How do you find the lengths of the curve y=int (sqrtt+1)^-2y=(t+1)2 from [0,x^2][0,x2] for the interval 0<=x<=10x1?

1 Answer
May 15, 2017

The arc length of the curve yy on a<=x<=baxb is given by:

L=int_a^bsqrt(1+(dy/dx)^2)dxL=ba1+(dydx)2dx

Here, to find dy/dxdydx, we have to apply the second Fundamental Theorem of Calculus.

y=int_0^(x^2)(sqrtt+1)^-2dty=x20(t+1)2dt

dy/dx=(sqrt(x^2)+1)^-2d/dx(x^2)=(x+1)^-2(2x)dydx=(x2+1)2ddx(x2)=(x+1)2(2x)

(dy/dx)^2=(4x^2)/(x+1)^4(dydx)2=4x2(x+1)4

So the arc length is given by:

L=int_0^1sqrt(1+(4x^2)/(x+1)^4)dxL=101+4x2(x+1)4dx

Putting this into a calculator or Wolfram Alpha:

L=1.07943L=1.07943