Can some one walk me through how to find the length of the curve given
This one is hard and I cant even start it. thanks in advance.
x=t^2
y=2t
0<= t <=2

x=t^2
y=2t
0<= t <=2

Length, C, of parametrized curve is given by:
C = INT.(a-->b)[sqrt[(dx/dt)^2 +(dy/dt)^2]]dt -----------(i)

x = t^2
dx/dt = 2t

y = 2t
dy/dt = 2

So,
C = INT.(0-->2)[sqrt[(2t)^2 +(2)^2]]dt
C = INT.(0-->2)[sqrt(4t^2 +4)]dt
C = (2)INT.(0-->2)[sqrt(t^2 +1)]dt

That is long to integrate, so we just use the Tables of Integrals, where it says:
INT.[sqrt(x^2 +a^2)]dx = (x/2)sqrt(x^2 +a^2) +(a^2 /2)*ln[x +sqrt(x^2 +a^2)]

So, with t = x, and a = 1,
C = (2){(t/2)sqrt(t^2 +1) +(1/2)*ln[t +sqrt(t^2 +1)]}|(0-->2)
C = (2){[(2/2)sqrt(4+1) +(1/2)ln[2 +sqrt(4+1)] -[0 +(1/2)ln[0 +sqrt(0 +1)]}
C = (2){[sqrt(5) +(1/2)ln(2 +sqrt(5))] -[0]}
C = 2{2.957886}
C = 5.916 units long -------------answer.