# Homework Help: Differential Geometry, curve length

1. Jan 14, 2012

### Stimpon

1. The problem statement, all variables and given/known data

2. Relevant equations

$L[c]:=\int_{a}^{b}(\sum_{i,j=1}^{2}g_{ij}(c(t))c_{i}'(t)c_{j}'(t))^{1/2}dt$

3. The attempt at a solution

So $g_{ij}(x,y)=0$ for $i{\neq}j$, $c_{1}'(t)=-Rsin(t)$, $c_{2}'(t)=Rcos(t)$

so $L[c]:=\int_{a}^{b}(\frac{1}{((Rsin(t))^{2}}R^{2}(sin^{2}(t)+cos^{2}(t))^{1/2}dt=\int_{a}^{b}\frac{1}{sin(t)}dt$

However the solutions has

$L[c]:=\int_{a}^{b}(\frac{1}{((Rcos(t))^{2}}R^{2}(sin^{2}(t)+cos^{2}(t))^{1/2}dt=\int_{a}^{b}\frac{1}{cos(t)}dt$

and he then goes on to use the given identity to find an antiderivative for $\frac{1}{cost}$

but I don't see how he has cost where I have sint.

Is he making a mistake or am I?

Last edited: Jan 14, 2012
2. Jan 14, 2012

### MathematicalPhysicist

I wonder how come the hint is right but the solution in the text isn't.