find the length of the curve

Question

x= 1/2t^2 , y= 1/9(6t+9)^3/2, t greater than or equal to 0 but less than or equal to 1

Answer 1

Let the arc length for a curve given in parametric form be

L = integral[a,b](ds)

where ds = sqrt(dx/dt^2 + dy/dt^2) dt

If x = 1/2t^2 and y = 1/9(6t + 9)^3/2

then

dx/dt = t and dy/dt = sqrt(6t + 9)

Plugging these into the arc length formula for parametric curves, we get

L = integral[a,b](sqrt(t^2 + 6t + 9)) dt

   = integral[a,b](sqrt((t + 3)^2)) dt

   = integral[a,b](t + 3) dt

   = [a,b](1/2t^2 + 3t)

Going from 0 to 1 gives

 L = 1/2(1)^2 + 3(1) - 0 = 7/2

so 7/2 is the length of the parametric curve given above from t = 0 to t = 1.