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SECTION 7.4

SECTION 7.4. ARC LENGTH. Definition 1. Thus, we define the length L of the curve C with equation y = f ( x ) , a ≤ x ≤ b, as the limit of the lengths of these inscribed polygons (if the limit exists):. THE ARC LENGTH FORMULA.

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SECTION 7.4

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  1. SECTION 7.4 ARC LENGTH

  2. Definition 1 • Thus, we define the length L of the curve C with equation y = f(x), a ≤x ≤ b, as the limit of the lengths of these inscribed polygons (if the limit exists): 7.4

  3. THE ARC LENGTH FORMULA • If we use Leibniz notation for derivatives, we can write the arc length formula as: If f ’is continuous on [a, b], then the lengthof the curve y = f(x), a ≤x ≤ b is: 7.4

  4. Example 1 • Find the length of the arc of the semicubical parabola y2 = x3between the points (1, 1)and (4, 8). (See Figure 5.) 7.4

  5. Example 1 SOLUTION • For the top half of the curve, we have: • Thus, the arc length formula gives: 7.4

  6. Example 1 SOLUTION • If we substitute u =1 +(9/4)x, then du =(9/4) dx. • When x =1, u =13/4. • When x =4, u =10. • Therefore, 7.4

  7. ARC LENGTH • If a curve has the equation x = g(y), c ≤y ≤ d, and g’(y)is continuous, then by interchanging the roles of x and y in Formula 2 or Equation 3, we obtain its length as: 7.4

  8. Example 2 • Find the length of the arc of the parabola y2 = x from (0, 0)to (1, 1). • SOLUTION • Since x = y2, we have dx/dy =2y. • Then,Formula 4 gives: 7.4

  9. Example 2 SOLUTION • We make the trigonometric substitutiony = ½ tan, which gives: dy =½ sec2dand • When y = 0, tan= 0;so = 0. • When y = 1, tan = 2;so = tan–1 2 = a. 7.4

  10. Example 2 SOLUTION • Thus, • We could have used Formula 21 in the Table of Integrals. 7.4

  11. Example 2 SOLUTION • As tan a = 2, we have: sec2 a = 1 + tan2 a = 5 • So, sec a = and 7.4

  12. Example 3 • Set up an integral for the length of the arc of the hyperbola xy =1 from the point (1, 1) to the point (2, ½). • Use Simpson’s Rule with n =10 to estimate the arc length. 7.4

  13. Example 3(a) SOLUTION • We have: • So, the arc length is: 7.4

  14. Example 3(b) SOLUTION • Using Simpson’s Rule with a = 1, b = 2, n = 10, Dx = 0.1 and , we have: 7.4

  15. Example 4 • Find the arc length function for the curve y = x2–⅛ln x taking P0(1, 1) as the starting point. • SOLUTION • If f ’(x)= x2– ⅛ln x, then 7.4

  16. Example 4 SOLUTION 7.4

  17. Example 4 SOLUTION • Thus, the arc length function is given by: 7.4

  18. Example 4 SOLUTION • For instance, the arc length along the curve from (1, 1) to (3, f(3)) is: 7.4

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