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Mathematics 8 Online
OpenStudy (anonymous):

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!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

*formula not equation

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