Arc Length and Curvature (3D vectors)

Quote:

Originally Posted by acg716

Let C be the curve of intersection of the parabolic cylinder x2 = 2y, and the surface 3z = xy. Find the exact length of C from the origin to the point (5, 25/2, 125/6).

My instructor told us to put those equations in parametric form so we could use the arc length formula, but I am not sure how to put those in that form.

I know that these are the equations I must use, but how do I solve for t from the 2 equations I'm given??
x = x_o - at
y= y_o - bt
z = z_o - ct

I assume x2 means x^2, that is, $x^2$ ....

The curve of intersection is found by solving $x^2 = 2y$ and $3z = xy$ simultaneously. Let $x = t$ . Then:

$x = t$ , $y = \frac{t^2}{2}$ and $z = \frac{t^3}{6}$ .

Substitute into the usual arclength formula and do the calculation. Since you're integrating from the point where x = 0 to the point where x = 5 the integration limits should be obvious.