# another curve lenght question

• Mar 15th 2008, 12:21 PM
lllll
another 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 PM
TheEmptySet
Quote:

Originally Posted by lllll
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.

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

$\displaystyle x=t$ 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.