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March 31, 2015

March 31, 2015

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

x=t^(2) + 1

y=2t - 3

0<t<1

- Calculus -
**Reiny**, Tuesday, April 16, 2013 at 11:05pmx = 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.

- Calculus -
**Steve**, Wednesday, April 17, 2013 at 11:01amds = √(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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