How to find the curvature K of the curve?

#r(t)=ti+t^2j+t^2k#

1 Answer
Apr 14, 2018

See below

Explanation:

Using a vector approach to curvature, #kappa#:

#kappa(t) = (|mathbf r' times mathbf r''|)/(|mathbf r'|^3)#

#mathbf r(t) = [t, t^2, t^2 ]^T#

#mathbf r'(t) = [1, 2t, 2t]^T#

#mathbf r''(t) = [0, 2, 2]^T#

#| mathbf r' times mathbf r'' | = | det ((hat x, hat y , hat z),(1, 2t, 2t),(0, 2, 2) ) | = | [0, -2, 2]^T| = 2 sqrt 2#

#|mathbf r'|^3 = (mathbf r' * mathbf r')^(3/2) = (1 + 8 t^2)^(3/2)#

#implies kappa = (2 sqrt 2)/( (1 + 8 t^2)^(3/2))#

Now, if you want to param in terms of the arc length, #s#, because #kappa(s) = (d mathbf(hat T(s)))/(ds)#, then:

#mathbf T(t) = mathbf r'(t)# and so #hat(mathbf T) = (mathbf r')/(|mathbf r'|)#

Plus:

#s = int_(Delta t) \ dot s \ dt = int_(Delta t) sqrt(mathbf r' * mathbf r') \ dt#

You can work from there.