I'm having trouble finding the lenght of the curve where

$\displaystyle x^2 = 2y \ \ \ 3z= xy$, from the origin to the the point (6,18,36). I have no idea how to approach this problem, so any help would be appreciated.

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- Mar 15th 2008, 12:21 PMlllllanother curve lenght question
I'm having trouble finding the lenght of the curve where

$\displaystyle x^2 = 2y \ \ \ 3z= xy$, from the origin to the the point (6,18,36). I have no idea how to approach this problem, so any help would be appreciated. - Mar 15th 2008, 05:41 PMTheEmptySet

So we need to find the parametric equation of the curve of intersection of the two surfaces.

lets start with

$\displaystyle x=t$ with this and equation one we get

$\displaystyle y=\frac{x^2}{2}=\frac{t^2}{2}$

Now we can use the 2nd equation and solve for z.

$\displaystyle z=\frac{xy}{3}=\frac{t^3}{6}$ so the vector equation is

$\displaystyle r(t)=<t,\frac{t^2}{2},\frac{t^3}{6}>$

Now with that done we need to calculate the arc length.

Note that $\displaystyle r(0)=<0,0,0>$ and $\displaystyle r(6)=<6,18,36>$

$\displaystyle L=\int_{t_0}^{t_1}\sqrt{ \left(\frac{\partial{x}}{\partial{t}} \right)^2+ \left(\frac{\partial{y}}{\partial{t}} \right)^2+\left(\frac{\partial{z}}{\partial{t}} \right)^2}dt$

so we get

$\displaystyle L=\int_0^6\sqrt{1+t^2+\frac{t^4}{4}}dt=\int_0^6\sq rt{\frac{t^4+4t^2+4}{4}}dt$ pulling out the 4 and factoring the numerator

$\displaystyle \frac{1}{2}\int_0^6\sqrt{(t^2+2)^2}dt=\frac{1}{2}\ int_0^6t^2+2dt$

It should be easy from here.

good luck.