Posted by Liz on Tuesday, April 16, 2013 at 10:18pm.
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

Calculus  Reiny, Tuesday, April 16, 2013 at 11:05pm
x = t^2 + 1
t^2 = x1
t = √(x1)
y = 2t  3
t = (y+3)/2
(y+3)/2 = √(x1)
y = 2√(x1)  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√(x1)  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/√(x1)
so (dy/dx)^2 = 1/(x1)
∫√( 1 + (dy/dx)^2 ) dx from 1 to 2
= ∫√( 1 + 1/(x1) ) 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.

Calculus  Steve, Wednesday, April 17, 2013 at 11:01am
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)
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