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).

Answer:Step-by-step explanation:Given is the functionxy =3Use product rule to find derivativexy'+y =0 OR [tex]y'=\frac{-y}{x}[/tex]Arc length = [tex]\int\limits^1_6 {\sqrt{1+\frac{y^2}{x^2} } } \, dx \\=\int\limits^1_6 {\sqrt{\frac{x^2+y^2}{x^2} } } \, dx \\\\[/tex]Substitute y = 3/xWE getarc length = [tex]\int\limits^1_6 {\sqrt{1+\frac{9}{x^4} } } \, dx \\\\=\int\limits^1_6 {\sqrt{\frac{x^4+9}{x^4} } } \, dx \\[/tex]Using simpson rule for 1/3 we havex (x^4+9)^1/2 f(x) 2 times or 4 times  1 3.16227766 3.16227766 3.16227766  1.5 3.75 1.666666667 6.666666667  2 5 1.25   2.5 6.932712312 1.10923397 4.43693588  3 9.486832981 1.054092553   3.5 12.61199826 1.029550878 4.118203512  4 16.2788206 1.017426287 2.034852575  4.5 20.4710161 1.010914375 4.0436575  5 25.17935662 1.007174265 2.01434853  5.5 30.39839634 1.004905664 4.019622656  6 36.12478374 1.003466215 2.00693243  6.5 42.35637496 1.002517751 4.010071002  7 49.09175083 1.001872466 2.003744932  7.5 56.32994319 1.001421212 4.005684849  8 64.07027392 1.00109803 2.00219606  8.5 72.31225691 1.000861687 4.00344675  9 81.05553652 1.000685636 2.001371272  9.5 90.29984773 1.000552329 4.002209318  10 100.0449899 1.000449899 2.000899798  10.5 110.2908088 1.000370148 4.00148059  11 121.0371844 1.000307309 2.000614618  11.5 132.2840221 1.000257256 4.001029023  12 144.0312466 1.00021699 4.82148E-05         52.27041363 8.711735605 Answer is 8.7117