Find the curvature at a point(vector function)

  • #1
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Homework Statement


Find the curvature of r(t)= <t^2, lnt, t lnt> at the point P(1,0,0)


Homework Equations


K(t) = |r'(t) x r''(t)|/(|r'(t)|^3)


The Attempt at a Solution


r'(t) = <2t, t^-1, lnt+1>
r''(t) = <2, -t^-2, t^-1>

|r'(t) x r''(t)| = sqrt[t^-4(4 + 4 lnt + ln^2t) + (4 ln^2t)]
|r'(t)| = sqrt[4t^2 + t^-2 + (ln^2t +2 lnt + 1)]

I don't know what value of (t) to sub into K(t) to get my final answer. I also have a feeling that my cross product is not right. Any help would be much appreciated.
 

Answers and Replies

  • #2
You want [itex](t^2, ln(t), tln(t))= (1, 0, 0)[/itex].

What value of t gives you that?

You are right that your cross product is wrong. Put your value of t into r' and r'' before calculating the cross product. That will simplify it a lot.
 
  • #3
Thanks a lot HallsofIvy, t = 1 provided I did the rest of the question right.
 
  • #4
t= 1 whether you did the rest of the problem right or not!
 

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