find the lenght of the curve at u= t^3/6, v= t where t is from -1 to 1

and E = 1, F=0 and G= 1+v^2.

do we find it from t =0 to 1 and then multiply it by 2? because if i were to integrate it from -1 to 1, then it would cancel each other out.

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- May 2nd 2012, 03:07 AMalexandrabel90finding length with first fundamental
find the lenght of the curve at u= t^3/6, v= t where t is from -1 to 1

and E = 1, F=0 and G= 1+v^2.

do we find it from t =0 to 1 and then multiply it by 2? because if i were to integrate it from -1 to 1, then it would cancel each other out. - May 4th 2012, 06:34 PMHallsofIvyRe: finding length with first fundamental
The length of a vector, <u, v>, on a surface with that first fundamental form would be

$\displaystyle \sqrt{\begin{bmatrix}u & v \end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 1+v^2\end{bmatrix}\begin{bmatrix}u \\ v\end{bmatrix}}$

$\displaystyle = \sqrt{u^2+ v^4+ v^2}= \sqrt{t^3/6+ t^4+ t^2}$

So you want to integrate

$\displaystyle \int_{-1}^1\sqrt{t^4+ t^3/6+ t^2}dt$

You can factor out a $\displaystyle t^2$ to get

$\displaystyle \int_{-1}^1 \sqrt{t^2+ t/6+ 1}tdt$

Because that, like any "metric" integral involves only even powers, the integral from -1 to 1**cannot**be 0. The integrals from -1 to 0 and from 0 to 1 will**add**, not cancel.