = 9 3 x , [ 1 , 3 ] Let y = 9 3 x y 0 =− 3, and s = Z 3 1 1 + 9 = 3 10 10 = 2 10. 7. y = x 3 / 2 , [ 1 , 2 ] Let y = x 3 / 2 y 0 = 3 2 x 1 / 2 ,and s = Z 2 1 r 1 + 9 4 xdx = 8 27 µ 1 + 9 4 x 3 / 2 ¯ ¯ ¯ ¯ 2 1 = 8 27 ± µ 11 2 3 / 2 µ 13 4 3 / 2 = 1 27 ³ 22 22 13 13 ´ . 8. y = 1 3 x 3 / 2 x 1 / 2 , [ 2 , 8 ] Let y = 1 3 x 3 / 2 x 1 / 2 y 0 = 1 2 x 1 / 2 1 2 x 1 / 2 , and 1 + ( y 0 ) 2 = 1 + µ 1 2 x 1 / 2 1 2 x 1 / 2 2 = 1 4 x + 1 2 + 1 4 x 1 = µ 1 2 x 1 / 2 + 1 2 x 1 / 2 2 . Hence, s = Z 8 2 q 1 + ( y 0 ) 2 = Z 8 2 s µ 1 2 x 1 / 2 + 1 2 x 1 / 2 2 = Z 8 2 ¯ ¯ ¯ ¯ 1 2 x 1 / 2 + 1 2 x 1 / 2 ¯ ¯ ¯ ¯
960 CHAPTER 8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS = Z 8 2 µ 1 2 x 1 / 2 + 1 2 x 1 / 2 dx since 1 2 x 1 / 2 + 1 2 x 1 / 2 > 0 = µ 1 3 x 3 / 2 + x 1 / 2 ¶ ¯ ¯ ¯ ¯ 8 2 = 17 2 3 .

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## chap08-et-instructor-solutions - FURT HER APPLICAT IONS 8...

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