Example 3 (a) set up the integral for the length of the arc of the hyperbola xy = 3 from the point (1, 3) to the point 6, 1 2 . (b) use simpson's rule with n = 10 to estimate the arc length. solution (a) we have y = 3 x dy dx = and so the arc length is l = 6 1 + dy dx 2 dx 1 = 6 1 + 9 x4 dx 1 = 6 x4 + 9 x2 dx 1 . (b) using simpson's rule with a = 1, b = 6, n = 10, δx = 0.5, and f(x) = , we have l = 6 1 + 9 x4 dx 1 ≈ δx 3 [f(1) + 4f(1.5) + 2f(2) + 4f(2.5) + ⋯ + 2f(5) + 4f(5.5) + f(6)] ≈ (rounded to four decimal places).

Step-by-step explanation:

Given is the function

xy =3

Use product rule to find derivative

xy'+y =0 OR

Arc length =

Substitute y = 3/x

WE get

arc length =

Using simpson rule for 1/3 we have

x(x^4+9)^1/2f(x)2 times or 4 times  

13.162277663.162277663.16227766  

1.53.751.6666666676.666666667  

251.25  

2.56.9327123121.109233974.43693588  

39.4868329811.054092553  

3.512.611998261.0295508784.118203512  

416.27882061.0174262872.034852575  

4.520.47101611.0109143754.0436575  

525.179356621.0071742652.01434853  

5.530.398396341.0049056644.019622656  

636.124783741.0034662152.00693243  

6.542.356374961.0025177514.010071002  

749.091750831.0018724662.003744932  

7.556.329943191.0014212124.005684849  

864.070273921.001098032.00219606  

8.572.312256911.0008616874.00344675  

981.055536521.0006856362.001371272  

9.590.299847731.0005523294.002209318  

10100.04498991.0004498992.000899798  

10.5110.29080881.0003701484.00148059  

11121.03718441.0003073092.000614618  

11.5132.28402211.0002572564.001029023  

12144.03124661.000216994.82148E-05  

 52.270413638.711735605

Answer is 8.7117