Question

find the curvature of r(t)= at the point (1,0,0). I know the denominator is |r'(t)|^3 but I don't know if I'm finding |r'(t)| correctly. for r'(t) I got <2t,1/t,lnt+1>. for |r'(t)| i've got (2t+1/t+lnt)sqrt(2lnt+1). This seems wrong please help clarify!

Answer 1

First figure out the value of \(t\) such that \[
\mathbf r(t) = \langle
0,0, 0
\rangle
\]

Answer 2

That way you can find the magnitude of a constant vector, rather than of a vector function.

Answer 3

So I plugged in 0 and then found the magnitude of a constant vector to be 1. What do I do with this? all I know right now is the equation |r'(t)xr''(t)|/|r'(t)|^3

Answer 4

*formula not equation