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<=10≤x≤1?
1 Answer
May 15, 2017
The arc length of the curve
L=int_a^bsqrt(1+(dy/dx)^2)dxL=∫ba√1+(dydx)2dx
Here, to find
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=∫10√1+4x2(x+1)4dx
Putting this into a calculator or Wolfram Alpha:
L=1.07943L=1.07943