# Find The Arc Length Of Y=2sqrt(x) With The Initial

 vjca From: USPosts: 10Votes: 10 Find the arc length of y=2sqrt(x) with the initial point (0,0) FamousSam From: USPosts: 58Votes: 116 Arc length same type x2 is x^2, right? dy/dx = 2x (dy/dx)^2 = 4x^2 Therefore the arc length for which we are looking is: 3 ? sqrt(1 + 4x^2) dx 0 Let u = arcsin(2x). Then 2x = sin(u) and 4x^2 = sin^2(u). In addition, since 2x = sin(u), 2dx = cos(u) du So now we have ? sqrt(1 + sin^2(u)) * cos(u) du. Then this problem is simply ? (1 + sin^2(u)) du which is the same as: ? ((1 + sin^2(u))^(1/2)) du = [(1 + sin^2(u))^(3/2)] / (3/2) = (2/3) * [(1 + sin^2(u))^(3/2)] and substituting back x = sin(u) / 2, we have: (2/3) * (1 + (x^2 / 4))^(3/2) from 0 to 3 At x = 3, this is (2/3) * (1 + 9/4))^(3/2)  WizRoger From: USPosts: 7Votes: 21 A simple way is to imagine that the equation is an equatin of a circle y = v(2 - x²) y² = 2 - x² x² + y² = 2 with a center at the origin and radius of v2 when x = 0, y = v2 when x = 1, y = 1 The angle formed with center as the angle between the two points is 45 degrees, or in other words, the arc length is 45/360 of the circumference of the circle arc length = 45/360 x 2 p r = (1/8)(2p)(v2)  Enter 