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
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.
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)
To find the arc length of a curve described by parametric equations over a given interval, we use the arc length formula:
L = ∫√(dx/dt)² + (dy/dt)² dt
In this case, we are given the parametric equations:
x=t^2 + 1
y=2t - 3
To find dx/dt, we differentiate x with respect to t:
dx/dt = d/dt (t^2 + 1) = 2t
To find dy/dt, we differentiate y with respect to t:
dy/dt = d/dt (2t - 3) = 2
Now that we have dx/dt and dy/dt, we can substitute them into the arc length formula:
L = ∫√(2t)² + 2² dt
L = ∫√(4t^2 + 4) dt
Next, we integrate this expression with respect to t over the given interval 0<t<1.