Question

Find the arc length of the curve described by the parametric equation over the given interval:
x=t^(2) + 1
y=2t - 3
0<t<1

Answer 1

x = t^2 + 1
t^2 = x-1
t = √(x-1)

y = 2t - 3
t = (y+3)/2

(y+3)/2 = √(x-1)
y = 2√(x-1) - 3

when t=0, x = 1, y=-3
when t = 1, x = 2, y = -1
so we want the length of the curve of
y = 2√(x-1) - 3 from the point (1,-3) to (2, -1)

If I recall, the length of a line segment
= ∫√( 1 + (dy/dx)^2 ) dx from 1 to 2

now dy/dx = 1/√(x-1)
so (dy/dx)^2 = 1/(x-1)

∫√( 1 + (dy/dx)^2 ) dx from 1 to 2
= ∫√( 1 + 1/(x-1) ) dx from 1 to 2
= ∫√( 1 + (dy/dx)^2 ) dx from 1 to 2
= ∫ ( (x+2)/(x+1) )^(1/2) dx from 1 to 2

now running into a brick wall, can't seem to integrate that.

Answer 2

ds = √(dx^2 + dy^2)

s = ∫[0,1] √[(2t)^2+(2)^2] dt
= ∫[0,1] √(4t^2+4) dt
= 2∫[0,1] √(t^2+1) dt
= t√(t^2+1) + arcsinh(t) [0,1]
= √2+arcsinh(1)