Differential Geometry exam confusion!

Right, I need to show that a(t)=(cost,sint,0)is not a geodesic of the plane.
But I'm not sure what I'm supposed to actually do!

Right I've been thinking about the following
a regular curve on a surface is a geodesic of the surface is
a''_(tan)=0?

So I need to get the unit tangent vector field to a(t)? which is
a'(t)/||a'(t)|| so
a'(t)=(-sint,cost,0) and ||a'(t)||=ROOT(sin^2(t)+cos^2(t))=1

so T(t)=(-sint,cost,0)

Now a''_tan(t)=(a''.T)T+(a''.(TXn)TXn)

n=(PHIu X PHIv)/||PHIu X PHIv||

I'm not sure where I'm going from here on in, I end up completely
confused with tons of definitions? Am I on the right track?
Any help much appreciated!

Quote:

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Originally Posted by musicmental85

Right, I need to show that a(t)=(cost,sint,0)is not a geodesic of the plane.
But I'm not sure what I'm supposed to actually do!

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Let us say from t = 0 to t=pi then a(t) is a semicircle from points (1,0,0) to (-1,0,0) with length int(0,pi)|a(t)|dt = pi while the line segment between these two points has length 1, and since 1 < pi it means the semicircle cannot be a geodesic.

Cool. That helps alot. I guess I don't need any of the stuff I was blabbing about then!

Just out of curiosity in the following question do I follow the usual calculus steps to get the max and min of a function:

The normal curvature k(u) for a unit tangent vector u satisfies
k(u)= A_1cos^2(0)+A_2sin^2(0) where A is alpha and 0 is theta
for some value of 0, where A_1 and A_2 are the principal curvatures. Given that 0 ranges over the interval [0,2PI) as u ranges over all possible unit tangent vectors, show that the max and min values taken by k(u) are precisely the principal curvatures

PS I know I can extend the range of 0 to [0,2PI] so that there will be a max and min.
I also see that when A_1=A_2 then k(u)=A_1=A_2